# Towards an AI assistant for human grid operators*

Marot Antoine  AI Lab
RTE France
Paris, France  Rozier Alexandre  AI Lab
RTE France
Paris, France  Dussartre Matthieu  R&D
RTE France
Paris, France  Crochepierre Laure  R&D RTE France
Paris, France
Université de Lorraine, CNRS, LORIA
Metz, France  Donnot Benjamin  AI Lab
RTE France
Paris, France

###### Abstract

Power systems are becoming more complex to operate in the digital age. As a
result, real-time decision-making is getting more challenging as the human
operator has to deal with more information, more uncertainty, more
applications and more coordination. While supervision has been primarily used
to help them make decisions over the last decades, it cannot reasonably scale
up anymore. There is a great need for rethinking the human-machine interface
under more unified and interactive frameworks. Taking advantage of the latest
developments in Human-machine Interactions and Artificial intelligence, we
share the vision of a new assistant framework relying on an hypervision
interface and greater bidirectional interactions. We review the known
principles of decision-making that drives the assistant design and supporting
assistance functions we present. We finally share some guidelines to make
progress towards the development of such an assistant.

###### Index Terms:

assistant, artificial intelligence, human-machine interaction, hypervision

##  I Introduction

From the beginning, power systems have been complex artificial systems to
operate. Over time, complexity has been rising and control room operators have
been using more and more applications setting side by side on multiple screens
to manage the system. While this simple approach to incrementally grow a
monitoring alarm-driven management system has been relatively effective until
now, we are now reaching a glass ceiling to add in any new application. This
is true in terms of physical space, but most importantly, in terms of
manageable cognitive load for any human operator. In that regard, human-
machine interfaces have indeed been identified as a risk factor to human error
[1] and should be considered more closely. In addition, latest research in
physchology [2] and neuroscience [3] have shed light on human decision-making
process and its limits, which could be a lead to improve in turn decision-
making.

Nowadays, the system complexity keeps rising given the advent of intermittent
renewable energies on the production side and of prosumers on the demand side,
coupled with the globalization of energy markets over a more and more
interconnected European grid. Grids are also aging, and grid developments
become more limited nowadays. Operators will hence need to operate a system
closer to its limits while dealing with additional automaton on the grid
inducing cyber-physical dynamics [4]. While there could eventually be a
temptation to develop a fully autonomous grid to cope with that complexity, it
fails short for such large critical system operations. Indeed coordination,
responsibility, accountability, and explainability are a must when operating
such a system and can only be reasonably achieved by humans today: human
operators remain key players. Lately, several works have proposed situation
awareness frameworks [5] [6] to help augment the operator’s comprehension of
safety-critical situations especially. In addition to better information
processing, it is also urgent to rethink the operator human-machine interface
[5] and interactions to assist the operator’s regular real-time decision
making. Rather than having operators adapt to the machine through a
technology-centered system engineering design, machine and operator’s could
co-adapt following human centered-design [7].

In terms of interfaces, we have seen tremendous innovations in other domains,
especially in consumer products such as smartphones, connected homes, social
media and social networks, search engines, recommendation systems. They have
been well-adopted for displaying the most relevant information to the user on
single screens through homogeneous format, while dealing in the background
with vast diverse amounts of information and user interactions. The
development of such interfaces has also been made possible thanks to the
latest developments of Artificial intelligence (AI) in the last decade. These
developments enabled more advanced and practical large scale real-time
information processing, such as in computer vision [8], image understanding
[9], natural language processing [10], recommendations [11]. This now
translates into an even more advanced form of interface: an assistant.
Assistants were found useful to both improve user’s performance on tasks and
enhance group collaboration on a common task [12]. In power systems, the
notion of an AI assistant was used lately in [13] [14].

In this paper, we present the design of an AI-infused assistant for the
control room operator based on the latest developments in human-computer
interaction design (HCID), AI, and decision-making science. We first review
the effectiveness and limits of humans and AI in decision making, highlighting
their complementarities. We further define an assistant. We then discuss the
specificities of an assistant for grid operators. Finally, we devise
guidelines for how researchers can develop and test such an assistant.

##  II Effectiveness and limits/bias of human and AI in taking decisions

###  II-A Human decision making

Human Decision-making is primarily a matter of attention and executive control
[3]. Taking proper decisions first involves paying attention to the right
information in the environment and making sense of it. It further implies
selecting relevant actions while inhibiting inappropriate ones, and eventually
executing one in a timely-manner. Following dual process theory [2], we can
describe 2 underlying fictitious operating and cooperative agents, called
System 1 and System 2. System 1 is the fast intuitive and heuristic agent
while System 2 is the slow and reasoning agent. System 2 is the one
responsible for decisions assisted by System 1 which continuously provides him
predictions for action. Most of the time, System 2 just lazily accepts System
1 proposal in usual situations without much more thinking, resulting in
successful, quick and cognitively effortless decisions. When confronted to
unusual situations however, System 2 can develop more explicit conscious
thinking, beyond System 1 predictions, to deliberate and come up with novel
and acceptable decisions while cognitively costly.

Young operators will rely more heavily on System 2 and can hence struggle
taking any good decision on time for several situations that still appear
complex and unusual for them. As they are very focused on trying to make sense
of it, they have a narrow attention and might miss important new information.
As they learn overtime through appropriate training and feedback becoming
expert, their intuitive System 1 grows for that field of expertise, enabling
them taking good and quick decisions even more often with ease. For an expert
operator, it has become a lot easier to operate a system intuitively, being
able to take more decisions as well as decisions in more difficult situations.
However, the downside can be overconfidence, overlooking unusual information
that would require more deliberation from System 2.

###  II-B Human biases and desirable assistance

Because it relies on fast heuristics and mostly jump to conclusions, System 1
indeed introduces several potential biases which can lead to human errors,
hence limiting the effectiveness of human decision-making. Cognitive biases
are beautifully summarized in the cognitive bias codex [15] and classified
through 4 problems they are trying to circumvent: a limited memory, the need
to act fast, the information overload and a lack of meaning. Among possibly
damaging biases, we can more specifically list:

  * •

anchoring bias: be over-reliant on the first piece of information we see.

  * •

confirmation bias: tend to pay attention only to information that confirms our
preconceptions.

  * •

overfidence bias: too confident about one’s abilities which causes to take
greater risks.

  * •

information bias: tendency to seek information when it does not affect action
(more information is not always better).

  * •

availability heuristic: overestimate the importance of information that is
available.

  * •

ostrich effect: ignore dangerous or negative information.

  * •

outcome bias: judge a decision based on the outcome rather than how exactly
the decision was made.

An assistant should hence help the operator avoid falling into those biases by
addressing the following needs:

  * •

augment its memory, knowledge retrieval and keep track of latest events

  * •

better filter or highlight information, enhancing attention

  * •

make sense of a situation and give feedback

  * •

make recommendations, possibly handle some tasks or alert on some undesirable
expected consequences

Let’s now consider what AI in its latest developments could bring in that
regard.

###  II-C AI potential for assistance

The latest deep learning revolution demonstrated some impressive practical
abilities of AI, being able to digest lots of information, memorize large
historical datasets, and learn by imitation to infer quickly effective actions
in context. Turing-award Yoshua Bengio recently described current deep-
learning AI as a System 1 kind of intelligence [16], while missing System 2
type. It is indeed presented as advanced pattern matching and recognition
machines like System 1 [2], being coined as artificial intuition [17]. It
however lacks the ability to reason about causality [18], hence lacking
understanding and some common sense. Yet Human and AI can be seen as
complementary heterogeneous intelligences that could achieve a superior
outcome when combined by co-evolving and learning from each other [19]. This
is best exemplified by Centaur’s chess [20], having humans play with machines,
and not against, to reach a superior performance. Used through an assistant,
AI seems capable of addressing the needs mentioned previously, providing a
complimentary and hence enhanced System 1 to the operator whose System 2
remains in charge of decisions overall.

To be used effectively by any human however, the AI assistant will need to
work along with a proper interface. Indeed, while human Systems 1 and 2 are
fully integrated into a cognitive system as a whole, AI and human are clearly
separated entities at first. Interactions, communication and shared
representations [21] need to be defined. Interpretable [22], explainable [23]
and trustworthy [24] AI are in that regard attributes that should get
integrated into the assistant. The AI should eventually be provably beneficial
to humans, pursuing preferably not a fixed pre-determined objective but
pursuing operator’s fuzzy objectives, by continuously learning its preferences
under uncertainty [25]. Moving away from these different maturing fields of AI
that should prove useful to create an assistant, let’s now define it more
precisely.

##  III Defining an artificial assistant

An assistant is an agent that helps in someone’s job, supporting him and
taking over on agreed tasks when possible.

###  III-A Assistant: balancing assistance, user control and Automation

To make things clearer, we should distinguish an assistant from single
assistance functions and from a whole automaton. An assistance function helps
the user gather or alert him about some new relevant information. It can take
in a user request in some expected format and compute a result to be
interpreted. Other ones can also monitor the user state and warn him about
risks it does not seem aware of. Situation awareness offers, in that sense,
advanced assistance functions. An assistant relies on assistance functions at
its core. But importantly, it also engages actively with the user. It offers a
unified interface and allows for dynamic bidirectional interactions with the
user to cooperate efficiently on task completion. However, in that
configuration, the user remains responsible for the proper operations of the
system. While some tasks could eventually become automated if they always get
delegated to the assistant, there is no explicit goal to automate any
particular task in the first place: this is mostly left to the user’s choice
over time.

Other industrial sectors have also defined different autonomy levels that we
can reflect on, like the one from the International Association of Public
Transport :

  * •

GoA0: Manual operation with no automatic protection

  * •

GoA1: Manual operation with automatic protection

  * •

GoA2: Semi-automatic operation

  * •

GoA3: Driverless operation

  * •

GoA4: Unattended operation

Many such fields, however, aims at a fully autonomous system without
operators. They diverge on that point from our goal of obtaining an augmented
operator through an assistant, which is closer to GoA2 level. Beyond that
point, GoA3 and GoA4 are targeting automation, and thus ends the comparison to
our probem here. GoA1 and GoA2 offers assistance functions discussed
previously, but without too much considerations of interface and interactions.
An assistant, as we illustrate on Figure1 and later discuss, is yet a
subsequent level not described there whose goal is to offer the right balance
between user control and autonomy [21] for enhanced decision-making. We will
now focus on the interface and interactions that more uniquely defines an
assistant.

Figure 1: The AI operator’s assistant: Hypervision interface, bidirectional
interactions and AI components running altogether in a coordinated and modular
fashion. Continuous revision is important to up to date shared representations

###  III-B Hypervision: smart interface & information management

Today’s supervision leaves the user the cognitive load to prioritize,
organize, and link every displayed information and alarms before considering
any decision. It can be regarded as a fragmented ecosystem from an operator
viewpoint. While it has been manageable for up to ten applications, it becomes
impractical with always more information and uncoordinated applications to
control under heterogeneous formats. Supervision gives access to the user to
every information available without much more processing. However, it does not
help deal with the information overload and lack of meaning problems that need
to be tackled for improved decision making: it dilutes the operator’s
attention. Let’s recall that humans can only take sequential decisions one
after another, with a limited working memory space of 4 information to
manipulate at a time.

To be effective at continuous decision-making, it is important to focus on one
task at a time, with the highest priority, and present only the most relevant
information to it. In that regard, we propose an “hypervision” framework to
bring the right information at the right time to the right person. It helps
overcome multiple biases, such as both information bias and anchoring bias, by
taking advantage of them rather than being influenced by them. Hypervision
relies on the definition of tasks created by processing and synthesizing the
necessary information. Those tasks do not have to be only-real time. However,
they are still preferably the ones anticipated to be completed or configured
ahead of time thanks to forecast, hence defining an expected trajectory that
might be adapted along the way. Reaching this higher level of information
enables the assistant to establish a simplified but relevant dialogue with the
operator, eventually providing him with diagnostics or even recommendations on
solutions. Hence, hypervision’s goal is to refocus the operator on task
completion rather than alarm monitoring, as illustrated in figure 2. It
creates the basis for more advance and effective bidirectional interactions
under shared representations of tasks.

Figure 2: Alarm monitoring with Supervision - Task completion with Hypervision

###  III-C Bidirectional interactions

While the choice of the interaction modalities (visual, audio, haptic, etc.),
as well as the form of the assistant avatar matters for enhanced interactions,
we will leave it open here and focus on the importance of bidirectional
interactions.

Human-Machine interactions have become a new scientific discipline in the
80’s, especially thanks to Lucy Suchman [26]. At a time AI was mostly centered
around expert systems with pre-defined rules, she shed light on the
ineffectiveness of such systems, mostly attributed to the lack of well-
designed interactions and learning loops beyond knowledge retrieval. She noted
that plans, similarly to predefined rules, are not prescriptive and not
something to follow exactly, because everything eventually depends on
circumstances and contingencies. Plans should rather be seen as heuristic and
available resources for actions that help focus one’s attention while
abstracting the details, but that should get updated through interactions to
take appropriate decisions. In the end, interfaces should not draw a dry
delimitation with its user but rather re-configures itself and conforms with
it.

Research [27] has shown an increased efficiency in Human-AI coupling when both
agents were able to initiate and respond to interactions. These were
historically mostly unidirectional, the assistant either asking a predefined
set of questions to build its context representation, or the user asking to
perform some predefined tasks. In a bidirectional relationship, the
interaction is collaborative, with neither the system nor the user in control
of the whole interaction. The assistant is capable of interacting with the
latter to refine its context representation (e.g. ask for a clarification when
ambiguities arise), thus improving its efficiency when asked to perform a
specific task. A good example of such bidirectional interaction is found in
[28], where when asked to find the shortest path to evacuate wounded people,
the assistant will for instance first ask which vehicles are available, then
react accordingly.

While new approaches let an assistant learn how and when to defer to an expert
[29], creating true human-computer partnerships becomes a practical reality
[30], [21] and are the object of study of the ”Cockpit and Bidirectional
Assistant” project [31]. We should now review the expected functions of such
an assistant.

##  IV Expected functions for an assistant

The paper [32] provides a ”unified set of design guidelines” to keep in mind
when designing AI-infused assistants, helpful for deciding which features
should receive a particular focus. We highlight below a relevant subset of
these guidelines:

  * •

Time services based on context \- Grid operators evolve in a time-constrained
environment where having the right information at the right time is paramount.
An assistant should engage interaction when the context allows it, based on
the operator’s cognitive load and impact of the interaction. An assistant
should logically engage the operator with a new task if more critical, while
not disturbing him from the current one if critical.

  * •

Show contextually relevant information \- While knowledge databases might
suggest to the operator some usual curative action for a given issue, the
context of a nearby maintenance, for instance, might make it inefficient. This
contextual event should be brought to the operator’s attention in that case.
Inversely, insensitive context should get filtered out.

  * •

Support efficient invocation and dismissal \- The number of actions an
operator can do in a time period is limited, and interacting with an assistant
should be as efficient as possible and not a burden. Should the assistant be
in charge of a task, it should not invoke subsequent interactions on it if
nothing significant has changed. In tense situations, the assistant should be
shorter and dismiss its interaction sooner.

  * •

Support efficient correction / Encourage granular feedback \- Grid operators
are well-trained experts, capable of evaluating the assistant’s answers and
provide feedback. Thus, the latter must be able to learn from them, for
instance, by understanding in hindsight that some additional context needs to
be considered to select a curative action or remembering that a line is under
maintenance during a defined period.

  * •

Inform the user about uncertainty in services provided \- If there is too much
uncertainty yet when considering some preventive actions, it is reasonable to
inform the operator about it so that he waits for the last instant to decide,
before the action opportunity expires. Also, the operator should be able to
assess any additional risks due to its actions in the next hours. Operators
should also get informed of possibly missing or bad quality data and hence
uncertain observability. The operator should finally know if any result is
deterministic or probabilistic.

  * •

Scope services when in doubt \- When not yet sure about some action
implementation because of uncertainty, the operator can first simply indicate
his intention of using such or such flexibility and later decide how he would
like to implement it: the assistant should be able to deal with different
levels of abstraction. This is also true when giving some contextual
information: ahead of time, it might be more relevant to only communicate
about aggregated loads in some areas, only sharing individual load values near
more certain real-time.

  * •

Learn from user behaviour / Remember recent interactions \- Operators often
have specific decision-making processes, some relying on numerous power-flow
simulations to assess a situation, others more akin to rely on their expertise
of the considered area, and a good assistant should adjust to these user-
specific behaviours. For instance, when supporting a user accustomed to
simulations, it could recompute them when the user is busy on the phone to
integrate the latest grid changes when he hangs up.

  * •

Convey the consequences of user actions \- As well as assistants should learn
over time, so should operators! It is often deplored that the consequences of
unary grid operations are poorly monitored, which in return, prevent operators
from valuable feedback. Assistants delivering a detailed report of how the
grid evolved after a specific action would tremendously speed up the way
operators acquire experience, and yield better grid management.
Interpretability and explainability are required here.

The assistant should eventually help the operator prioritize his tasks thanks
to these functions and the hypervision interface. We will now propose some
initial guidelines to more concretely design, implement and test such an
assistant.

##  V Guidelines for designing, implementing and testing an assistant

Designing an assistant in practice might still seem complex beyond the
discussed framework and principles. We devise here some pragmatic guidelines
to start simple on a common but modular ground, listing some already available
building blocks as well.

###  V-A Grounded Design Considerations

####  V-A1 Modeling Tasks as shared representations

It should be noted that in other industrial sectors such as aeronautics, tasks
in processes have been codified more precisely at a granular level, which
gives the operator a clearer framework to work and coordinate with, as for the
assistant. We should aim at such explicit modeling.

A task is first defined by the problem it needs to solve specifically, such as
a safety problem - an overload over a line, its priority and the residual time
to complete it. It should then contain relevant context to understand the root
of the problem, what might be already known about it, recent related events or
tasks, as well as the persons involved. It should further come with some
suggestions about available actions to the operators, and their expected
effectiveness. It should finally retain a decision for completion and meta-
attributes about it. Task categories and attributes should be more
exhaustively establish through future works.

Also, opposite to traditional approaches in power systems that mainly tries to
focus on the most critical situations we ever have to solve, we suggest here
to start studying tasks in regular situations and gradually increasing the
number of needed bidirectional interactions. To operators it should prove
useful to start with the most basic but sometimes time-consuming tasks with
often low added value. That way, building trust in the first place should be
easier while still helping ease their cognitive load.

####  V-A2 Simple situational use case as a sandbox

We offer a simple interesting use case to highlight key difficulties in daily
grid management through the interplay here of preventive and curative
decisions under uncertainty. This makes us think about how the operator-agent
interaction should take place.

An operator starts monitoring a grid composed of two smaller areas at 7:00am.
Forecasts show that an incident should occur around 9:00am in area 1, with
three available corrective actions after simulations, each being able to be
executed just before 9:00am. Another incident should happen around 8:30am in
area 2, leaving only a couple of minutes to execute the only preventive
action.

Figure 3: A simple scenario where incidents are forecasted in multiple parts
of a grid, and several corrective actions with different setup durations are
possible. Green dots each represent an available action, and the blue arrows
the time it needs to be effective. For instance, starting a nuclear power
plant takes longer than applying topological grid changes, and would have a
longer blue arrow.

A couple of questions arises:

  * •

Which decisions have priority ? It seems that a preventive action on area 2
should be urgently taken, but maybe the forecast isn’t that reliable. The
assistant should be able to provide the operator with these uncertainties.

  * •

How should the result of the simulations be presented to the operator ? We can
see that the last simulation of 8:00am in area 1 shows that the forecasted
incident should not happen anymore. Is it a simulation artifact, or has the
situation improved with more recent measurements? Maybe the assistant should
run successive simulations over time and alert the operator as soon as the
forecasted situation evolves.

  * •

How does applying a corrective action on area 1 reflect on area 2 ? Would it
lead to a less secure grid state ? What coordination is required ? Maybe
there’s a new maintenance operation in this area that isn’t taken into account
by the simulation.

  * •

Which of the three corrective action in area 1 should be taken ? The operator
has to mediate between economical, practical and safety arguments, each with a
degree of uncertainty. How could the assistant be helpful there ?

Our objective here is not to provide any viable solution, but rather to
demonstrate that grid operators are confronted with complex decisions even on
apparently simple cases, in which context-dependent trade-offs always have to
be made. Future works could build a library of such canonical cases to be
studied in the community.

###  V-B Unified Interface & Data collection as an industrial stack

The hypervision framework relies on a generic and single interface that should
be able to integrate any kind of tasks, and apply to different industrial
systems for instance. While previous supervised applications would still run
in the background, hypervision is a master process responsible for displaying
the right information in a proper and standardized format. An example of such
an existing framework is the open-source Operator Fabric [33] which could be
used both by industrial and researchers as a unified interface for decision-
making processes. Such a framework is also a corner-stone to digitalize the
decision-making process, centralize every necessary information and hence
capitalize on them. This historical data-collection is essential for
continuous improvement, experiments, as well as for creating the datasets from
which AI can learn recommendations. Data should get labelled and its quality
properly monitored. These developments should create a necessary technical
stack for an assistant.

###  V-C Power system AI modules for assistant functions

Recent surveys list interesting developments of AI for power systems [34],
[35]. For an assistant, AI can today be used to make corrective action
recommendations to an operator through adaptive interpretable expert system
[14], imitation learning [13] or reinforcement learning [36]. It can learn
from user behaviour and help convey the consequences of operator’s action by
comparison. Exhaustive risk assessment [37] also helps in prioritizing tasks.
Further, automatic hierarchical and contextual representations of the grid
[38] enable scope services and give greater flexibility to convey the right
context and interpret a situation. [39] also lets an AI learn interpretable
and physically-consistent contextual indicators associated with a particular
operator’s task. Finally, [40],[41] and [42] let operators explore
interactively and iteratively historical explainable factors across similar
situations and decisions for augmenting and keeping up-to-date the system
knowledge and proper labels. This is an illustrative sample of today’s AI
potential to provide effective assistance functions. New developments are
needed to augment the assistant functions continuously.

###  V-D Assistant evaluation & development of shared benchmarks

In order to assess the relevance of an assistant in a real-world scenario, and
eventually compare multiple assistants, it is necessary to set up repeatable
evaluation protocols and define common benchmark tasks.

####  V-D1 Evaluation

As for now, there is not yet a standard testing protocol to evaluate
artificial assistants. However, we could draw insights from other domains such
as interpretable machine-learning [22] or interactive visualization [43]. As
done in [22], we could come to structured and step-by-step experimental
practices to evaluate candidate-assistants on incrementally difficult task.

Moreover, several Virtual-Assistant (VA) related studies have also tried to
define custom evaluation criteria. For instance in [44], authors compare their
VA against both a simpler interactive data-exploration scheme and a non-
interactive solution-search. They assess the use of their assistant on three
factors :

  * •

Performance \- Is an operator more efficient with a VA ? Here, task-specific
performance metrics are used.

  * •

Human-learning \- Is a virtually-assisted operator learning as much about the
problem and it’s underlying model than without the assistant ? This factor is
assessed using questions and tests at the end of each task.

  * •

Usability \- Is a user able to able to handle the use of the assistant ? This
evaluation is performed using a standard System Usability Scale (SUS) [45].

####  V-D2 Benchmarks

Because of confidentiality issues, it is often hard to share real-world data
on decision-making problems. Thus, we should aim at developing synthetic but
realistic environments from which to extract representative and relevant
scenarios. To define these scenarios, we could rely on already existing
decision-making assessment frameworks, such as [46] originally designed for
road safety. Moreover, synthetic and realistic environments for sequential
decision-making have recently been developed for power systems [47] and could
be interactively further studied with grid2viz study tool [48].

##  VI Conclusion

In this paper, we have presented the framework and principles for designing an
AI assistant for grid operators, opening new research directions for augmented
decision-making. We have provided initial guidelines and already available
materials in power systems to start exploring this rich and promising new
field of human-machine partnerships.

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# Further insights into the mysteries of the values of zeta functions at
integers

Ján Mináč* and Tung T. Nguyen** and Nguy$\tilde{\text{\^{e}}}$n Duy Tân  *
Department of Mathematics
Western University
London, Ontario N6A 5B7
CANADA<EMAIL_ADDRESS>** Brain and Mind Institute and Department of Mathematics
Western University
London, Ontario N6A 5B7
CANADA<EMAIL_ADDRESS>*** School of Applied Mathematics and Informatics
Hanoi University of Science and Technology
01 Dai Co Viet Road, Hanoi
VIETNAM<EMAIL_ADDRESS>
###### Abstract.

We present a remarkably simple and surprisingly natural interpretation of the
values of zeta functions at negative integers and zero. Namely we are able to
relate these values to areas related to partial sums of powers. We apply these
results to further interpretations of values of $L$-functions at negative
integers. We hint in a very brief way at some expected connections of this
work with other current efforts to understand the mysteries of the values of
zeta functions at integers.

###### Key words and phrases:

Hurwitz zeta functions, Bernoulli polynomials, Generalized Bernoulli numbers,
$L$-functions

###### 2010 Mathematics Subject Classification:

Primary 11M35, 11M06, 11B68

J.M. is partially supported by the Natural Sciences and Engineering Research
Council of Canada (NSERC) grant R0370A01. He gratefully acknowledges the
Western University Faculty of Science Distinguished Professorship in 2020-2021
and support of Western Academy for Advanced Research as Western Fellow. N.D.T.
is partially supported by the Ministry of Education and Training of Vietnam
via the project ”On the arithmetic of algebraic groups and homogeneous spaces
over local and global fields and their extensions”

To the memory of Goro Shimura with gratitude and admiration

##  1\. Introduction

Ever since Euler made stunning discoveries connecting some values of zeta
functions with powers of $\pi$, there has been a tremendous effort of
mathematicians to comprehend well the “true reasons behind these connections”
and to further extend these results to other values and other zeta functions.
This work belongs to one of the most dramatic and exciting parts of
mathematics and some parts of theoretical physics. Some extraordinary
progress, conjectures, and various deep concepts have been obtained. Yet, the
air of tremendous mystery, excitement, and strong desire to make further
progress has never been altered nor slowed down.

Let us begin with two very naive questions which a child might ask:

  1. (1)

How many natural numbers are there?

  2. (2)

What is the sum of all natural numbers?

The obvious answers to both questions namely: “infinity” are correct but they
lack a certain depth, subtlety, and connections with combinatorial properties
of natural numbers. L. Euler was able to offer some more subtle and more
interesting answers:

  1. (1)

$1+1+1+\cdots=-\dfrac{1}{2}.$

  2. (2)

$1+2+3+\cdots=-\dfrac{1}{12}.$

One has to be cautious when handling these beautiful Euler’s equalities in a
formal way. As we shall see below that equalities can be explained in a subtle
way as values of Rieman’s zeta function. However, in order to do it properly,
the Riemann zeta function defined below as a convergent series for complex
values $s=\sigma+\mathrm{i}t$ for $\sigma>1$ has to be analytically extended
to the entire complex plane except when $s=1$. Therefore one can not formally
handle the equalities $(1)$ and $(2)$ in the usual way as when we consider
just finite sums. For example if we just add formally both equalities we shall
obtain equality:

$2+3+4+\cdots=-\dfrac{7}{12}.$

However if we just subtract from the equation $(2)$ the number one we obtain

$2+3+4+\cdots=-\dfrac{13}{12}$

which would contradict the first equality.

Therefore below we consider rigorous proper modern interpretation of Euler’s
equalities using values of Riemann zeta function.

We now write the above-mentioned equalities of Euler using the values of the
Riemann zeta function as:

  1. (1)

$\zeta(0)=-\dfrac{1}{2}.$

  2. (2)

$\zeta(-1)=-\dfrac{1}{12}.$

In fact, S. Ramanujan rediscovered these statements and for example formula
$(2)$ appeared in exactly the same form in Ramanujan’s letter to G. Hardy
addressed in 16 January, 1913 (see [1, page 351]).

Recall that the Riemann zeta function $\zeta(s)$ with $s=\sigma+it$ is the
analytical continuation of the function

$f(s)=\sum_{n=1}^{\infty}\frac{1}{n^{s}},$

with $\sigma>1$ to the whole complex plane except when $s=1$ where $\zeta(s)$
has a simple pole (see [40]). Euler’s answers are deep and the assigned
numbers as a summation of these divergent series have various interesting
interpretations. For interested readers, we just refer to some articles and
surveys related to these topics (see for examples [14], [18], [22], [25],
[26], [29], [32]).

In paper [30], Mináč raised and answered the following naive question:

######  Question 1.1.

Suppose $a\in\mathbb{N}$ or $a=0$. Are there any connections between the
values $\zeta(-a)$ and the partial sums

$S_{a}(M)=\sum_{n=1}^{M-1}n^{a}?$

Here $S_{a}(M)$ is considered as a polynomial in $M$.

The answer is the fact on page 1 in [30]

$\zeta(-a)=\int_{0}^{1}S_{a}(x)dx.$

Hence it is observed that

$\zeta(0)=\int_{0}^{1}(x-1)dx=-\frac{1}{2}.$

Thus $\zeta(0)$ is equal to $(-1)$ times the area of the right triangle with
acute angles $45^{o}$ and legs of length $1$. This is because the graph of
$x-1$ over the interval $[0,1]$ has the shape of a right triangle when taken
together with the x-axis and y-axis (see the displayed Figure 1 below).

Figure 1. Graph of $x-1$ with $x\in[0,1]$

Similarly $\zeta(-1)$ is $(-1)$ time area of the red region in figure 2 below.

$\zeta(-1)=\int_{0}^{1}\frac{x(x-1)}{2}dx=-\frac{1}{12}.$

Figure 2. Graph of $\frac{x-x^{2}}{2}$ with $x\in[0,1]$

Observe also that

$\zeta(-2)=\int_{0}^{1}\frac{x(x-1)(2x-1)}{6}dx=0.$

This fact also has a geometrical meaning as if

$f(x):=\frac{x(x-1)(2x-1)}{6},$

then

$f(1-x)=-\frac{x(x-1)(2x-1)}{6}=-f(x).$

In fact, the graph displayed function of $f(x)$ in the interval $[0,1]$ in
figure $3$, the red region bounded by the horizontal axis
$x\in[0,\frac{1}{2}]$ and $f(x)$, the blue region bounded by the horizontal
axis $x\in[\frac{1}{2},1]$ and $f(x)$ show that in the evaluation of the
integral $\int_{0}^{1}f(x)dx$, the two areas considered with proper signs will
cancel each other.

Figure 3. Graph of $\frac{x(x-1)(2x-1)}{6}$ with $x\in[0,1]$

In fact, in general for all $a\in\mathbb{N}$

$\int_{0}^{1}S_{a}(x)dx=\int_{0}^{1}\frac{B_{a+1}(x)-B_{a}(x)}{a+1}dx,$

where $B_{a+1}(x)$ is the $(a+1)$-th Bernoulli polynomial (see [33, Section 2,
Chapter 1]). Hence if $a=2n$ and $n\in\mathbb{N}$, we see that

$B_{2n+1}=B_{2n+1}(1)=B_{2n+1}(0)=B_{2n+1}\left(\frac{1}{2}\right)=0.$

From the general property

$B_{m}(1-x)=(-1)^{m}B_{m}(x),$

we see immediately in general, in the evaluation

$\zeta(-2n)=\int_{0}^{1}\frac{B_{2n+1}(x)-B_{2n+1}}{2n+1}dx,$

we obtain the cancellation of two signed areas as in the special case with
$\zeta(-2)$. Hence we see a nice geometric explanation $\zeta(-2n)=0$ for all
$n\in\mathbb{N}$.

Because $B_{2n}(x)-B_{2n}$ does not change sign between $0$ and $1$ (see [33,
Section 3, Chapter 1]), we also see that all values
$\zeta(-1),\zeta(-3),\zeta(-5),\ldots$ are in fact signed areas of regions by
the horizontal axis $x\in[0,1]$ and functions
$\dfrac{B_{2}(x)-B_{2}}{2},\dfrac{B_{4}(x)-B_{4}}{4},\dfrac{B_{6}(x)-B_{6}}{6},\cdots$
These values are also equal to
$-\dfrac{B_{2}}{2},-\dfrac{B_{4}}{4},-\dfrac{B_{6}}{6},\ldots$ where
$B_{2},B_{4},B_{6},\ldots$ are just Bernoulli numbers whose signs alternate.

Interested readers are encouraged to also see articles [39], [17]. This paper
is meant as the first paper of a series of papers related to special values of
$L$-functions. It is beyond the scope of this paper to provide a comprehensive
historical overview of the study of the special values of $L$-functions and
their arithmetic significance. Therefore, we merely point out several
highlights of these investigations which also motivate our investigations.
Some of these investigations are rather sophisticated and they involve the
most advanced topics of current number theory. We hope that our elementary
point of view and possibly unexpected direct connections with partial sums and
signed areas will bring additional insight and inspiration to understand
special values of $L$-functions. We also hope to find a bridge between these
elementary and more sophisticated studies of $L$-functions in the subsequent
papers.

One of the strongest motivations for studying special values of $L$-functions
is the class number formula for residue of the Dedekind zeta function
$\zeta_{K}(s)$ of an algebraic number field $K$ expressed via basic invariants
of $K$ like the regulator $R_{K}$, the class number $h_{K}$, the number of
roots of unity $\omega_{K}$ contained in $K$, discriminant $D_{K}$, the number
of real places $r_{1}$, and the number of complex places $r_{2}$. This
remarkable formula has the following form

${\displaystyle\lim_{s\to
1}(s-1)\zeta_{K}(s)={\frac{2^{r_{1}}\cdot(2\pi)^{r_{2}}\cdot R_{K}\cdot
h_{K}}{w_{K}\cdot{\sqrt{|D_{K}|}}}}}.$

See for example [9, Chapter 5, Section 2.1]. This formula provides a bridge
between the algebraic and the analytic world.

Observe in particular that $\zeta(s)=\zeta_{\mathbb{Q}}(s)$. Because
$\lim_{s\to 1}(s-1)\zeta(s)=1$ we see that the class number formula in view of
$r_{1}=1,r_{2}=0,\omega_{\mathbb{Q}}=2,|D_{\mathbb{Q}}|=1$ amounts to
$h_{\mathbb{Q}}=1$ and we recover the unique factorization of integers into
products of primes. In fact, these strong immediate connections between
arithmetic and analysis were already used and exploited by Euler.

In the subsequent work of H. Poincaré, L. Mordell, E. Artin, H. Hasse, A. Weil
and others, these ideas have been generalized to higher dimensional objects
beginning with elliptic curves and studies of their related $L$-functions.
These generalizations not only enriched and empowered number theory and its
reach but they also provided a new broad perspective on the original
investigations of L. Euler on special values of various $L$-functions. Using
early computer experimentation, and beautiful elementary heuristic arguments,
B. Birch and P. Swinnerton-Dyer proposed their remarkable conjecture relating
the rank of a given elliptic curve $E$ over $\mathbb{Q}$ with the Taylor
expansion of $L(E,s)$ at $s=1$. We refer readers to more details of the Birch
and Swinnerton-Dyer conjecture and its various refinements and developments to
a beautiful short exposition of A. Wiles in [42]. There readers will also find
important further references which include work of A. Beilinson, S. Bloch, G.
Faltings, J. Tate, and others. Here we only want to point out that S. Bloch,
A. Beilinson, and P. Deligne were able to relate $L$-functions attached to
various algebraic varieties with motivic cohomology. However, their work
predicts $L$-values only up to an undetermined rational factor. On the other
hand, L. Euler found values of $\zeta(2n)$ for each $n\in\mathbb{N}$

$\zeta(2n)=(-1)^{n+1}\frac{(2\pi)^{2n}}{2(2n)!}B_{2n}.$

This formula has a considerable significance because it not only highlights
the important transcendental factor $\pi^{2n}$ but also the very important
rational factor of great arithmetic and combinatorial significance. In fact,
Euler was able to essentially determine his “real version of functional
equation” of later Riemann’s complex valued zeta functions (see [27] and
[41]).

Applying his consideration, Euler was thus able to relate values of
$\zeta(2n)$ with values $\zeta(1-2n)$ we considered above. In agreement with
Euler considerations, S. Bloch and K. Kato were able to propose their well-
known conjecture on special values of $L$-functions (see [7], [14], [19],
[21], [22], [23], [24], [29], [31], [32], [38], [43], [44], [45]). Namely S.
Bloch and K. Kato proposed a beautiful insightful formula for these
corresponding $L$-functions using suitably Tamagawa measure. The details are
beyond the scope of this paper and we just refer readers to [7] and subsequent
papers.

For us an important motivation for this project where this paper is our first
contribution is that these special values of $L$-functions are related to
certain volumes and that they have a deep arithmetic and combinatorial
meaning.

We have been further influenced by the remarkable Elkies’s paper [15] and
related papers [3], [4], [6], [20], [36], and [37] as well as other papers
where the elegant and clever calculations of related polytopes is used
together with very interesting combinatorial interpretation of some related
numbers is used to provide quite stunning insight into some special values of
$L$-functions and zeta values in general.

We now discuss very shortly the outline of our paper. In section $2$ we
provide they key generalizations of some results mentioned in the Introduction
from Riemann zeta values to Hurwitz zeta values. In section 3, we apply these
results to get a similar insight into some values of $L$-functions at negative
integers. In section $4$, we further generalize our considerations to study
values of twisted $L$-functions. Finally, in section 5 we consider a special
case of zeta functions namely $L(s,\chi_{4})$. This is somewhat complementary
insight to inspiring results in [15]. We conclude our paper with Appendix A
containing some specific examples, formulas, graphs, tables as well as some
further considerations and comments on a specific literature.

##  2\. Hurwitz zeta function

The Hurwitz zeta function is defined for complex arguments $s$ with ${\rm
Re}(s)>1$ and real $a$ with $0<a\leq 1$ by

$\zeta(s,a)=\sum_{n=0}^{\infty}\dfrac{1}{(n+a)^{s}}.$

This series is absolutely convergent for the given values of $s$ and $a$ and
can be extended to a meromorphic function defined for all $s\not=1$. The
Riemann zeta function $\zeta(s)=\zeta(s,1)$.

The Bernoulli polynomials $B_{n}(x)$ and Bernoulli numbers $B_{n}$ are defined
as

$\dfrac{te^{xt}}{e^{t}-1}=\sum_{n=0}^{\infty}B_{n}(x)\dfrac{t^{n}}{n!}$

and

$B_{n}(x)=\sum_{k=0}^{n}\binom{n}{k}B_{k}x^{n-k}.$

For every $n\geq 1$, the Bernoulli polynomial $B_{n}(x)$ satisfies the
difference equation

$B_{n}(x+1)-B_{n}(x)=nx^{n-1}.$

In particular $B_{n}(1)=B_{n}(0)$. We also have

$B_{n}^{\prime}(x)=nB_{n-1}(x).$

Bernoulli numbers can be defined recursively by

$B_{0}=1,B_{1}=-\dfrac{1}{2},B_{n}=\sum_{k=0}^{n}{n\choose k}B_{k},n\geq 2.$

######  Theorem 2.1.

(See [2, Theorem 12.13]) For every integer $n\geq 0$, we have

$\zeta(-n,a)=-\dfrac{B_{n+1}(a)}{n+1}.$

For any $a\in(0,1]$, $n\in\mathbb{N}$ we define $S_{n,a}(x)$ to be the unique
polynomial such that for every integer $M\geq 2$ one has

$S_{n,a}(M)=\sum_{k=0}^{M-2}(k+a)^{n}.$

We also set $S_{n}(x)=S_{n,1}(x)$, so that $S_{n}(x)$ is the unique polynomial
sastifying that

$S_{n}(M)=\sum_{k=1}^{M-1}k^{n},\text{ for every positive integer $M$}.$

Existence of these polynomials follows from Proposition 2.2 below. The
uniqueness is clear because no two distinct polynomials with complex
coefficients can have the same values at infinitely many arguments. Observe
that our polynomials $S_{n,a}(x)$ has rational coefficients.

######  Proposition 2.2.

$S_{n,a}(x)=\dfrac{B_{n+1}(x+a-1)-B_{n+1}(a)}{n+1}.$

###### Proof.

For every integer $M\geq 2$, one has

$\displaystyle S_{n,a}(M)$ $\displaystyle=\sum_{k=0}^{M-2}(k+a)^{n}$
$\displaystyle=\sum_{k=0}^{M-2}\dfrac{B_{n+1}(k+a+1)-B_{n+1}(k+a)}{n+1}$
$\displaystyle=\dfrac{B_{n+1}(M+a-1)-B_{n+1}(a)}{n+1}.$

∎

We have the following theorem which generalizes the fact on [30, page 459].

######  Theorem 2.3.

$\int_{1-a}^{2-a}S_{n,a}(x)dx=\zeta(-n,a).$

We will give three different proofs for this theorem. First, let us explain
our first proof using the relation between $S_{n,a}(x)$ and Bernoulli
polynomials.

###### First proof of Theorem 2.3.

One has

$\displaystyle\int_{1-a}^{2-a}S_{n,a}(x)dx$
$\displaystyle=\int_{1-a}^{2-a}\dfrac{B_{n+1}(x+a-1)-B_{n+1}(a)}{n+1}dx$
$\displaystyle=\dfrac{1}{n+1}\dfrac{B_{n+2}(x+a-1)}{n+2}\bigg{\rvert}_{1-a}^{2-a}-\dfrac{B_{n+1}(a)}{n+1}x\bigg{\rvert}_{1-a}^{2-a}$
$\displaystyle=\dfrac{B_{n+2}(1)-B_{n+2}(0)}{(n+1)(n+2)}-\dfrac{B_{n+1}(a)}{n+1}$
$\displaystyle=\zeta(-n,a).\qed$

We provide another proof using the relation between values of the Hurwitz zeta
functions and values of the classical zeta function.

By definition when $n=0$, $S_{0,a}(x)=x-1$. In this case, the above
proposition is true because

$\int_{1-a}^{2-a}(x-1)dx=\frac{1}{2}-a=\zeta(0,a).$

From now on, we assume $n>0$. We recall the following proposition which
expresses the Hurwitz zeta functions in term of the Riemann zeta function.

######  Proposition 2.4 (See [26], Theorem 2.2).

For $0<a\leq 1$, we have

$\zeta(s,a)=\frac{1}{a^{s}}+\sum_{k=0}^{\infty}(-1)^{k}\frac{\Gamma(s+k)}{\Gamma(s)k!}a^{k}\zeta(s+k).$

###### Second proof of Theorem 2.3.

We recall that that $\zeta(s)$ has a simple pole at $s=1$ with residue $1$ and
$\Gamma(s)$ has a simple pole at $-m$ for $m\in\mathbb{N}$ with residue
$\dfrac{(-1)^{m}}{m!}$. By plugging in $s=-n$ and using the above formula and
these two facts, we have the following relations

$\zeta(-n,a)=a^{n}-\frac{a^{n+1}}{n+1}+\sum_{k=0}^{n}{n\choose
k}a^{k}\zeta(k-n).$ (2.1)

Let us consider the sum

$\displaystyle S_{n,a}(M)$ $\displaystyle=\sum_{m=0}^{M-2}(a+m)^{n}$
$\displaystyle=\sum_{m=0}^{M-2}\left(\sum_{k=0}^{n}{n\choose
k}a^{k}m^{n-k}\right)$ $\displaystyle=\sum_{k=0}^{n}{n\choose
k}a^{k}\left(\sum_{m=0}^{M-2}m^{n-k}\right)$
$\displaystyle=a^{m}S_{0}(x)+\sum_{k=0}^{n-1}{n\choose k}a^{k}S_{n-k}(M-1).$

Since this is true for all $M\in\mathbb{N}$, we conclude that

$S_{n,a}(x)=a^{n}S_{0}(x)+\sum_{k=0}^{n-1}{n\choose k}a^{k}S_{n-k}(x-1).$
(2.2)

By integrating both sides from $1-a$ to $2-a$ we get

$\int_{1-a}^{2-a}S_{n,a}(x)dx=a^{m}\int_{1-a}^{2-a}S_{0}(x)+\sum_{k=0}^{n-1}{n\choose
k}a^{k}\int_{1-a}^{2-a}S_{n-k}(x-1)dx.$ (2.3)

To compute the right hand side, we need the following lemma.

######  Lemma 2.5.

For $k\geq 1$, we have

$\int_{1-a}^{2-a}S_{k}(x-1)dx=\frac{(-a)^{k+1}}{k+1}+\zeta(-k).$

###### Proof.

First, we recall that

$S_{k}(x)=\frac{B_{k+1}(x)-B_{k+1}(1)}{k+1}.$

Therefore, we have

$\begin{split}\int_{1-a}^{2-a}S_{n}(x-1)dx&=\int_{-a}^{1-a}S_{n}(x)dx\\\
&=\int_{-a}^{1-a}\frac{B_{k+1}(x)-B_{k+1}(1)}{k+1}\\\
&=\frac{(-a)^{k+1}}{k+1}-\frac{B_{k+1}(1)}{k+1}\\\
&=\frac{(-a)^{k+1}}{k+1}+\zeta(-k).\end{split}$

Here we use the fact that $\zeta(-k)=-\dfrac{B_{k+1}(1)}{k+1}$. ∎

Note that when $k=0$, the above statement is not true. Instead, we have

$\int_{1-a}^{2-a}S_{0}(x)dx=\zeta(0,a)=\frac{1}{2}-a=\zeta(0)+(1-a).$

By Lemma 2.5 and the above remark, equation 2.3 becomes

$\displaystyle\int_{1-a}^{2-a}S_{n,a}(x)dx$
$\displaystyle=a^{n}(\zeta(0)+(1-a))+\sum_{k=0}^{n-1}{n\choose
k}a^{k}\left(\frac{(-a)^{n-k+1}}{n-k+1}+\zeta(n-k)\right)$
$\displaystyle=a^{n}(1-a)+\sum_{k=0}^{n}{n\choose
k}a^{k}\zeta(k-n)+\frac{a^{n+1}}{n+1}\sum_{k=0}^{n-1}(-1)^{n-k+1}{{n+1}\choose
k}.$

Note that

$\sum_{k=0}^{n+1}(-1)^{n-k+1}{{n+1}\choose k}=(1-1)^{n+1}=0.$

Therefore

$\sum_{k=0}^{n-1}(-1)^{n-k+1}{{n+1}\choose k}=-(-(n+1)+1)=n.$

Consequently, we have

$\displaystyle\int_{1-a}^{2-a}S_{n,a}(x)dx$
$\displaystyle=a^{n}(1-a)+\frac{na^{n+1}}{n+1}+\sum_{k=0}^{n}{n\choose
k}a^{k}\zeta(k-n)$
$\displaystyle=a^{n}-\frac{a^{n+1}}{n+1}+\sum_{k=0}^{n}{n\choose
k}a^{k}\zeta(k-n).$

Compare this equation and equation 2.1, we conclude that

$\zeta(-n,a)=\int_{1-a}^{2-a}S_{n,a}(x)dx.\qed$

Finally, we provide another proof without using the Bernoulli polynomials.
Instead, we use the relation between values of the Hurwitz zeta functions to
prove this theorem. Note that our approach is very close to the second
approach discussed in [30].

###### Third proof of Theorem 2.3.

For each $n\geq 1$ we have

$\displaystyle\frac{1}{(n+a-1)^{s-1}}-\frac{1}{(n+a)^{s-1}}$
$\displaystyle=\frac{1}{(n+a)^{s}}\left((1-\frac{1}{n+a})^{-(s-1)}-1\right)$
$\displaystyle=\frac{1}{(n+a)^{s-1}}\left(\frac{s-1}{(n+a)}+\frac{(s-1)s}{2!}\frac{1}{(n+a)^{2}}+\cdots\right)$
$\displaystyle=\sum_{k=0}^{\infty}\frac{\Gamma(s+k)}{\Gamma(s-1)(k+1)!}\frac{1}{(a+n)^{s+k}}.$

Summing over all $n$ we have

$\displaystyle\frac{1}{a^{s-1}}$
$\displaystyle=\sum_{n=1}^{\infty}\left[\sum_{k=0}^{\infty}\frac{\Gamma(s+k)}{\Gamma(s-1)(k+1)!}\frac{1}{(a+n)^{s+k}}\right]$
$\displaystyle=\sum_{k=0}^{\infty}\frac{\Gamma(s+k)}{\Gamma(s-1)(k+1)!}\left[\sum_{n=1}^{\infty}\frac{1}{(a+n)^{s+k}}\right]$
$\displaystyle=\sum_{k=0}^{\infty}\frac{\Gamma(s+k)}{\Gamma(s-1)(k+1)!}(\zeta(s+k,a)-\frac{1}{a^{s+k}}).$

We remark that $\zeta(s,a)$ is a meromorphic function with the only simple
pole at $s=1$. Furthermore, the residue at $s=1$ is $1$. Furthermore, for each
$m\in\mathbb{N}$, $\Gamma(s)$ has simple pole at $-m$ with residue
$\dfrac{(-1)^{m}}{m!}$. By plugging in $s=-n$ and using these two facts, we
have the following

$a^{n+1}=\sum_{k=0}^{n}(-1)^{k+1}{{n+1}\choose{k+1}}(\zeta(k-n,a)-a^{n-k})+\frac{(-1)^{n+1}}{n+2}.$
(2.4)

Next, we show that the integrals $\int_{1-a}^{2-a}S_{n,a}(x)dx$ satisfy a
similar relation. To do so, we use the following collapsing sum

$\displaystyle(M+a-2)^{n+1}$
$\displaystyle=a^{n+1}+\sum_{k=1}^{M-2}[(k+a)^{n+1}-(k+a-1)^{n+1}]$
$\displaystyle=a^{n+1}+\sum_{k=1}^{M-2}\left[\sum_{m=1}^{n+1}{{n+1}\choose
m}(-1)^{m+1}(k+a)^{n+1-m}\right]$
$\displaystyle=a^{n+1}+\sum_{m=1}^{n+1}\left[(-1)^{m+1}{{n+1}\choose
m}\sum_{k=1}^{M-2}(k+a)^{n-m+1}\right]$
$\displaystyle=a^{n+1}+\sum_{m=1}^{n+1}(-1)^{m+1}{{n+1}\choose
m}(S_{n-m+1}(M)-a^{n-m+1})$
$\displaystyle=a^{n+1}+\sum_{k=0}^{n}(-1)^{k}{{n+1}\choose
k+1}(S_{n-k}(M)-a^{n-k}).$

The last equality is obtained by substituting $m=k+1$.

By integrating both sides of this equation, we have

$\int_{1-a}^{2-a}(x+a-2)^{n+1}dx=a^{n+1}+\sum_{k=0}^{n}(-1)^{k}{{n+1}\choose
k+1}\left(\int_{1-a}^{2-a}S_{n-k}(x)dx-a^{n-k}\right).$ (2.5)

We have

$\int_{1-a}^{2-a}(x+a-2)^{n+1}dx=\int_{-1}^{0}x^{n+1}dx=\frac{(-1)^{n+1}}{n+2}.$

Therefore, equation 2.5 can be written as

$a^{n+1}=\sum_{k=0}^{n}(-1)^{k+1}{{n+1}\choose
k+1}\left(\int_{1-a}^{2-a}S_{n-k}(x)dx-a^{n-k}\right)+\frac{(-1)^{n+1}}{n+2}.$
(2.6)

We know that $\int_{1-a}^{2-a}S_{0,a}(x)dx=\dfrac{1}{2}-a=\zeta(0,a)$. By
equations 2.4 and 2.6, we can conclude that

$\int_{1-a}^{2-a}S_{a,n}(x)dx=\zeta(-n,a).\qed$

As a corollary of Theorem 2.3, one obtains

$\zeta(-n)=\zeta(-n,1)=\int_{0}^{1}S_{n,1}(x)dx=\int_{0}^{1}S_{n}(x)dx.$

Let us provide some concrete examples for Theorem 2.3.

######  Examples 2.6.

Let $0<a\leq 1$. Then we have

$S_{0,a}(x)=x-1.$ $S_{1,a}(x)=\sum_{k=0}^{x-2}(a+k)=\frac{(2a+x-2)(x-1)}{2}.$

Inductively, we can compute $S_{n,a}(M)$ using the recursive relation 2.2

$S_{n,a}(x)=a^{n}S_{0}(x)+\sum_{k=0}^{n-1}{n\choose k}a^{k}S_{n-k}(x-1).$

For example, for $n=2$ we have

$S_{2,a}(x)=\sum_{k=0}^{x-2}(a+k)^{2}=(x-1)a^{2}+a(x-2)(x-1)+\frac{(x-2)(x-1)(2x-3)}{6}.$

For $n=3$ we have

$S_{3,a}(x)=a^{3}(x-1)+3a^{2}\frac{(x-1)(x-2)}{2}+3a\frac{(x-2)(x-1)(2x-3)}{6}+\frac{((x-1)(x-2))^{2}}{4}.$

######  Examples 2.7.

Let us consider the special case when $a=\frac{1}{2}$. In this case, the
Hurwitz zeta function is given by

$\zeta(s,\frac{1}{2})=\sum_{n=0}^{\infty}\frac{1}{(n+\frac{1}{2})^{s}}=2^{s}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{s}}$

By the above calculations, we have

$S_{0,\frac{1}{2}}(x)=x-1,$

and

$S_{1,\frac{1}{2}}(x)=\frac{(x-1)^{2}}{2}.$

By Theorem 2.3, we have

$\zeta(0,\frac{1}{2})=\int_{\frac{1}{2}}^{\frac{3}{2}}(1-x)dx=0.$

We also have

$\zeta(1,\frac{1}{2})=\int_{\frac{1}{2}}^{\frac{3}{2}}\frac{(x-1)^{2}}{2}dx=\frac{1}{24}.$

######  Corollary 2.8.

If $n=2k+1$ is an odd positive integer, then

$\zeta(-2k-1)=(-1)^{k+1}(area),$

here $(area)$ is the area of a region bounded by the $x$-axis and the graph of
the function $\dfrac{B_{2k+2}(x)-B_{2k+2}}{2k+2}$ on the interal $[0,1]$.

###### Proof.

One has

$\zeta(-2k-1)=\int_{0}^{1}S_{2k+1}(x)dx=\int_{0}^{1}\dfrac{B_{2k+2}(x)-B_{2k+2}}{2k+2}dx.$

On the other hand,

$\zeta(-2k-1)=-\dfrac{B_{2k+2}}{2k+2}=(-1)^{k+1}\dfrac{(-1)^{k}B_{2k+2}}{2k+2}.$

Hence

$\dfrac{(-1)^{k}B_{2k+2}}{2k+2}=(-1)^{k+1}\zeta(-2k-1))=\int_{0}^{1}(-1)^{k+1}\dfrac{B_{2k+2}(x)-B_{2k+2}}{2k+2}dx.$

By [33, Page 8], $B_{2k+2}(x)-B_{2k+2}$ does not change sign between $0$ and
$1$ and $(-1)^{k}B_{2k+2}>0$. Therefore
$\phi_{k}(x):=(-1)^{k+1}(B_{2k+2}(x)-B_{2k+2})/(2k+2)>0$ on $(0,1)$. Hence
$\int_{0}^{1}\phi_{k}(x)dx$ is the area of the region bounded by the $x$-axix
and the graph of $\phi_{k}(x)$ on the interval $[0,1]$, and

$\zeta(-2k-1)=(-1)^{k+1}\int_{0}^{1}\phi_{k}(x)dx=(-1)^{k+1}(area).\qed$

######  Proposition 2.9.

$S_{n}(x)+(-1)^{n}S_{n}(1-x)=0.$

###### Proof.

We first observe that $S_{n}(x+1)-S_{n}(x)=x^{n}$ and $S_{n}(1)=S_{n}(0)=0$.
Now let us consider the polynomial

$p_{n}(x)=S_{n}(x)+(-1)^{n}S_{n}(1-x).$

We have

$\displaystyle p_{n}(x+1)-p_{n}(x)$
$\displaystyle=[S_{n}(x+1)-S_{n}(x)]-(-1)^{n}[S_{n}(-x)-S_{n}(1-x)]$
$\displaystyle=x^{n}-(-1)^{n}(-x)^{n}$ $\displaystyle=0.$

Hence $p_{n}(x)$ must be a constant. Therefore

$p_{n}(x)=S_{n}(0)+(-1)^{n}S_{n}(1)=0.\qed$

######  Proposition 2.10.

Suppose $n$ is an even positive integer. Then

$\zeta(-n)=\int_{0}^{1}S_{n}(x)dx=0$

and

$\int_{0}^{1/2}S_{n}(x)dx=\dfrac{(2^{-n-1}-2)}{(n+1)(n+2)}B_{n+2}.$

###### Proof.

Since $n$ is even, $S_{n}(x)+S_{n}(1-x)=0$. Hence

$\displaystyle\zeta(-n)$
$\displaystyle=\int_{0}^{1}S_{n}(x)dx=\int_{0}^{1/2}S_{n}(x)dx+\int_{1/2}^{1}S_{n}(x)dx$
$\displaystyle=\int_{0}^{1/2}S_{n}(x)dx-\int_{1/2}^{0}S_{n}(1-x)dx$
$\displaystyle=\int_{0}^{1/2}S_{n}(x)dx+\int_{0}^{1/2}S_{n}(1-x)dx$
$\displaystyle=0.$

For the second statement, one has

$\displaystyle\int_{0}^{1/2}S_{n}(x)dx$
$\displaystyle=\int_{0}^{1/2}\dfrac{B_{n+1}(x)-B_{n+1}}{n+1}dx$
$\displaystyle=\dfrac{1}{n+1}\dfrac{B_{n+2}(x)}{n+2}\bigg{\rvert}_{0}^{1/2}$
$\displaystyle=\dfrac{B_{n+2}(1/2)-B_{n+2}(0)}{(n+1)(n+2)}$
$\displaystyle=\dfrac{(2^{-n-1}-2)}{(n+1)(n+2)}B_{n+2},$

here we have used the formula that $B_{k}(1/2)=(2^{1-k}-1)B_{k}$ for every
$k\geq 0$. ∎

##  3\. Special values of $L$-functions

Suppose $\chi$ is a Dirichlet character of conductor $k$. The $L$-fuction
attached to $\chi$ is defined by

$L(s,\chi)=\sum_{n=1}^{\infty}\dfrac{\chi(n)}{n^{s}},\quad{\rm Re}(s)>1.$

The function $L(s,\chi)$ can be continued analytically to the whole complex
plane, except for a simple pole at $s=1$ when $\chi=1$. One has

$L(s,\chi)=\sum_{r=1}^{k}\chi(r)k^{-s}\zeta(s,\dfrac{r}{k}).$

We define

$S_{n,\chi}(x)=k^{n}\sum_{r=1}^{k}\chi(r)S_{n,r/k}(x+1-r/k).$

######  Lemma 3.1.

If $\chi$ is nontrivial then $S_{n,\chi}(x)=-\dfrac{B_{n+1,\chi}}{n+1}$.

###### Proof.

One has

$\displaystyle S_{n,\chi}(x)$
$\displaystyle=k^{n}\sum_{r=1}^{k}\chi(r)S_{n,r/k}(x+1-r/k)$
$\displaystyle=k^{n}\sum_{r=1}^{k}\chi(r)\dfrac{B_{n+1}(x)-B_{n+1}(r/k)}{n+1}$
$\displaystyle=k^{n}\sum_{r=1}^{k}\chi(r)\dfrac{B_{n+1}(x)}{n+1}-k^{n}\sum_{r=1}^{k}\chi(r)\dfrac{B_{n+1}(r/k)}{n+1}=-\dfrac{B_{n+1,\chi}}{n+1},$

here we used the formula that $\displaystyle
B_{n+1,\chi}=k^{n}\sum_{r=1}^{k}\chi(r)B_{n+1}(r/k)$ and that for non-trivial
$\chi$, $\sum\limits_{r=1}^{k}\chi(r)=0$. ∎

######  Corollary 3.2.

$L(-n,\chi)=\int_{0}^{1}S_{n,\chi}(x)dx=-\dfrac{B_{n+1,\chi}}{n+1}.$

###### Proof.

We have

$\displaystyle L(-n,\chi)$
$\displaystyle=k^{n}\sum_{r=1}^{k}\chi(r)\zeta(-n,r/k)$
$\displaystyle=k^{n}\sum_{r=1}^{k}\chi(r)\int_{1-r/k}^{2-r/k}S_{n,r/k}(x)dx$
$\displaystyle=k^{n}\sum_{r=1}^{k}\chi(r)\int_{0}^{1}S_{n,r/k}(x+1-r/k)dx$
$\displaystyle=\int_{0}^{1}k^{n}\sum_{r=1}^{k}\chi(r)S_{n,r/k}(x+1-r/k)dx$
$\displaystyle=\int_{0}^{1}S_{n,\chi}(x)dx.\qed$

Recall that $\chi$ is a Dirichlet character of conductor $k$. We define
$P_{n,\chi}(x)$ to be the unique polynomial such that for every integer $M\geq
1$ one has

$P_{n,\chi}(M)=\sum_{r=1}^{Mk}\chi(r)r^{n}.$

######  Lemma 3.3.

$P_{n,\chi}(x)=\dfrac{B_{n+1,\chi}(xk)-B_{n+1,\chi}}{n+1}.$

###### Proof.

For every integer $M\geq 2$, one has

$P_{n,\chi}(M)=\sum_{r=0}^{Mk}\chi(r)r^{n}=\dfrac{B_{n+1,\chi}(Mk)-B_{n+1,\chi}}{n+1}.\qed$

Recall that character $\chi$ is said to be even (respectively, odd) if
$\chi(-1)=\chi(-1)$ (respectively, $\chi(-1)=-\chi(1)$).

######  Lemma 3.4.

Suppose either $\chi$ is even and $n$ even or $\chi$ is odd and $n$ is odd.
Then $B_{n,\chi}(k/2)=B_{n,\chi}(-k/2)$.

###### Proof.

For any $x$, one has

$B_{n,\chi}(x+k)-B_{n,\chi}(x)=n\sum_{r=1}^{k}\chi(r)(r+x)^{n-1}.$

By substituting $x=-k/2$, one obtains

$B_{n,\chi}(\dfrac{k}{2})-B_{n,\chi}(-\dfrac{k}{2})=n\sum_{r=1}^{k}\chi(r)(r-\dfrac{k}{2})^{n-1}.$

We treat the case that $\chi$ is even and $n$ is even. The other case can be
treated similarly. Since $\chi$ is even and $n$ is even, for each $r$ coprime
to $k$, one has

$\displaystyle\chi(r)(r-\dfrac{k}{2})^{n-1}+\chi(k-r)(\dfrac{k}{2}-r)^{n-1}$
$\displaystyle=\chi(r)(r-\dfrac{k}{2})^{n-1}+\chi(-r)(\dfrac{k}{2}-r)^{n-1}$
$\displaystyle=(\chi(r)-\chi(-r))(r-\dfrac{k}{2})^{n-1}=0.$

Thus

$B_{n,\chi}(\dfrac{k}{2})-B_{n,\chi}(-\dfrac{k}{2})=n\sum_{r=1}^{k}\chi(r)(r-\dfrac{k}{2})^{n-1}=0.\qed$

######  Proposition 3.5.

Suppose either $\chi$ is even and $n$ even or $\chi$ is odd and $n$ is odd. We
have

$\int_{-1/2}^{1/2}P_{n,\chi}(x)dx=L(-n,\chi).$

###### Proof.

One has

$\displaystyle\int_{-1/2}^{1/2}P_{n,\chi}(x)dx$
$\displaystyle=\int_{-1/2}^{1/2}\dfrac{B_{n+1,\chi}(xk)-B_{n+1,\chi}}{n+1}dx$
$\displaystyle=\int_{-k/2}^{k/2}\dfrac{B_{n+1,\chi}(y)dy}{k(n+1)}-\dfrac{B_{n+1,\chi}}{n+1}$
$\displaystyle=\dfrac{1}{k(n+1)}\dfrac{B_{n+2,\chi}(y)}{n+2}\bigg{\rvert}_{-k/2}^{k/2}-\dfrac{B_{n+1,\chi}}{n+1}$
$\displaystyle=\dfrac{B_{n+2,\chi}(k/2)-B_{n+2,\chi}(-k/2)}{k(n+1)(n+2)}-\dfrac{B_{n+1,\chi}}{n+1}$
$\displaystyle=L(-n,\chi).\qed$

##  4\. Special values of twisted $L$-functions

Suppose $\chi$ is a Dirichlet character of conductor $k$. For each $0<a\leq
1$, the $L$-fuction attached to $\chi$ is defined by

$L(s,a,\chi)=\sum_{n=0}^{\infty}\dfrac{\chi(n+1)}{(n+a)^{s}},\quad{\rm
Re}(s)>1.$

Note that when $a=1$, this definition agrees with the classical $L$-function
associated with $\chi$.

We have

$\displaystyle L(s,a,\chi)$
$\displaystyle=\displaystyle{\sum_{n=0}^{\infty}\dfrac{\chi(n+1)}{(n+a)^{s}}}$
$\displaystyle=\sum_{r=0}^{k-1}\left(\sum_{n\equiv
r(k)}^{\infty}\frac{\chi(n+1)}{(n+a)^{s}}\right)$
$\displaystyle=\sum_{r=0}^{k-1}\left(\sum_{m=0}^{\infty}\frac{\chi(mk+r+1)}{(mk+r+a)^{s}}\right)$
$\displaystyle=\sum_{r=0}^{k-1}\chi(r+1)k^{-s}\left(\sum_{m=0}^{\infty}\frac{1}{(m+\frac{r+a}{k})^{s}}\right)$
$\displaystyle=\sum_{r=0}^{k-1}\chi(r)k^{-s}\zeta(s,\frac{r+a}{k})$
$\displaystyle=\sum_{r=1}^{k}\chi(r)k^{-s}\zeta(s,\frac{r+a-1}{k}).$

From this expression, we can see that the function $L(s,a,\chi)$ can be
continued analytically to the whole complex plane, except for a simple pole at
$s=1$ when $\chi=1$. We define

$S_{n,a,\chi}(x)=k^{n}\sum_{r=1}^{k}\chi(r)S_{n,\frac{r+a-1}{k}}(x+1-\frac{r+a-1}{k}).$

We also define the following modified generalized Bernoulli polynomials
$\tilde{B}_{n,\chi}(x)$

$\tilde{B}_{n,\chi}(x)=B_{n,\chi}(x-1),$

where $B_{n,\chi}(x)$ is the generalized Bernoulli polynomial defined in [13,
Equation 1.2.7]. Our justification for this modified definition is that when
$\chi$ is the trivial character, we have $\tilde{B}_{n,\chi}(x)=B_{n}(x)$. We
have the following lemma.

######  Lemma 4.1.

If $\chi$ is nontrivial then
$S_{n,a,\chi}(x)=-\dfrac{\tilde{B}_{n+1,\chi}(a)}{n+1}.$

###### Proof.

One has

$\displaystyle S_{n,a,\chi}(x)$
$\displaystyle=k^{n}\sum_{r=1}^{k}\chi(r)S_{n,\frac{r+a-1}{k}}(x+1-\frac{r+a-1}{k})$
$\displaystyle=k^{n}\sum_{r=1}^{k}\chi(r)\dfrac{B_{n+1}(x)-B_{n+1}(\frac{r+a-1}{k})}{n+1}$
$\displaystyle=k^{n}\sum_{r=1}^{k}\chi(r)\dfrac{B_{n+1}(x)}{n+1}-k^{n}\sum_{r=1}^{k}\chi(r)\dfrac{B_{n+1}(\frac{r+a-1}{k})}{n+1}=-\dfrac{\tilde{B}_{n+1,\chi}(a)}{n+1},$

here we used the formula that
$\tilde{B}_{n+1,\chi}(a)=k^{n}\sum\limits_{r=1}^{k}\chi(r)B_{n+1}(\frac{r+a-1}{k})$
(see [13, Equation 1.2.14]). ∎

######  Corollary 4.2.

Let $\chi$ be a non-trivial primitive Dirichlet character. We have

$L(-n,a,\chi)=\int_{0}^{1}S_{n,\chi}(x)dx=-\dfrac{\tilde{B}_{n+1,\chi}(a)}{n+1}.$

###### Proof.

We have

$\displaystyle L(-n,a,\chi)$
$\displaystyle=k^{n}\sum_{r=1}^{k}\chi(r)\zeta(-n,\frac{r+a-1}{k})$
$\displaystyle=k^{n}\sum_{r=1}^{k}\chi(r)\int_{1-\frac{r+a-1}{k}}^{2-\frac{r+a-1}{k}}S_{n,\frac{r+a-1}{k}}(x)dx$
$\displaystyle=k^{n}\sum_{r=1}^{k}\chi(r)\int_{0}^{1}S_{n,\frac{r+a-1}{k}}(x+1-\frac{r+a-1}{k})dx$
$\displaystyle=\int_{0}^{1}k^{n}\sum_{r=1}^{k}\chi(r)S_{n,\frac{r+a-1}{k}}(x+1-\frac{r+a-1}{k})dx$
$\displaystyle=\int_{0}^{1}S_{n,\chi,a}(x)dx.$

∎

##  5\.  The zeta function $L(s,\chi_{4})$

In this section, we use the studies in the previous sections to study the zeta
function $L(s,\chi_{4})$ where $\chi_{4}$ is the quadratic character of
conductor $4$ given by

$\chi(m)=\begin{cases}0&\mbox{if }2|m\\\ 1&\mbox{if }m\equiv 1\pmod{4}\\\
-1&\mbox{if }m\equiv 3\pmod{4}.\end{cases}$

###  5.1. The first integral representation of $L(-n,\chi_{4})$

By Corollary 3.2, we have

$L(-2n,\chi_{4})=-\frac{B_{2n+1,\chi_{4}}}{2n+1}=-\frac{4^{2n}}{2n+1}\left(B_{2n+1}\left(\frac{1}{4}\right)-B_{2n+1}\left(\frac{3}{4}\right)\right).$

By [33, page 7, formula (2.8)]), we have $B_{n}(x)=(-1)^{n}B_{n}(1-x)$, we
have

$B_{2n+1}\left(\frac{3}{4}\right)=-B_{2n+1}\left(\frac{1}{4}\right).$

Therefore, we have

$L(-2n,\chi_{4})=-\frac{4^{2n+1}B_{2n+1}\left(\frac{1}{4}\right)}{2(2n+1)}.$

By Theorem 2.1, we know that

$\zeta(-2n,\frac{1}{4})=-\frac{B_{2n+1}\left(\frac{1}{4}\right)}{2n+1}.$

Combining these two facts we have

$L(-2n,\chi_{4})=\frac{4^{2n+1}}{2}\zeta(-2n,\frac{1}{4}).$

By Theorem 2.3, we have the following

######  Proposition 5.1.

Let $n$ be a positive integer. Then

  1. (1)

If $n$ is odd then $L(-n,\chi_{4})=0$.

  2. (2)

If $n$ is even then

$L(-n,\chi_{4})=\frac{4^{n+1}}{2}\int_{\frac{3}{4}}^{\frac{7}{4}}S_{n,\frac{1}{4}}(x)dx.$

######  Examples 5.2.

Let us give a concrete example for Proposition 5.1. By Example 2.6, we have

$S_{2,\frac{1}{4}}(x)=\frac{1}{16}(x-1)+\frac{1}{4}(x-1)(x-2)+\frac{(x-1)(x-2)(2x-3)}{6}.$

Therefore, by the above proposition we have

$L(-2,\chi_{4})=\frac{4^{3}}{2}\int_{\frac{3}{4}}^{\frac{7}{4}}\left[\frac{1}{16}(x-1)+\frac{1}{4}(x-1)(x-2)+\frac{(x-1)(x-2)(2x-3)}{6}\right]dx=-\frac{1}{2}.$

###  5.2. The second integral representation of $L(-n,\chi_{4})$

In this subsection, we provide another integral representation for
$L(-n,\chi_{4})$. Our presentation here is strongly influenced by [34]. First,
let us recall the Lerch zeta function and its special values studied in [34].

$\zeta(s,a,\gamma)=\sum_{n=0}^{\infty}\frac{\gamma^{n}}{(n+a)^{s}},$

where $0<a\in\mathbb{R}$ and $0<|\gamma|\leq 1$. When $\gamma=\exp(-2\pi
i\alpha)$ with $\alpha\in\mathbb{R}\backslash\mathbb{Z}$, we have the
following theorem.

######  Theorem 5.3.

[34, Theorem 0.2] For $0<k\in\mathbb{Z}$, $\Re(a)>0$, the value
$\zeta(1-k,a,\gamma)$ is given by

$\zeta(1-k,a,\gamma)=E_{c,k-1}(a)/(1+c^{-1}),$

where $c=\gamma^{-1}$ and $E_{c,k-1}(t)$ is the Euler polynomial defined by
[34, Equation 0.3].

When $c=1$ (equivalently $\gamma=-1)$, $E_{1,n}(t)$ is the classical Euler
polynomial of degree $n$. Here are some important properties of Euler
polynomials that we will use later on (see [35, Page 25]).

$E_{c,n}=2^{n}E_{c,n}\left(\frac{1}{2}\right).$ (5.1)
$(d/dt)E_{c,n}(t)=nE_{c,n-1}(t)\quad(n>0).$ (5.2)
$E_{c,n}(t+1)+cE_{c,n}(t)=(1+c)t^{n}.$ (5.3)

We have the following identity

$L(s,\chi_{4})=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{s}}=\frac{1}{2^{s}}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+\frac{1}{2})^{s}}=2^{-s}\zeta(s,\frac{1}{2},-1).$
(5.4)

By Theorem 5.2 we have

$\zeta(-n,\frac{1}{2},-1)=\frac{E_{1,n}(\frac{1}{2})}{2}.$

By equation 5.4, we deduce that

$L(-n,\chi_{4})=\frac{2^{n}E_{1,n}(\frac{1}{2})}{2}=\frac{E_{1,n}}{2}.$

Let us now give an integral representation of $L(-n,\chi_{4})$. From equation
5.3, we can see that

$E_{1,n}(t+2)-E_{1,n}(t)=2((t+1)^{n}-t^{n}).$

In particular, $E_{1,n}(2)-E_{1,n}(0)=2$ for all $n$. By equation 5.2 we have

$\int_{0}^{2}E_{1,n}(x)dx=\left[\frac{E_{1,n+1}(x)}{n+1}\right]_{0}^{2}=\frac{E_{1,n+1}(2)-E_{n+1}(0)}{n+1}=\frac{2}{n+1}.$

Consequently, we have

$\displaystyle-\frac{1}{4}\int_{0}^{2}(E_{1,n}(x)-E_{1,n}-\frac{1}{n+1})dx$
$\displaystyle=\frac{E_{1,n}}{2}-\frac{1}{4}\left[\int_{0}^{2}E_{1,n}(x)dx-\frac{2}{n+1}\right]$
$\displaystyle=\frac{E_{1,n}}{2}=L(-n,\chi_{4}).$

In summary, we have just proved the following

######  Proposition 5.4.

For all $n\geq 0$ we have

$L(-n,\chi_{4})=-\frac{1}{4}\int_{0}^{2}(E_{1,n}(x)-E_{1,n}-\frac{1}{n+1})dx.$

##  Appendix A Some formulas and figures

In this appendix, we provide the formulas of $S_{n}(x)$ for small $n$ and
their graphs on the interval $[0,1]$. First, we have the following formulas
for $S_{n}(x)$ with $1\leq n\leq 6$.

$S_{1}(x)=\frac{(x-1)x}{2},S_{2}(x)=\frac{(x-1)x(2x-1)}{6},$
$S_{3}(x)=\frac{(x(x-1))^{2}}{4},S_{4}(x)=\frac{1}{6}(x^{6}-3x^{5}+\frac{5}{2}x^{4}-\frac{1}{2}x^{2}),$
$S_{5}(x)=\frac{1}{6}(x^{6}-3x^{5}+\frac{5}{2}x^{4}-\frac{1}{2}x^{2},S_{6}=\frac{1}{7}(x^{7}-\frac{7}{2}x^{6}+\frac{7}{2}x^{5}-\frac{7}{6}x^{3}+\frac{1}{6}x).$

We plot the graphs of $S_{n}(x)$ for $n\in\\{1,3,5\\}$.

Here are the graphs of $S_{n}(x)$ for $n\in\\{2,4,6\\}$.

Finally, we plot the region bounded by $S_{n}(x)$ and $y=0$ over
$[0,\frac{1}{2}]$.

Next, we compute certain generalized Bernoulli numbers $B_{n,\chi}$ where
$\chi$ is a quadratic character of conductor $p$. Concretely, this is the
character given by Legendre symbol
$\chi(a)=\left(\frac{a}{p}\right),a\in\mathbb{Z}.$ First, we provide a table
of $B_{n,\chi}$ for $p\leq 11$.

$n$ | $B_{n}$ | $B_{n,\chi_{3}}$ | $B_{n,\chi_{4}}$ | $B_{n,\chi_{5}}$ | $B_{n,\chi_{7}}$ | $B_{n,\chi_{11}}$
---|---|---|---|---|---|---
$0$ | 1 | 0 | 0 | 0 | 0 | 0
$1$ | -1/2 | -1/3 | -1/2 | 0 | -1 | -1
$2$ | 1/6 | 0 | 0 | 4/5 | 0 | 0
$3$ | 0 | 2/3 | 3/2 | 0 | 48/7 | 18
$4$ | -1/30 | 0 | 0 | -8 | 0 | 0
$5$ | 0 | -10/3 | -25/2 | 0 | -160 | -12750/11
$6$ | 1/42. | 0 | 0 | 804/5 | 0 | 0
$7$ | 0 | 98/3 | 427/2 | 0 | 8176 | 152082
$8$ | -1/30 | 0 | 0 | -5776 | 0 | 0.
$9$ | 0 | -1618/3 | -12465/2 | 0 | -5086656/7 | -33743250
$10$ | 5/66 | 0 | 0 | 1651004/5 | 0 | 0
$11$ | 0 | 40634/3 | 555731/2 | 0 | 99070928 | 11392546506
$12$ | -691/2730 | 0 | 0 | -27622104 | 0 | 0

Next, we provide a table of $B_{n,\chi}$ for $13\leq p\leq 23$.

$n$ | $B_{n,\chi_{13}}$ | $B_{n,\chi_{17}}$ | $B_{n,\chi_{19}}$ | $B_{n,\chi_{23}}$
---|---|---|---|---
$0$ | 0 | 0 | 0 | 0
$1$ | 0 | 0 | -1 | -3
$2$ | 4 | 8 | 0 | 0
$3$ | 0 | 0 | 66 | 144
$4$ | -232 | -656 | 0 | 0
$5$ | 0 | 0 | -13450 | -34080
$6$ | 401556/13 | 138984 | 0 | 0
$7$ | 0 | 0 | 5303074 | 18665136
$8$ | -7482704 | -958428704/17 | 0 | 0
$9$ | 0 | 0 | -66751985430/19 | -17895000384
$10$ | 2890943420 | 37040430040 | 0 | 0
$11$ | 0 | 0 | 3539203405562 | 605747775717744/23
$12$ | -1634752049016 | -35766492971568 | 0 | 0

While working on these numerical data, we observed that in our examples, it is
always the case that $v_{p}(B_{\frac{p-1}{2},\chi})=-1$. Using the work of
Ernvall (see [16]), we are able to prove a more general statement.

######  Remark A.1.

After finding a direct proof and searching further the literature, we also
found Carlitz’s result (see [11, Theorem 3]) that implies the above
observation. The announcement of results in [11] is in [10]. Observe however
that there is a misprint in Theorem 1, line 4 in [10]. The relevant part of
[11, Theorem 3] states in our special case where values of character are just
$1$ or $-1$ the following: “Let $g$ be a primitive root mod $p$. If the
conductor $f$ of the primitive character mod $f$ is a prime number $p$, $p>2$
,then $\frac{B_{n},\chi}{n}$ is integer unless $p$ and $1-\chi(g)g^{n}$ are
not coprime in which case …” This statement does imply that
$v_{p}(B_{\frac{p-1}{2},\chi})=-1$ and also the more general version of our
Proposition A.1. In this case (see [11, Theorem 3] for the relevant notations)
$\nu=0$ and using that $g$ is a primitive root mod $p$ we see that that
$\gcd(p,1-\chi(g)g^{n})=\gcd(p,1+g^{n})>1$ iff
$n\equiv\frac{p-1}{2}\pmod{(p-1)}$, and therefore we have that
$pB_{\frac{p-1}{2},\chi}\equiv-1\pmod{p}.$ For the reader’s convenience here
we provide a direct short proof of this very interesting statement.

######  Proposition A.2.

Let $\chi$ be a quadratic character of conductor $p$ where $p$ is an odd prime
number and $n\geq 0$. Then

  1. (1)

If $n\not\equiv\frac{p-1}{2}\pmod{(p-1)}$ then $B_{n,\chi}\in\mathbb{Z}$.

  2. (2)

If $n\equiv\frac{p-1}{2}\pmod{(p-1)}$ then $B_{n,\chi}=\frac{a}{p}$ where
$a\in\mathbb{Z}$ and $a\equiv-1\pmod{p}$.

###### Proof.

By [16, Theorem 1.2], we know that if $n\not\equiv\delta_{\chi}\pmod{2}$ then
$B_{n,\chi}=0$. Furthermore, we remark that for all odd primes $p$

$\frac{p-1}{2}\equiv\delta_{\chi}\pmod{2}.$

Therefore, it is is sufficient to prove this proposition when
$n\equiv\delta_{\chi}\pmod{2}$.

By [16, Theorem 1.4], we know that $pB_{n,\chi}$ is $p$-integral.
Additionally, by [16, Theorem 1.5], $B_{n,\chi}$ is $q$-integral for all
primes $q\neq p$. Combining these two facts, we can conclude that
$pB_{n,\chi}\in\mathbb{Z}$ for all $n\geq 0$. Furthermore, by [16, Theorem
1.6], we know that

$p^{2}B_{n,\chi}\equiv\sum_{a=1}^{p^{2}}\chi(a)a^{n}\pmod{p^{2}}.$

We have the following congruences modulo $p^{2}$

$\displaystyle\sum_{a=1}^{p^{2}}\chi(a)a^{n}$
$\displaystyle=\sum_{a=1}^{p}\left(\sum_{k=0}^{p-1}\chi(kp+a)(kp+a)^{n}\right)$
$\displaystyle=\sum_{a=1}^{p}\chi(a)\sum_{k=0}^{p-1}\left(a^{n}+n(kp)a^{n-1}\right)$
$\displaystyle=\sum_{a=1}^{p}\chi(a)\left(pa^{n}+npa^{n-1}\sum_{k=1}^{p-1}k\right)$
$\displaystyle=p\sum_{a=1}^{p}\chi(a)a^{n}+\frac{np^{2}(p-1)}{2}\sum_{a=1}^{p}\chi(a)a^{n-1}$
$\displaystyle=p\sum_{1}^{p}\chi(a)a^{n}\pmod{p^{2}}.$

The second congruence comes from the expansion of $(a+kp)^{n}$ leaving out
terms which are divisible by $p^{2}$. The last congruence comes from the
identity

$\sum_{k=1}^{p-1}k=\frac{p(p-1)}{2}.$

We have the following well-known simple lemma.

######  Lemma A.3.

Let $r$ be a natural number. Then

$\sum_{a=1}^{p}a^{r}=\begin{cases}0&\mbox{if }p-1\nmid r\\\ -1&\mbox{if
}p-1|r\end{cases}.$

Using this lemma and the fact that $\chi(a)\equiv a^{\frac{p-1}{2}}\pmod{p}$
we have the following congruences modulo $p$

$\sum_{a=1}^{p}\chi(a)a^{n}\equiv\sum_{a=1}^{p}a^{n+\frac{p-1}{2}}\equiv\begin{cases}-1&\mbox{if
}n\equiv\frac{p-1}{2}\pmod{(p-1)}\\\ 0&\mbox{else. }\end{cases}$

Consequently, we have the following congruences modulo $p^{2}$

$p^{2}B_{n,\chi}\equiv\begin{cases}-p&\mbox{if
}n\equiv\frac{p-1}{2}\pmod{(p-1)}\\\ 0&\mbox{else. }\end{cases}$

Combining this congruence and the fact that $pB_{n,\chi}\in\mathbb{Z}$, the
proposition follows easily. ∎

######  Acknowledgements.

Over the years after publication of the short note [30], Mináč received a
number of encouraging correspondences and discussions concerning further
generalizations and explorations of possible values of other zeta functions.
We are extremely grateful for our correspondents including but not limited to
S. Chebolu, R. Dwilewicz, G. Everest, E. Frenkel, A. Granville, J. Merzel, L.
Muller, Š. Porubský, P. Ribenboim, Ch. Rottger, A. Schultz, M.Z Spivey, B.
Sury. Further, we are are grateful to K. Dilcher and R. Ernvall for helping us
to obtain some related references including the very nice R. Ernvall’s thesis
[16]. We are also grateful to an anonymous referee for his/her careful reading
of the manusript, and for providing us with helpful comments and valuable
suggestions. Last but not least, we are grateful to Leslie Hallock for her
careful proofread.

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