# Efficient and Accurate Co-Visible Region Localization with Matching Key-
Points Crop (MKPC): A Two-Stage Pipeline for Enhancing Image Matching
Performance

Hongjian Song*, Yuki Kashiwaba*, Shuai Wu, Canming Wang

###### Abstract

Image matching is a classic and fundamental task in computer vision. In this
paper, under the hypothesis that the areas outside the co-visible regions
carry little information, we propose a matching key-points crop (MKPC)
algorithm. The MKPC locates, proposes and crops the critical regions, which
are the co-visible areas with great efficiency and accuracy. Furthermore,
building upon MKPC, we propose a general two-stage pipeline for image
matching, which is compatible to any image matching models or combinations. We
experimented with plugging SuperPoint + SuperGlue into the two-stage pipeline,
whose results show that our method enhances the performance for outdoor pose
estimations. What’s more, in a fair comparative condition, our method
outperforms the SOTA on Image Matching Challenge 2022 Benchmark, which
represents the hardest outdoor benchmark of image matching currently.††*
denotes contributing equally to this work.

##  1 Introduction

Image matching is a fundamental and classic task in computer vision, playing a
crucial role in various downstream tasks such as Simultaneous Localization and
Mapping (SLAM), Structure-from-Motion (SfM), and Localization. In the task of
image matching, the objective is to establish the relationship between two
given images by employing methods such as feature point extraction [1; 2],
feature point description [1; 2], and feature point matching [3]. This
relationship can be described using a fundamental matrix, an essential matrix,
or a homography matrix. Recently, deep learning models have been developed to
implement end-to-end keypoint matching, capable of outputting matched key-
points by simply inputting two images [4; 5; 6; 7].

While most current works focus on interest point detectors [1; 2], descriptors
[1; 2], matching, or implementing in an end-to-end way [4; 5; 6; 7]. we found
that the input images have a significant impact on the performance of image
matching. Essentially, improving the performance of image matching primarily
involves reducing the number of incorrect matches while increasing the number
of correct matches. Under the assumption that areas outside the co-visible
regions contain little valuable information and the matching points outside
these regions are incorrect matches, we define the co-visible area of two
images as the "critical region" (Figure 1). By focusing on the critical region
as the matching target, we aim to enhance image matching performance.
Identifying the critical regions shared between two images and applying fine-
grained matching within these regions offers two significant advantages: (i)
it refines the matching of critical regions and (ii) it rejects the majority
of outliers or noisy points. Based on these considerations, we propose a
simple but effective algorithm in image matching, called Matching Key-Points
Crop (MKPC). The essence of MKPC is a DBSCAN clustering algorithm. The MKPC
algorithm leverages matching key-points obtained from arbitrary matching
models to efficiently and accurately crop the critical regions of both images
with great efficiency and remarkable accuracy, while the key-points produced
by upstream models are representative.

Building upon MKPC, we propose a two-stage pipeline for image matching that
can be easily extended for any image matching models while preserving superior
performance. In the first stage, arbitrary image matching models generate
matching points for the two images. Subsequently, the MKPC algorithm clusters
these matching points and proposes a Region of Interest (ROI) area, which
represents the critical regions for both images. A naive array slicing
operation crops the critical regions, which are then fed into arbitrary image
matching models after necessary pre-processing. This constitutes the second
stage of our two-stage pipeline. Our experiments demonstrate that the two-
stage pipeline with MKPC as the core consistently improves image matching
performance on the PhotoTourism dataset, which is a subset of YFCC100M dataset
[8] with ground truth poses. The experimental results not only validate the
effectiveness of our proposed method, including MKPC and the two-stage
pipeline but also support the theoretical foundation that the key to enhancing
image matching performance lies in accurately locating the co-visible areas.
Although our method is also applicable to indoor scenarios, it requires higher
computational complexity. Furthermore, our approach outperforms the state-of-
the-art (SOTA) on the Image Matching Challenge 2022 benchmark, which is
considered the most challenging outdoor pose estimation dataset.

The contributions of this paper are as follows: (i) We propose that the key to
enhancing the performance of image matching lies in identifying critical
regions, or the co-visible areas in both images. (ii) We introduce Matching
Key-points Crop(MKPC), a critical region finder that efficiently crops these
areas with remarkable accuracy, and (iii) Building upon MKPC, we present a
two-stage pipeline that enhances the effectiveness of arbitrary image matching
models for outdoor pose estimations. In addition, our proposed two-stage
pipeline achieved a higher score than the SOTA score on the Image Matching
Challenge 2022 benchmark, which represents the most difficult benchmark for
outdoor estimation currently.

Figure 1: An example of critical regions of two images

##  2 Method

Defining the co-visible area between two images as the critical region, we
propose that the key to improve the effect of image matching is to find the
critical region with better accuracy. We propose this point based on the
assumption that the co-visible area does not contain information useful for
image matching outside that region. Therefore, we propose a matching key-
points crop(MKPC) algorithm, which is capable to locate the critical region
effectively and accurately. Building upon MKPC, we further propose a two-stage
pipeline which supports being extended arbitrarily.

###  2.1 Matching Key-points Crop (MKPC)

The MKPC algorithm crops critical regions by clustering the matching key-
points between two images outputted by arbitrary image matching models. The
workflow is shown is Figure 2. Before applying MKPC, we already have two
images $I_{1}$ and $I_{2}$, and matched key-points on both images with any
model (or any combinations of models), which is defined as $X^{1}_{1}$ and
$X^{1}_{2}$, respectively. $X^{1}_{i}$ denotes the $i_{th}$ image in the first
stage of the two-stage pipeline which will be proposed in the next subsection.
It consists of the following three steps: (i) Clustering the matching key-
points($X^{1}_{1}$ and $X^{1}_{2}$) of two images($I_{1}$ and $I_{2}$) with
DBSCAN. (ii) Generate a bounding box by filtering and gathering those clusters
(iii) Crop the area covered by the Bounding box. The Algorithm 1 describes the
MKPC algorithm flow in detail. With the input of two images ($I_{1}$, $I_{2}$)
and corresponding matched key-points in stage-one ($X^{1}_{1}$ and
$X^{1}_{2}$), the MKPC outputs the respective cropped critical regions.

We choose DBSCAN, instead of other cluster algorithms such as K-means, to
implement MKPC for some of its structural advantages. (i) DBSCAN is a density-
based method, which intuitively suitable for this work. That is, areas with
high-density matching key-points must be important, while places with low
density may be not so important or even outliers. (ii) DBSCAN is robust to
outliers, and the one of the motivations of DBSCAN is rejecting outliers.
(iii) we never needs to specify the number of clusters before applying a
DBSCAN, while we needs before applying other clusters such as K-means.

Algorithm 1 The Workflow of Matching key-points Crop (MKPC)

0: Two images to be matched $I_{1}$ and $I_{2}$; The matched key-points on
both images with any models (or combinations of models) $X^{1}_{1}$ and
$X^{1}_{2}$; A cluster selection threshold $T$, which is set to $0.05$ in our
experiment;

0: (1) Clustering on $X^{1}_{1}$ and $X^{1}_{2}$ with a DBSCAN to get
collections of clusters $C_{1}$ and $C_{2}$ for both images; (2) Sorted the
clusters in $C_{1}$ and $C_{2}$ in descending order according to the number of
points in each cluster. Supposing clusters $C_{1}$ and $C_{2}$ contained $k1$
and $k2$ clusters, respectively, we define
$C_{1}=\\{c^{1}_{1},c^{2}_{1},\cdots,c^{k1}_{1}\\}$ and
$C_{2}=\\{c^{1}_{2},c^{2}_{2},\cdots,c^{k2}_{2}\\}$, in which $c^{1}_{1}$ and
$c^{1}_{2}$ denotes the "largest" cluster in $C_{1}$ and $C_{2}$,
respectively; (3) Define two empty candidate clustering sets $C^{\prime}_{1}$
and $C^{\prime}_{2}$ for each of the two images; (4) Select the largest size
cluster $c^{1}_{1}$ for $C_{1}$. Append $c^{1}_{1}$ into $C^{\prime}_{1}$; (5)
Define the current selected index as $i=2$, Select the next largest cluster
$c_{next}=c^{i}_{1}$; Get the ratio $r$ between $c_{next}$ and $c^{1}_{1}$,
$r_{1}=\frac{c_{next}}{c^{1}_{1}}$; while ($r>=T$ and $i<=k1$) do Append
$c^{next}$ into $C^{\prime}_{1}$; $i=i+1$; Select the next largest cluster
$c_{next}=c^{i}_{1}$; Get the ratio $r$ between $c_{next}$ and $c^{1}_{1}$,
$r_{1}=\frac{c_{next}}{c^{1}_{1}}$; (6) Apply the same operations as step (4)
and step (5) for another image $I_{2}$ and obtain the candidate clusters set
$C^{\prime}_{2}$; (7) Take the 4 points $P_{1}$ in the plane coordinate system
that are the most upper-left, the most lower-left, the most upper-right, and
the most lower-right of all points in the all the clusters in
$C^{\prime}_{1}$. Apply the same operations for $C^{\prime}_{2}$ to get
$P_{2}$;(8) Use $P_{1}$ to crop $I_{1}$ to get $I^{\prime}_{1}$. Use $P_{1}$
to crop $I_{2}$ to get $I^{\prime}_{2}$; (9) Return $I^{\prime}_{1}$ and
$I^{\prime}_{2}$;

Figure 2: The workflow of matching key-points crop(MKPC)

Take Figure 3 as an example. After clustering, $X^{1}_{1}$ is split into four
clusters (red with 235 points, green with 894 points, yellow with 471 points,
and pink with 9 points) and outliers (blue), and $X^{1}_{2}$ is split into two
clusters(red, green) and outliers (blue). For $I_{2}$, it’s obvious that the
red and green clusters will be gather and the outliers will be rejected. For
$I_{1}$, in the first loop, MKPC selects the green one with the largest
clusters and the yellow clusters as the second largest cluster, and both are
extended into the candidate clustering sets $C^{\prime}_{1}$ after the
judgment. Afterwards, the red one is extended while the pink one is rejected.
Hence, after the loop in Algorithm 1, the green, yellow, and red clusters are
gathered, while the pink one and the blue one are rejected.

Figure 3: A demo of matching key-points crop(MKPC).After the loop in 1, the
green, yellow, and red cluster will be gathered, while the pink one and the
blue one will be rejected. If the pink one is included, more areas without
useful information will be included, making rejected the pink cluster makes
sense.

###  2.2 Two-stage Pipeline of Image Matching

With MKPC as the core, we propose an arbitrarily extensible two-stage pipeline
for image matching. As shown in Figure 4, The two-stages pipeline support
plugging any image matching models or recipe of models into both stages, such
that the models set for stage 1 is
$M^{1}=\\{m^{1}_{1},m^{1}_{2},\cdots,m^{1}_{n_{1}}\\}$, and the models set for
stage 2 is $M^{2}=\\{m^{2}_{1},m^{2}_{2},\cdots,m^{2}_{n_{2}}\\}$. Each of the
models in stage-one produce a set of key-points for both images. The
concatenated key-points of stage-one $X_{1}^{1}$ and $X_{2}^{1}$ are then fed
into MKPC algorithm. Based on the two images $I_{1}$ and $I_{2}$ and the key-
points of stage-one $X_{1}^{1}$ and $X_{2}^{1}$, The MKPC generates the
cropped critical regions for both images ($I^{\prime}_{1}$ and
$I^{\prime}_{2}$). With the cropped regions of both images, each of the models
in stage-two outputs the matched key-points for stage-two ($X_{1}^{2}$ and
$X_{2}^{2}$). Afterwards, the models in stage-two matches on the cropped area,
producing the key-points for the second stage $X^{2}_{1}$ and $X^{2}_{2}$.
Concatenating the key-points of both stages, i.e. getting the matched key-
points for the first image $X_{1}=Cat(X^{1}_{1},X^{2}_{1})$ and the matched
key-points for the second image $X_{2}=Cat(X^{1}_{2},X^{2}_{2})$, two sets of
matching key-points for both images are generated. Both of the sets are then
fed into a RANSAC algorithm to maintain the F matrix, E matrix, or H matrix,
as illustrated in Algorithm 2. Including the key-points of stage-one may seem
contradictory to our assumption that little valuable information is outside
the critical regions. However, MKPC suffers a risk of removing small co-
visible regions because it discards clusters with few matching points.
Although including the key-points in stage-1 potentially absorbs outliers, the
advantage of encompassing the matching points that stage-2 may miss is
greater. The percentage of correct correspondence points increases thanks to
MKPC and the matching key-points in stage-2, making it easier for RANSAC to
handle the noisy points generated by stage-1.

Figure 4: The workflow of matching key-points crop(MKPC) Algorithm 2 The
workflow of the two-stage pipeline for image matching

0: Two images to be matched $I_{1}$ and $I_{2}$; A set of models for stage-one
$M^{1}=\\{m^{1}_{1},m^{1}_{2},\cdots,m^{1}_{n_{1}}\\}$; A set of models for
stage-two $M^{2}=\\{m^{2}_{1},m^{2}_{2},\cdots,m^{2}_{n_{2}}\\}$;

0: (1) Define two empty sets of points $X^{1}_{1}$ and $X^{1}_{2}$ for both
images in stage-one; for ($m^{1}_{i}$ in $M^{1}$ ; $0<i<n_{1}+1$) do Input
$I_{1}$ and $I_{2}$ into $m^{1}_{i}$; Get the matching key-points of $I_{1}$
and $I_{2}$ as $p^{1}_{1}$ and $p^{1}_{2}$; Extend $p^{i}_{1}$ into
$X^{1}_{1}$, extend $p^{i}_{2}$ into $X^{1}_{2}$; (2) Input $I_{1}$,
$X^{1}_{1}$, $I_{2}$, and $X^{1}_{2}$ into MKPC algorithm and get the cropped
critical regions $I^{\prime}_{1}$ and $I^{\prime}_{2}$; (3) Define two empty
sets of points $X^{2}_{1}$ and $X^{2}_{2}$ for both images in stage-one; for
($m^{2}_{i}$ in $M^{2}$ ; $0<i<n_{2}+1$) do Feed $I^{\prime}_{1}$ and
$I^{\prime}_{2}$ into $m^{2}_{i}$; Get the matching key-points of
$I^{\prime}_{1}$ and $I^{\prime}_{2}$ as $p^{\prime 1}_{1}$ and $p^{\prime
1}_{2}$; Extend $p^{\prime i}_{1}$ into $X^{2}_{1}$, extend $p^{\prime i}_{2}$
into $X^{2}_{2}$; (4) Scale the coordinates of $X^{2}_{1}$ into the scale of
$I_{1}$; Scale the coordinates of $X^{2}_{2}$ into the scale of $I_{2}$; (5)
Concatenate $X^{1}_{1}$ and $X^{2}_{1}$ to get $X_{1}$; Concatenate
$X^{1}_{2}$ and $X^{2}_{2}$ to get $X_{2}$; (6) Calculate the F matrix / E
matrix / H matrix by feeding $X_{1}$ and $X_{2}$ into RANSAC; (7) Return F
matrix / E matrix / H matrix

###  2.3 Crop Area Expansion (CAE)

Due to the structural problem of image matching models, the matching points
may not always occur at the edges of objects. Therefore, cutting out only the
outer perimeter of the cluster may miss some co-visible areas, wasting
information as shown in Figure 5(a). To this end, we designed a padding
operations called Crop Area Expansion (CAE) to expand the critical regions
obtained by MKPC. The CAE is controlled by two parameter, which is the
horizontal expand factor $e_{h}$ and the vertical expand factor $e_{f}$. In
Figure 5(b), we set $e_{h}=1.05$ and $e_{v}=1.0$ for Image Matching Challenge
2022 benchmark, as there are usually informative regions in horizontal
direction while the areas in vertical direction(buildings) are frequently
information-poor(sky, ground, etc.). Thus, the CAE saves the area abandoned by
MKPC incorrectly.

Figure 5: The effect of Crop Area Expansion (CAE). The CAE saves the area
abandoned by MKPC incorrectly.

##  3 Experiments

As shown in Figure 6, The two-stage pipeline greatly reject the outliers,
constrain area of matching and refine the matching in critical regions. We
conduct experiment by plugging SuperPoint + SuperGlue into our two-stage
pipeline on YFCC100M and ScanNet for outdoor and indoor pose estimation,
respectively. What’s more, we try plugging combinations of models into the
two-stage pipeline, without any training, our method outperforms the SOTA of
IMC2022 benchmark.

Figure 6: The Effect comparison of LoFTR and two-stage pipeline on two demo
images pairs. The two-stage pipeline greatly reject the outliers, constrain
regions of matching and refine the matching in critical areas.

###  3.1 MKPC with single model for outdoor scenario

#### Dataset and Metrics.

For outdoor pose estimation, we follow the experiments setting in SuperGlue
[3], evaluation on the PhotoTourism dataset, which is a subset of YFCC100M
dataset [8] with ground truth poses. For the metrics, as [9; 10; 11], We take
the AUC of the pose error at the thresholds ($5^{\circ}$, $10^{\circ}$,
$20^{\circ}$) where the maximum of the angular errors evaluates the pose error
in rotation and translation. The experimental results are recorded in Table 1.
where we try RANSAC and USAC_MAGSAC (supported by OpenCV) to get the E matrix.
As shown in Table 1, the two-stage pipeline refine the performance of
Superpoint + SuperGlue under the same sets of resolutions and the same
parameters estimation algorithm by small but stable margin. Please note that
we simply uses the pretrained weights on MegaDepth [12] for SuperPoint and
SuperGlue. More specifically, under the same sets of resolutions and inference
times, the two-stage pipeline is always better than ensemble directly. Without
the constraint of keeping the computational complexity no larger than
baselines, our approach brings a better enhancement.

Table 1: Use SuperPoint + SuperGlue with MKPC on YFCC. "Res" means resolution, which indicates the longest dimension after being resized. AUC@$5^{\circ}$ means the AUC of the pose error at the thresholds$=5^{\circ}$. "Params" means parameters. Res for stage-1 | Res for stage-2 | Params estimation | AUC@5 | AUC@10 | AUC@20
---|---|---|---|---|---
1600 | None | RANSAC | 37.73 | 58.03 | 74.57
1600 | None | USAC_MAGSAC | 38.29 | 58.33 | 74.58
1200, 1600 | None | USAC_MAGSAC | 40.64 | 60.67 | 76.37
1200 | 1600 | USAC_MAGSAC | 40.88 | 60.58 | 76.26
1600 | 1200 | USAC_MAGSAC | 40.94 | 60.78 | 76.26
1200, 1400, 1600 | None | USAC_MAGSAC | 41.51 | 61.22 | 76.73
1200, 1400 | 1600 | USAC_MAGSAC | 41.66 | 61.48 | 76.72
1200, 1600 | 1400 | USAC_MAGSAC | 41.95 | 61.52 | 76.80
1600 | 1200, 1400 | USAC_MAGSAC | 42.20 | 61.65 | 76.95
840, 1200, 1600 | None | USAC_MAGSAC | 41.25 | 60.93 | 76.40
840, 1200, 1400, 1600 | None | USAC_MAGSAC | 41.30 | 61.13 | 76.65
1600 | 840, 1200, 1400 | USAC_MAGSAC | 42.10 | 61.42 | 76.79
840, 1200 | 1400, 1600 | USAC_MAGSAC | 42.08 | 61.60 | 76.92
1200, 1600 | 840, 1400 | USAC_MAGSAC | 42.23 | 61.92 | 77.20
840, 1200, 1600 | 1400 | USAC_MAGSAC | 42.30 | 62.08 | 77.17
840, 1200, 1600 | 1200, 1400, 1600 | USAC_MAGSAC | 43.18 | 62.59 | 77.59

###  3.2 MKPC with multiple models for outdoor pose estimation

Image Matching Challenge 2022 benchmark is an outdoor dataset, containing
images pairs with variable degrees, which is often larger than those in other
outdoor dataset. Each pair in IMC2022 benchmark is collected never less than a
24-hours interval, which is usually months or years apart, causing a domain
gap. Bridging this domain gap is the main challenge in this benchmark. It
requires researchers and competitors getting the fundamental matrix and uses
mean Average Accuracy (mAA), which computes the error in terms of
rotation(degrees) and translation(meters) for evaluation. As shown in Table 2,
our method outperforms the previous SOTA, which currently ranks the first
place on both private leaderboard and public leaderboard of the benchmark.
Please note that this challenge is held on Kaggle, which constraints the
submission time within 9 hours. The winners of IMC 2022 all ensembles with
multiple models, making their submission time close to the time limit, which
means that the comparison between our method and theirs are fair. As shown in
Table 2, the Top-3 in IMC benchmark have significant advantages to other
winners. However, our method outperforms the SOTA by 0.289% on private
leaderboard and 0.312% on public leaderboard. On private leaderboard, the
margin between our methods and #rank1 is larger than that between #rank1 and
#rank2. On public leaderboard, the margin is larger than that of #rank1 and
#rank3. What’s more, the margin between ours and #rank4 is about two times to
that between #rank4 and #rank11.

Our method brings more gain on IMC2022 Benchmark than SuperPoint + SuperGlue
on PhotoTourism dataset because of three reasons. (i) The larger degrees with
greater variance between images in IMC2022 maximizing the potential of our
method, because as the degree increases, the necessity for extracting the co-
visible area improves. (ii) The domain gap cause by the various collection
time and environments constrain single model from fully capturing the common
features between two images, making it’s vital to ensemble multiple model for
a better score on IMC 2022. Meanwhile, our approach is suitable for model
ensemble naturally (iii) The comparison in Table 2 includes multiple model. As
illustrated in the last reason, our model is significantly stronger when
applied to ensemble. Appropriate ensemble in the first stage improves the
quality of clipping in the ROI region, leading to better rejection of outliers
and an increase in the upper limit in the stage-2. In stage-2, proper ensemble
enhances the effectiveness of the models for fine matching within critical
regions, ultimately producing better metrics.

Finally, Both of the comparisons (the comparison between our method and other
solutions on IMC2022 benchmark and the comparison between our method on
IMC2022 benchmark and PhotoTourism) show not only the great effectiveness but
also the strong generalization and robustness of our method.

Table 2: Our method on Image Matching Challenge 2022 Benchmark. We submit our method after the competition ends, the baselines are the winners solution on the final leaderboard in IMC 2022. The #Rank1 represents the champion of IMC 2022, which is the SOTA on this benchmark. The #Rank 1 to 11 were awarded gold medals by Kaggle, which are chosen as baselines. The public leaderboard takes 49 % of the total IMC benchmark, while the private leaderboard takes 51 %, while the private one is more an important, for which represents the ability of generalization of methods. The metric is mean Average Accuracy(mAA) Methods | mAA on Public Leaderboard | mAA on Private Leaderboard
---|---|---
SuperPoint+SuperGlue | 0.67627 | 0.67054
DKM | 0.66731 | 0.68651
LoFTR | 0.75478 | 0.75606
QuadTree | 0.81309 | 0.81223
AspanFormer | 0.79437 | 0.81438
#Rank11 | 0.84783 | 0.84562
#Rank10 | 0.83833 | 0.84611
#Rank9 | 0.84803 | 0.84786
#Rank8 | 0.84675 | 0.84871
#Rank7 | 0.84287 | 0.84884
#Rank6 | 0.85270 | 0.85023
#Rank5 | 0.84824 | 0.85092
#Rank4 | 0.85238 | 0.85225
#Rank3 | 0.86100 | 0.85866
#Rank2 | 0.86000 | 0.86210
#Rank1 (previous SOTA) | 0.86367 | 0.86343
Ours | 0.86679 | 0.86634

###  3.3 MKPC with single model for indoor pose estimation

ScanNet is a dataset for indoor pose estimation. Our experiments follow the
set of SuperGlue [3], whose results are in Table 3. Under the same sets of
resolutions and inference time, our approach can hardly surpass the naive
ensemble. We think it’s caused by the difference between the various
morphology and texture distribution in indoor scenario (ScanNet) and outdoor
scenario (YFCC). The indoor scenes always contain object with less texture,
which is hard for image matching model. The effect of our two-stage pipeline
strongly relies on the quality of the key-points generated by the modes in
stage-one. Without key-points of great quality, MKPC is unable to propose the
critical region with high accuracy. Moreover, the low scores of SuperPoint +
SuperGlue on ScanNet reflects that only using SuperPoint + SuperGlue for
stage-one may output key-points with lower quality, resulting in bad ROIs
produced by MKPC. A poor ROI may instead have a negative impact on the result,
rather than enhancement.

Table 3: Use SuperPoint + SuperGlue with MKPC on ScanNet. "Res" means resolution, which indicates the longest dimension after being resized. AUC@k means the AUC of top-k. Res for stage-1 | Res for stage-2 | AUC@5 | AUC@10 | AUC@20
---|---|---|---|---
640 | None | 16.18 | 34.11 | 52.47
840 | None | 16.56 | 34.8 | 53.44
640, 840 | None | 18.02 | 37.18 | 55.98
640, 840, 1024 | None | 19.59 | 38.37 | 57.47
640, 840, 1024, 1280 | None | 19.84 | 38.93 | 57.00
640, 840, 1024, 1280, 1536 | None | 20.33 | 39.04 | 57.15
840, 1024, 1280 | 840, 1280 | 20.18 | 38.76 | 56.94

###  3.4 The Time-consumption of the MKPC and the Two-stage Pipeline

Although our method requires models to infer for multiple times, the time-
consumption is not as large as it looks like for the following reasons. (i)
The MKPC is a plug-and-play module and the two-stage pipeline is campatible to
any image matching models, while users don’t require to re-train or fine-tune
any image matching models. This advantage makes it a convenient choice for
improving the effect of image model without any training. (ii) The parameters
estimations algorithm such as RANSAC filter outliers and find the global
optimized F/E/H matrix within a maximum iterations, which will stop if the
matching error decreases below the threshold in early stage. In stage-two, the
critical regions are areas proposed by MKPC, making the key-points generated
by the models in stage-two have a much higher quality. Hence, the quality of
sets of key-points generated by our two-stage pipeline are usually better than
those produced by single model or naive ensemble, making RANSAC may stop in an
earlier stage (iii) The DBSCAN consumes little time and computational
resource, which is almost negligible. Please note that the time-consumption
various from datasets, which is essentially influenced by the distribution of
the original images.

##  4 Conclusion

In this paper, we first introduce an hypothesis that the area outside the co-
visible regions of both images contain little information. To narrowing the
restrain space and reject outliers, we propose a plug-and-play but remarkable
algorithm called matching key-points crop (MKPC). Based on MKPC, we further
present a two-stage pipeline which is compatible to any image matching models
and able to strengthen any image models without re-training or fine-tuning.
The experiments indicate that our method has a small but promising effect for
outdoor pose estimation, while it still works on indoor scenario but requires
more computational complexity. What’s more, our method surpassing the SOTA of
Image Matching Challenge 2022 Benchmark, indicating the potential and benefits
of our approach when implemented for ensemble, and the effectiveness and
generalization of our method when applying for multiple models.

##  5 Future Work

In future work, we aim to address the limitations and challenges observed in
our experiments on various datasets. Specifically, we plan to (i) explore the
potential of our method in model fusion by incorporating multiple image
matching models with our approach, particularly on benchmarks like IMC2022
where image pairs with larger angle deviations and domain gaps, as our
approach seems particularly suitable for these scenarios. (ii) Extending our
two-stage pipeline to newer and more advanced models beyond
SuperPoint+SuperGlue, as the accuracy of the first stage is crucial for our
method. We expect that conducting further experiments on PhotoTourism(subset
of YFCC100M) and ScanNet datasets using stronger baselines or ensemble of
models will help us better understand our method and enhance the performance
across various scenarios.

## References

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#  Relations between values of arithmetic Gevrey series, and applications to
values of the Gamma function

S. Fischler and T. Rivoal

(August 27, 2024)

###### Abstract

We investigate the relations between the rings E, G and D of values taken at
algebraic points by arithmetic Gevrey series of order either $-1$
($E$-functions), $0$ (analytic continuations of $G$-functions) or $1$
(renormalization of divergent series solutions at $\infty$ of $E$-operators)
respectively. We prove in particular that any element of G can be written as
multivariate polynomial with algebraic coefficients in elements of E and D,
and is the limit at infinity of some $E$-function along some direction. This
prompts to defining and studying the notion of mixed functions, which
generalizes simultaneously $E$-functions and arithmetic Gevrey series of order
1. Using natural conjectures for arithmetic Gevrey series of order 1 and mixed
functions (which are analogues of a theorem of André and Beukers for
$E$-functions) and the conjecture ${\bf D}\cap{\bf E}=\overline{\mathbb{Q}}$
(but not necessarily all these conjectures at the same time), we deduce a
number of interesting Diophantine results such as an analogue for mixed
functions of Beukers’ linear independence theorem for values of $E$-functions,
the transcendance of the values of the Gamma function and its derivatives at
all non-integral algebraic numbers, the transcendance of Gompertz constant as
well as the fact that Euler’s constant is not in E.

##  1 Introduction

A power series
$\sum_{n=0}^{\infty}\frac{a_{n}}{n!}x^{n}\in\overline{\mathbb{Q}}[[x]]$ is
said to be an $E$-function when it is solution of a linear differential
equation over $\overline{\mathbb{Q}}(x)$ (holonomic), and $|\sigma(a_{n})|$
(for any $\sigma\in\textup{Gal}(\overline{\mathbb{Q}}/\overline{\mathbb{Q}})$)
and the least common denominator of $a_{0},a_{1},\ldots,a_{n}$ grow at most
exponentially in $n$. They were defined and studied by Siegel in 1929, who
also defined the class of $G$-functions: a power series
$\sum_{n=0}^{\infty}a_{n}x^{n}\in\overline{\mathbb{Q}}[[x]]$ is said to be a
$G$-function when $\sum_{n=0}^{\infty}\frac{a_{n}}{n!}x^{n}$ is an
$E$-function. In this case,
$\sum_{n=0}^{\infty}n!a_{n}z^{n}\in\overline{\mathbb{Q}}[[z]]$ is called an
-function, following the terminology introduced by André in [1]. $E$-functions
are entire, while $G$-functions have a positive radius of convergence, which
is finite except for polynomials. Here and below, we see
$\overline{\mathbb{Q}}$ as embedded into $\mathbb{C}$. Following André again,
$E$-functions, $G$-functions and -fonctions are exactly arithmetic Gevrey
series of order $s=-1,0,1$ respectively. Actually André defines arithmetic
Gevrey series of any order $s\in\mathbb{Q}$, but the set of values at
algebraic points is the same for a given $s\neq 0$ as for $s/|s|$ using [1,
Corollaire 1.3.2].

-functions are divergent series, unless they are polynomials. Given an -function $\mathfrak{f}$ and any $\theta\in\mathbb{R}$, except finitely many values mod $2\pi$ (namely anti-Stokes directions of $\mathfrak{f}$), one can perform Ramis’ 1-summation of $\mathfrak{f}(1/z)$ in the direction $\theta$, which coincides in this setting with Borel-Laplace summation (see [14] or [9]). This provides a function denoted by $\mathfrak{f}_{\theta}(1/z)$, holomorphic on the open subset of $\mathbb{C}$ consisting in all $z\neq 0$ such that $\theta-\frac{\pi}{2}-\varepsilon<\arg z<\theta+\frac{\pi}{2}+\varepsilon$ for some $\varepsilon>0$, of which $\mathfrak{f}(1/z)$ is the asymptotic expansion in this sector (called a large sector bisected by $\theta$). Of course $\mathfrak{f}(1/z)$ can be extended further by analytic continuation, but this asymptotic expansion may no longer be valid. When an -function is denoted by $\mathfrak{f}_{j}$, we shall denote by $\mathfrak{f}_{j,\theta}$ or $\mathfrak{f}_{j;\theta}$ its 1-summation and we always assume (implicitly or explicitly) that $\theta$ is not an anti-Stokes direction.

In [8], [9] and [10, §4.3], we have studied the sets G, E and D defined
respectively as the sets of all the values taken by all (analytic
continuations of) $G$-functions at algebraic points, of all the values taken
by all $E$-functions at algebraic points and of all values
$\mathfrak{f}_{\theta}(1)$ where $\mathfrak{f}$ is an -function ($\theta=0$ if
it is not an anti-Stokes direction, and $\theta>0$ is very small otherwise.)
These three sets are countable sub-rings of $\mathbb{C}$ that all contain
$\overline{\mathbb{Q}}$; conjecturally, they are related to the set of periods
and exponential periods, see §3. (The ring D is denoted by in [10].)

We shall prove the following result in §3.

######  Theorem 1.

Every element of ${\bf G}$ can be written as a multivariate polynomial (with
coefficients in $\overline{\mathbb{Q}}$) in elements of E and D.

Moreover, ${\bf G}$ coincides with the set of all convergent integrals
$\int_{0}^{\infty}F(x)dx$ where $F$ is an $E$-function, or equivalently with
the set of all finite limits of $E$-functions at $\infty$ along some
direction.

Above, a convergent integral $\int_{0}^{\infty}F(x)dx$ means a finite limit of
the $E$-function $\int_{0}^{z}F(x)dx$ as $z\to\infty$ along some direction;
this explains the equivalence of both statements.

We refer to Eq. (3.2) in §3 for an expression of $\log(2)$ as a polynomial in
elements in E and D; the number $\pi$ could be similarly expressed by
considering $z$ and $iz$ instead of $z$ and $2z$ there. Examples of the last
statement are the identities (see [12] for the second one):

$\int_{0}^{+\infty}\frac{\sin(x)}{x}dx=\frac{\pi}{2}\quad\textup{and}\quad\int_{0}^{+\infty}J_{0}(ix)e^{-3x}dx=\frac{\sqrt{6}}{96\pi^{3}}\Gamma\Big{(}\frac{1}{24}\Big{)}\Gamma\Big{(}\frac{5}{24}\Big{)}\Gamma\Big{(}\frac{7}{24}\Big{)}\Gamma\Big{(}\frac{11}{24}\Big{)}.$

It is notoriously difficult to prove/disprove that a given element of G is
transcendental; it is known that a Siegel-Shidlovskii type theorem for
$G$-functions can not hold mutatis mutandis. Theorem 1 suggests that an
alternative approach to the study of the Diophantine properties of elements of
G can be through a better understanding of joint study of the elements of E
and D, modulo certain conjectures to begin with. Our applications will not be
immediately directed to the elements of G but rather to the understanding of
the (absence of) relations between the elements of E and D.

It seems natural (see [9, p. 37]) to conjecture that ${\bf E}\;\cap\;{\bf
G}=\overline{\mathbb{Q}}$, and also that ${\bf G}\;\cap\;{\bf
D}=\overline{\mathbb{Q}}$, though both properties seem currently out of reach.
In this paper, we suggest (see §2) a possible approach towards the following
analogous conjecture.

######  Conjecture 1.

We have ${\bf E}\;\cap\;{\bf D}=\overline{\mathbb{Q}}$.

In §2 we shall make a functional conjecture, namely Conjecture 3, that implies
Conjecture 1. We also prove that Conjecture 1 has very important consequences,
as the following result shows.

######  Theorem 2.

Assume that Conjecture 1 holds. Then $\Gamma^{(s)}(a)$ is a transcendental
number for any rational number $a>0$ and any integer $s\geq 0$, except of
course if $s=0$ and $a\in\mathbb{N}$.

One of the aims of this paper is to show that combining \- and $E$-functions
may lead to very important results in transcendental number theory. Let us
recall now briefly the main known results on $E$-functions.

Point $(i)$ in the following result is due to André [2] for $E$-functions with
rational Taylor coefficients, and to Beukers [6] in the general case. André
used this property to obtain a new proof of the Siegel-Shidlovskii Theorem,
and Beukers to prove an optimal refinement of this theorem (namely, $(ii)$
below).

######  Theorem A.

$(i)$ $[$André, Beukers$]$ If an $E$-function $F(z)$ is such that $F(1)=0$,
then $\frac{F(z)}{z-1}$ is an $E$-function.

$(ii)$ $[$Beukers$]$ Let $\underline{F}(z):={}^{t}(f_{1}(z),\ldots,f_{n}(z))$
be a vector of $E$-functions solution of a differential system
$\underline{F}^{\prime}(z)=A(z)\underline{F}(z)$ for some matrix $A(z)\in
M_{n}(\overline{\mathbb{Q}}(z))$.

Let $\xi\in\overline{\mathbb{Q}}^{*}$ which is not a pole of a coefficient of
$A$. Let $P\in\overline{\mathbb{Q}}[X_{1},\ldots,X_{n}]$ be a homogeneous
polynomial such that

$P(f_{1}(\xi),\ldots,f_{n}(\xi))=0.$

Then there exists $Q\in\overline{\mathbb{Q}}[Z,X_{1},\ldots,X_{n}]$,
homogeneous in the $X_{i}$, such that

$Q(z,f_{1}(z),\ldots,f_{n}(z))=0\mbox{ identically and
}P(X_{1},\ldots,X_{n})=Q(\xi,X_{1},\ldots,X_{n}).$

In particular, we have

${\rm trdeg}_{\overline{\mathbb{Q}}}(f_{1}(\xi),\ldots,f_{n}(\xi))={\rm
trdeg}_{\overline{\mathbb{Q}}(z)}(f_{1}(z),\ldots,f_{n}(z)).$

The Siegel-Shidlovskii Theorem itself is the final statement about equality of
transcendence degrees.

In this paper we state conjectural analogues of these results, involving
-functions. The principal difficulty is that these functions are divergent
power series, and the exact analogue of Theorem A is meaningless. André
discussed the situation in [2] and even though he did not formulate exactly
the following conjecture, it seems plausible to us. From it, we will show how
to deduce an analogue of the Siegel-Shidlovskii theorem for -functions.
Ferguson [7, p. 171, Conjecture 1] essentially stated this conjecture when
$\mathfrak{f}(z)$ has rational coefficients and when $\theta=0$.

######  Conjecture 2.

Let $\mathfrak{f}(z)$ be an -function and $\theta\in(-\pi/2,\pi/2)$ be such
that $\mathfrak{f}_{\theta}(1)=0$. Then $\frac{\mathfrak{f}(z)}{z-1}$ is an
-function.

In other words, the conclusion of this conjecture asserts that
$\frac{z}{1-z}\mathfrak{f}(1/z)$ is an -function in $1/z$; this is equivalent
to $\frac{\mathfrak{f}(1/z)}{z-1}$ being an -function in $1/z$ (since we have
$\frac{\mathfrak{f}(1/z)}{z-1}=O(1/z)$ unconditionally as $|z|\to\infty$).

Following Beukers’ proof [6] yields the following result (see [3, §4.6] for a
related conjecture).

######  Theorem 3.

Assume that Conjecture 2 holds.

Let
$\underline{\mathfrak{f}}(z):={}^{t}(\mathfrak{f}_{1}(z),\ldots,\mathfrak{f}_{n}(z))$
be a vector of -functions solution of a differential system
$\underline{\mathfrak{f}}^{\prime}(z)=A(z)\underline{\mathfrak{f}}(z)$ for
some matrix $A(z)\in M_{n}(\overline{\mathbb{Q}}(z))$. Let
$\xi\in\overline{\mathbb{Q}}^{*}$ and
$\theta\in(\arg(\xi)-\pi/2,\arg(\xi)+\pi/2)$ ; assume that $\xi$ is not a pole
of a coefficient of $A$, and that $\theta$ is anti-Stokes for none of the
$\mathfrak{f}_{j}$.

Let $P\in\overline{\mathbb{Q}}[X_{1},\ldots,X_{n}]$ be a homogeneous
polynomial such that

$P(\mathfrak{f}_{1,\theta}(1/\xi),\ldots,\mathfrak{f}_{n,\theta}(1/\xi))=0.$

Then there exists $Q\in\overline{\mathbb{Q}}[Z,X_{1},\ldots,X_{n}]$,
homogeneous in the $X_{i}$, such that

$Q(z,\mathfrak{f}_{1}(z),\ldots,\mathfrak{f}_{n}(z))=0\mbox{ identically and
}P(X_{1},\ldots,X_{n})=Q(1/\xi,X_{1},\ldots,X_{n}).$

In particular, we have

${\rm
trdeg}_{\overline{\mathbb{Q}}}(\mathfrak{f}_{1,\theta}(1/\xi),\ldots,\mathfrak{f}_{n,\theta}(1/\xi))={\rm
trdeg}_{\overline{\mathbb{Q}}(z)}(\mathfrak{f}_{1}(z),\ldots,\mathfrak{f}_{n}(z)).$

As an application of Theorem 3, we shall prove the following corollary. Note
that under his weaker version of Conjecture 2, Ferguson [7, p. 171, Theorem 2]
proved that Gompertz’s constant is an irrational number.

######  Corollary 1.

Assume that Conjecture 2 holds. Then for any $\alpha\in\overline{\mathbb{Q}}$,
$\alpha>0$, and any $s\in\mathbb{Q}\setminus\mathbb{Z}_{\geq 0}$, the number
$\int_{0}^{\infty}(t+\alpha)^{s}e^{-t}dt$ is a transcendental number.

In particular, Gompertz’s constant $\delta:=\int_{0}^{\infty}e^{-t}/(t+1)dt$
is a transcendental number.

In this text we suggest an approach towards Conjecture 1, based on the new
notion of mixed functions which enables one to consider $E$\- and -functions
at the same time. In particular we shall state a conjecture about such
functions, namely Conjecture 3 in §2, which implies both Conjecture 1 and
Conjecture 2. The following result is a motivation for this approach.

######  Proposition 1.

Assume that both Conjectures 1 and 2 hold. Then neither Euler’s constant
$\gamma:=-\Gamma^{\prime}(1)$ nor $\Gamma(a)$ (with
$a\in\mathbb{Q}^{+}\setminus\mathbb{N}$) are in E.

It is likely that none of these numbers is in G, but (as far as we know) there
is no “functional” conjecture like Conjecture 3 that implies this. It is also
likely that none is in D as well, but we don’t know if this can be deduced
from Conjecture 3.

The structure of this paper is as follows. In §2 we define and study mixed
functions, a combination of $E$\- and -functions. Then in §3 we express any
value of a $G$-function as a polynomial in values of $E$\- and -functions,
thereby proving Theorem 1. We study derivatives of the $\Gamma$ function at
rational points in §4, and prove Theorem 2 and Proposition 1. At last, §5 is
devoted to adapting Beukers’ method to our setting: this approach yields
Theorem 3 and Corollary 1.

##  2 Mixed functions

###  2.1 Definition and properties

In view of Theorem 1, it is natural to study polynomials in $E$\- and
-functions. We can prove a Diophantine result that combines both Theorems
A$(ii)$ and 3 but under a very complicated polynomial generalization of
Conjecture 2. We opt here for a different approach to mixing $E$\- and
-functions for which very interesting Diophantine consequences can be deduced
from a very easy to state conjecture which is more in the spirit of Conjecture
2. We refer to §2.3 for proofs of all properties stated in this section
(including Lemma 1 and Proposition 2), except Theorem 4.

######  Definition 1.

We call mixed (arithmetic Gevrey) function any formal power series

$\sum_{n\in\mathbb{Z}}a_{n}z^{n}$

such that $\sum_{n\geq 0}a_{n}z^{n}$ is an $E$-function in $z$, and
$\sum_{n\geq 1}a_{-n}z^{-n}$ is an -function in $1/z$.

In other words, a mixed function is defined as a formal sum
$\Psi(z)=F(z)+\mathfrak{f}(1/z)$ where $F$ is an $E$-function and
$\mathfrak{f}$ is an -function. In particular, such a function is zero if, and
only if, both $F$ and $\mathfrak{f}$ are constants such that
$F+\mathfrak{f}=0$; obviously, $F$ and $\mathfrak{f}$ are uniquely determined
by $\Psi$ upon assuming (for instance) that $\mathfrak{f}(0)=0$. The set of
mixed functions is a $\overline{\mathbb{Q}}$-vector space stable under
multiplication by $z^{n}$ for any $n\in\mathbb{Z}$. Unless $\mathfrak{f}(z)$
is a polynomial, such a function $\Psi(z)=F(z)+\mathfrak{f}(1/z)$ is purely
formal: there is no $z\in\mathbb{C}$ such that $\mathfrak{f}(1/z)$ is a
convergent series. However, choosing a direction $\theta$ which is not anti-
Stokes for $\mathfrak{f}$ allows one to evaluate
$\Psi_{\theta}(z)=F(z)+\mathfrak{f}_{\theta}(1/z)$ at any $z$ in a large
sector bisected by $\theta$. Here and below, such a direction will be said not
anti-Stokes for $\Psi$ and whenever we write $\mathfrak{f}_{\theta}$ or
$\Psi_{\theta}$ we shall assume implicitly that $\theta$ is not anti-Stokes.

Definition 1 is a formal definition, but one may identify a mixed function
with the holomorphic function it defines on a given large sector by means of
the following lemma.

######  Lemma 1.

Let $\Psi$ be a mixed function, and $\theta\in\mathbb{R}$ be a non-anti-Stokes
direction for $\Psi$. Then $\Psi_{\theta}$ is identically zero (as a
holomorphic function on a large sector bisected by $\theta$) if, and only if,
$\Psi$ is equal to zero (as a formal power series in $z$ and $1/z$).

Any mixed function $\Psi(z)=F(z)+\mathfrak{f}(1/z)$ is solution of an
$E$-operator. Indeed, this follows from applying [1, Theorem 6.1] twice: there
exist an $E$-operator $L$ such that $L(\mathfrak{f}(1/z))=0$, and an
$E$-operator $M$ such that $M(L(F(z)))=0$ (because $L(F(z))$ is an
$E$-function). Hence $ML(F(z)+\mathfrak{f}(1/z))=0$ and by [1, p. 720, §4.1],
$ML$ is an $E$-operator.

We formulate the following conjecture, which implies both Conjecture 1 and
Conjecture 2.

######  Conjecture 3.

Let $\Psi(z)$ be an mixed function, and $\theta\in(-\pi/2,\pi/2)$ be such that
$\Psi_{\theta}(1)=0$. Then $\frac{\Psi(z)}{z-1}$ is an mixed function.

The conclusion of this conjecture is that $\Psi(z)=(z-1)\Psi_{1}(z)$ for some
mixed function $\Psi_{1}$. This conclusion can be made more precise as
follows; see §2.3 for the proof.

######  Proposition 2.

Let $\Psi(z)=F(z)+\mathfrak{f}(1/z)$ be an mixed function, and
$\theta\in(-\pi/2,\pi/2)$ be such that $\Psi_{\theta}(1)=0$. Assume that
Conjecture 3 holds for $\Psi$ and $\theta$.

Then both $F(1)$ and $\mathfrak{f}_{\theta}(1)$ are algebraic, and
$\frac{\mathfrak{f}(1/z)-\mathfrak{f}_{\theta}(1)}{z-1}$ is an -function.

Of course, in the conclusion of this proposition, one may assert also that
$\frac{F(z)-F(1)}{z-1}$ is an $E$-function using Theorem A$(i)$.

Conjecture 3 already has far reaching Diophantine consequences: Conjecture 2
and Theorem 2 stated in the introduction, and also the following result that
encompasses Theorem 3 in the linear case.

######  Theorem 4.

Assume that Conjecture 3 holds.

Let ${\bf\Psi}(z):={}^{t}(\Psi_{1}(z),\ldots,\Psi_{n}(z))$ be a vector of
mixed functions solution of a differential system
${\bf\Psi}^{\prime}(z)=A(z){\bf\Psi}(z)$ for some matrix $A(z)\in
M_{n}(\overline{\mathbb{Q}}(z))$. Let $\xi\in\overline{\mathbb{Q}}^{*}$ and
$\theta\in(\arg(\xi)-\pi/2,\arg(\xi)+\pi/2)$ ; assume that $\xi$ is not a pole
of a coefficient of $A$, and that $\theta$ is anti-Stokes for none of the
$\Psi_{j}$.

Let $\lambda_{1},\ldots,\lambda_{n}\in\overline{\mathbb{Q}}$ be such that

$\sum_{i=1}^{n}\lambda_{i}\Psi_{i,\theta}(\xi)=0.$

Then there exist $L_{1},\ldots,L_{n}\in\overline{\mathbb{Q}}[z]$ such that

$\sum_{i=1}^{n}L_{i}(z)\Psi_{i}(z)=0\mbox{ identically and
}L_{i}(\xi)=\lambda_{i}\mbox{ for any }i.$

In particular, we have

${\rm
rk}_{\overline{\mathbb{Q}}}(\Psi_{1,\theta}(\xi),\ldots,\Psi_{n,\theta}(\xi))={\rm
rk}_{\overline{\mathbb{Q}}(z)}(\Psi_{1}(z),\ldots,\Psi_{n}(z)).$

The proof of Theorem 4 follows exactly the linear part of the proof of Theorem
3 (see §5.1), which is based on [6, §3]. The only difference is that
-functions have to be replaced with mixed functions, and Conjecture 2 with
Conjecture 3. In particular Proposition 4 stated in §5.1 remains valid with
these modifications.

However a product of mixed functions is not, in general, a mixed function.
Therefore the end of [6, §3] does not adapt to mixed functions, and there is
no hope to obtain in this way a result on the transcendence degree of a field
generated by values of mixed functions.

As an application of Theorem 4, we can consider the mixed functions
$1,e^{\beta z}$ and $\mathfrak{f}(1/z):=\sum_{n=0}^{\infty}(-1)^{n}n!z^{-n}$,
where $\beta$ is a fixed non-zero algebraic number. These three functions are
linearly independent over $\mathbb{C}(z)$ and form a solution of a
differential system with only 0 for singularity (because
$(\mathfrak{f}(1/z))^{\prime}=(1+1/z)f(1/z)-1$), hence for any
$\alpha\in\overline{\mathbb{Q}}$, $\alpha>0$ and any
$\varrho\in\overline{\mathbb{Q}}^{*}$, the numbers
$1,e^{\varrho},\mathfrak{f}_{0}(1/\alpha):=\int_{0}^{\infty}e^{-t}/(1+\alpha
t)dt$ are $\overline{\mathbb{Q}}$-linearly independent (for a fixed $\alpha$,
take $\beta=\varrho/\alpha$).

###  2.2 Values of mixed functions

We denote by ${\bf M}_{G}$ the set of values $\Psi_{\theta}(1)$, where $\Psi$
is a mixed function and $\theta=0$ if it is not anti-Stokes, $\theta>0$ is
sufficiently small otherwise. This set is obviously equal to ${\bf E}+{\bf
D}$.

######  Proposition 3.

For every integer $s\geq 0$ and every $a\in\mathbb{Q}^{+}$, $a\neq 0$, we have
$\Gamma^{(s)}(a)\in e^{-1}{\bf M}_{G}$.

This results follows immediately from Eq. (4.4) below (see §4.2), written in
the form

$\Gamma^{(s)}(a)=e^{-1}\big{(}(-1)^{s}es!E_{a,s+1}(-1)+\mathfrak{f}_{a,s+1;0}(1)\big{)},$

because $e^{z}E_{a,s+1}(-z)$ is an $E$-function and
$\mathfrak{f}_{a,s+1;0}(1)$ is the 1-summation in the direction $0$ of an
-function.

It would be interesting to know if $\Gamma^{(s)}(a)$ belongs to ${\bf M}_{G}$.
We did not succeed in proving it does, and we believe it does not. Indeed, for
instance if we want to prove that $\gamma\in{\bf M}_{G}$, a natural strategy
would be to construct an $E$-function $F(z)$ with asymptotic expansion of the
form $\gamma+\log(z)+\mathfrak{f}(1/z)$ in a large sector, and then to
evaluate at $z=1$. However this strategy cannot work since there is no such
$E$-function (see the footnote in the proof of Lemma 1 in §2.3).

###  2.3 Proofs concerning mixed functions

To begin with, let us take Proposition 2 for granted and prove that Conjecture
3 implies both Conjecture 1 and Conjecture 2. Concerning Conjecture 2 it is
clear. To prove that it implies Conjecture 1, let $\xi\in{\bf D}$, i.e.
$\xi=\mathfrak{f}_{\theta}(1)$ is the 1-summation of an -function
$\mathfrak{f}(z)$ in the direction $\theta=0$ if it is not anti-Stokes, and
$\theta>0$ close to 0 otherwise. Assume that $\xi$ is also in ${\bf E}$: we
have $\xi=F(1)$ for some $E$-function $F(z)$. Therefore,
$\Psi(z)=F(z)-\mathfrak{f}(1/z)$ is a mixed function such that
$\Psi_{\theta}(1)=0$. By Conjecture 3 and Proposition 2, we have
$\xi=\mathfrak{f}_{\theta}(1)\in\overline{\mathbb{Q}}$. This concludes the
proof that Conjecture 3 implies Conjecture 1.

Let us prove Proposition 2 now. Assuming that Conjecture 3 holds for $\Psi$
and $\theta$, there exists a mixed function
$\Psi_{1}(z)=F_{1}(z)+\mathfrak{f}_{1}(1/z)$ such that
$\Psi(z)=(z-1)\Psi_{1}(z)$. We have

$F(z)-(z-1)F_{1}(z)+\mathfrak{f}(1/z)-(z-1)\mathfrak{f}_{1}(1/z)=0$ (2.1)

as a formal power series in $z$ and $1/z$. Now notice that
$z-1=z(1-\frac{1}{z})$, and that we may assume $\mathfrak{f}$ and
$\mathfrak{f}_{1}$ to have zero constant terms. Denote by $\alpha$ the
constant term of $\mathfrak{f}(1/z)-z(1-\frac{1}{z})\mathfrak{f}_{1}(1/z)$.
Then we have

$F(z)-(z-1)F_{1}(z)+\alpha+\mathfrak{f}_{2}(1/z)=0$

for some -function $\mathfrak{f}_{2}$ without constant term, so that
$\mathfrak{f}_{2}=0$, $F(z)=(z-1)F_{1}(z)-\alpha$ and
$F(1)=-\alpha\in\overline{\mathbb{Q}}$. This implies
$\mathfrak{f}_{\theta}(1)=\alpha$, and
$\frac{\mathfrak{f}(1/z)-\mathfrak{f}_{\theta}(1)}{z-1}=\mathfrak{f}_{1}(1/z)$
is an -function since $\mathfrak{f}_{2}=0$. This concludes the proof of
Proposition 2.

At last, let us prove Lemma 1. We write $\Psi(z)=F(z)+\mathfrak{f}(1/z)$ and
assume that $\Psi_{\theta}$ is identically zero. Modifying $\theta$ slightly
if necessary, we may assume that the asymptotic expansion $-\mathfrak{f}(1/z)$
of $F(z)$ in a large sector bisected by $\theta$ is given explicitly by [9,
Theorem 5] applied to $F(z)-F(0)$; recall that such an asymptotic expansion is
unique (see [9]). As in [9] we let $g(z)=\sum_{n=1}^{\infty}a_{n}z^{-n-1}$
where the coefficients $a_{n}$ are given by
$F(z)-F(0)=\sum_{n=1}^{\infty}\frac{a_{n}}{n!}z^{n}$. For any
$\sigma\in\mathbb{C}\setminus\\{0\\}$ there is no contribution in $e^{\sigma
z}$ in the asymptotic expansion of $F(z)$, so that $g(z)$ is holomorphic at
$\sigma$. At $\sigma=0$, the local expansion of $g$ is of the form
$g(z)=h_{1}(z)+h_{2}(z)\log(z)$ with $G$-functions $h_{1}$ and $h_{2}$, and
the coefficients of $h_{2}$ are related to those of $\mathfrak{f}$; however we
shall not use this special form (111Actually we are proving that the
asymptotic expansion of a non-polynomial $E$-function is never a
$\mathbb{C}$-linear combination of functions
$z^{\alpha}\log^{k}(z)\mathfrak{f}(1/z)$ with $\alpha\in\mathbb{Q}$,
$k\in\mathbb{N}$ and -functions $\mathfrak{f}$: some exponentials have to
appear.). Now recall that $g(z)=G(1/z)/z$ where $G$ is a $G$-function; then
$G$ is entire and has moderate growth at infinity (because $\infty$ is a
regular singularity of $G$), so it is a polynomial due to Liouville’s theorem.
This means that $F(z)$ is a polynomial in $z$. Recall that asymptotic
expansions in large sectors are unique. Therefore both $F$ and $\mathfrak{f}$
are constant functions, and $F+\mathfrak{f}=0$. This concludes the proof of
Lemma 1.

##  3 Proof of Theorem 1: values of $G$-functions

In this section we prove Theorem 1. Let us begin with an example, starting
with the relation proved in [15, Proposition 1] for
$z\in\mathbb{C}\setminus(-\infty,0]$:

$\gamma+\log(z)=zE_{1,2}(-z)-e^{-z}\mathfrak{f}_{1,2;0}(1/z)$ (3.1)

where $E_{1,2}$ is an $E$-function, and $\mathfrak{f}_{1,2}$ is an -function,
both defined below in §4.2.

Apply Eq. (3.1) at both $z$ and $2z$, and then substract one equation from the
other. This provides a relation of the form

$\log(2)=F(z)+e^{-z}\mathfrak{f}_{1;0}(1/z)+e^{-2z}\mathfrak{f}_{2;0}(1/z)$
(3.2)

valid in a large sector bisected by $0$, with an $E$-function $F$ and
-functions $\mathfrak{f}_{1}$ and $\mathfrak{f}_{2}$. Choosing arbitrarily a
positive real algebraic value of $z$ yields an explicit expression of
$\log(2)\in{\bf G}$ as a multivariate polynomial in elements of ${\bf E}$ and
D. But this example shows also that a polynomial in $E$\- and -functions may
be constant eventhough there does not seem to be any obvious reason. In
particular, the functions $1$, $F(z)$, $e^{-z}\mathfrak{f}_{1;0}(1/z)$, and
$e^{-2z}\mathfrak{f}_{2;0}(1/z)$ are linearly dependent over $\mathbb{C}$.
However we see no reason why they would be linearly dependent over
$\overline{\mathbb{Q}}$. This could be a major drawback to combine in $E$\-
and -functions, since functions that are linearly dependent over $\mathbb{C}$
but not over $\overline{\mathbb{Q}}$ can not belong to any Picard-Vessiot
extension over $\overline{\mathbb{Q}}$.

Let us come now to the proof of Theorem 1. We first prove the second part,
which runs as follows (it is reproduced from the unpublished note [16]).

From the stability of the class of $E$-functions by $\frac{d}{dz}$ and
$\int_{0}^{z}$, we deduce that the set of convergent integrals
$\int_{0}^{\infty}F(x)dx$ of $E$-functions and the set of finite limits of
$E$-functions along some direction as $z\to\infty$ are the same. Theorem
2$(iii)$ in [9] implies that if an $E$-function has a finite limit as
$z\to\infty$ along some direction, then this limit must be in ${\bf G}$.
Conversely, let $\beta\in{\bf G}$. By Theorem 1 in [8], there exists a
$G$-function $G(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ of radius of convergence
$\geq 2$ (say) such that $G(1)=\beta$. Let
$F(z)=\sum_{n=0}^{\infty}\frac{a_{n}}{n!}z^{n}$ be the associated
$E$-function. Then for any $z$ such that $\textup{Re}(z)>\frac{1}{2}$, we have

$\frac{1}{z}G\Big{(}\frac{1}{z}\Big{)}=\int_{0}^{+\infty}e^{-xz}F(x)dx.$

Hence, $\beta=\int_{0}^{+\infty}e^{-x}F(x)dx$ where $e^{-z}F(z)$ is an
$E$-function.

We shall now prove the first part of Theorem 1. In fact, we shall prove a
slightly more general result, namely Theorem 5 below. We first recall a few
notations. Denote by ${\bf S}$ the ${\bf G}$-module generated by all
derivatives $\Gamma^{(s)}(a)$ (with $s\in\mathbb{N}$ and
$a\in\mathbb{Q}\setminus\mathbb{Z}_{\leq 0}$), and by ${\bf V}$ the ${\bf
S}$-module generated by ${\bf E}$. Recall that ${\bf G}$, ${\bf S}$ and ${\bf
V}$ are rings. Conjecturally, ${\bf G}=\mathcal{P}[1/\pi]$ and ${\bf
V}=\mathcal{P}_{e}[1/\pi]$ where $\mathcal{P}$ and $\mathcal{P}_{e}$ are the
ring of periods and the ring of exponential periods over
$\overline{\mathbb{Q}}$ respectively (see [8, §2.2] and [10, §4.3]). We have
proved in [10, Theorem 3] that ${\bf V}$ is the ${\bf S}$-module generated by
the numbers $e^{\rho}\chi$, with $\rho\in\overline{\mathbb{Q}}$ and $\chi\in$
D.

######  Theorem 5.

The ring ${\bf V}$ is the ring generated by ${\bf E}$ and D. In particular,
all values of $G$-functions belong to the ring generated by ${\bf E}$ and D.

In other words, the elements of ${\bf V}$ are exactly the sums of products
$ab$ with $a\in{\bf E}$ and $b\in{\bf D}$.

###### Proof of Theorem 5.

We already know that ${\bf V}$ is a ring, and that it contains ${\bf E}$ and
D. To prove the other inclusion, denote by $U$ the ring generated by ${\bf E}$
and D. Using Proposition 3 proved in §2.2 and the functional equation of
$\Gamma$, we have $\Gamma^{(s)}(a)\in U$ for any $s\in\mathbb{N}$ and any
$a\in\mathbb{Q}\setminus\mathbb{Z}_{\leq 0}$. Therefore for proving that ${\bf
V}\subset U$, it is enough to prove that ${\bf G}\subset U$.

Let $\xi\in{\bf G}$. Using [11, Theorem 3] there exists an $E$-function $F(z)$
such that for any for any $\theta\in[-\pi,\pi)$ outside a finite set, $\xi$ is
a coefficient of the asymptotic expansion of $F(z)$ in a large sector bisected
by $\theta$. As the proof of [11, Theorem 3] shows, we can assume that $\xi$
is the coefficient of $e^{z}$ in this expansion.

Denote by $L$ an $E$-operator of which $F$ is a solution, and by $\mu$ its
order. André has proved [1] that there exists a basis
$(H_{1}(z),\ldots,H_{\mu}(z))$ of formal solutions of $L$ at infinity such
that for any $j$, $e^{-\rho_{j}z}H_{j}(z)\in{\rm
NGA}\\{1/z\\}_{1}^{\overline{\mathbb{Q}}}$ for some algebraic number
$\rho_{j}$. We recall that elements of ${\rm
NGA}\\{1/z\\}_{1}^{\overline{\mathbb{Q}}}$ are arithmetic Nilsson-Gevrey
series of order 1 with algebraic coefficients, i.e.
$\overline{\mathbb{Q}}$-linear combinations of functions $z^{k}(\log
z)^{\ell}\mathfrak{f}(1/z)$ with $k\in\mathbb{Q}$, $\ell\in\mathbb{N}$ and
-functions $\mathfrak{f}$. Expanding in this basis the asymptotic expansion of
$F(z)$ in a large sector bisected by $\theta$ (denoted by $\widetilde{F}$),
there exist complex numbers $\kappa_{1}$, …, $\kappa_{d}$ such that
$\widetilde{F}(z)=\kappa_{1}H_{1}(z)+\ldots+\kappa_{\mu}H_{\mu}(z)$. Then we
have $\xi=\kappa_{1}c_{1}+\ldots+\kappa_{\mu}c_{\mu}$, where $c_{j}$ is the
coefficient of $e^{z}$ in $H_{j}(z)\in e^{\rho_{j}z}{\rm
NGA}\\{1/z\\}_{1}^{\overline{\mathbb{Q}}}$. We have $c_{j}=0$ if $\rho_{j}\neq
1$, and otherwise $c_{j}$ is the constant coefficient of $e^{-z}H_{j}(z)$: in
both cases $c_{j}$ is an algebraic number. Therefore to conclude the proof
that $\xi\in U$, it is enough to prove that $\kappa_{1},\ldots,\kappa_{\mu}\in
U$.

For simplicity let us prove that $\kappa_{1}\in U$. Given solutions
$F_{1},\ldots,F_{\mu}$ of $L$, we denote by $W(F_{1},\ldots,F_{\mu})$ the
corresponding wronskian matrix. Then for any $z$ in a large sector bisected by
$\theta$ we have

$\kappa_{1}=\frac{\det W(F(z),H_{2,\theta}(z),\ldots,H_{\mu,\theta}(z))}{\det
W(H_{1,\theta}(z),\ldots,H_{\mu,\theta}(z))}$

where $H_{j,\theta}(z)$ is the 1-sommation of $H_{j}(z)$ in this sector. The
determinant in the denominator belongs to $e^{az}{\rm
NGA}\\{1/z\\}_{1}^{\overline{\mathbb{Q}}}$ with
$a=\rho_{1}+\ldots+\rho_{\mu}\in\overline{\mathbb{Q}}$. As the proof of [10,
Theorem 6] shows, there exist $b,c\in\overline{\mathbb{Q}}$, with $c\neq 0$,
such that

$\det W(H_{1,\theta}(z),\ldots,H_{\mu,\theta}(z))=cz^{b}e^{az}.$

We take $z=1$, and choose $\theta=0$ if it is not anti-Stokes for $L$ (and
$\theta>0$ sufficiently small otherwise). Then we have

$\kappa_{1}=c^{-1}e^{-a}\Big{(}\det
W(F(z),H_{2,\theta}(z),\ldots,H_{\mu,\theta}(z))\Big{)}_{|z=1}\in U.$

This concludes the proof. ∎

######  Remark 1.

The second part of Theorem 1 suggests the following comments. It would be
interesting to have a better understanding (in terms of E, G and D) of the set
of convergent integrals $\int_{0}^{\infty}R(x)F(x)dx$ where $R$ is a rational
function in $\overline{\mathbb{Q}}(x)$ and $F$ is an $E$-function, which are
thus in G when $R=1$ (see [16] for related considerations). Indeed, classical
examples of such integrals are
$\int_{0}^{+\infty}\frac{\cos(x)}{1+x^{2}}dx=\pi/(2e)\in\pi{\bf E}$, Euler’s
constant $\int_{0}^{+\infty}\frac{1-(1+x)e^{-x}}{x(1+x)}dx=\gamma\in{\bf
E}+e^{-1}{\bf D}$ (using Eq. (3.1) and [20, p. 248, Example 2]) and Gompertz
constant $\delta:=\int_{0}^{+\infty}\frac{e^{-x}}{1+x}dx\in{\bf D}$. A large
variety of behaviors can thus be expected here.

For instance, using various explicit formulas in [13, Chapters 6.5–6.7], it
can be proved that

$\int_{0}^{+\infty}R(x)J_{0}(x)dx\in{\bf G}+{\bf E}+\gamma{\bf
E}+\log(\overline{\mathbb{Q}}^{\star}){\bf E}$

for any $R(x)\in\overline{\mathbb{Q}}(x)$ without poles on $[0,+\infty)$,
where $J_{0}(x)=\sum_{n=0}^{\infty}(ix/2)^{2n}/n!^{2}$ is a Bessel function.

A second class of examples is when $R(x)F(x)$ is an even function without
poles on $[0,+\infty)$ and such that $\lim_{|x|\to\infty,{\rm Im}(x)\geq
0}x^{2}R(x)F(x)=0$. Then by the residue theorem,

$\int_{0}^{+\infty}R(x)F(x)dx=i\pi\sum_{\rho,\,{\rm
Im}(\rho)>0}\textup{Res}_{x=\rho}\big{(}R(x)F(x)\big{)}\in\pi{\bf E}$

where the summation is over the poles of $R$ in the upper half plane.

##  4 Derivatives of the $\Gamma$ function at rational points

In this section we prove Theorem 2 and Proposition 1 stated in the
introduction, dealing with $\Gamma^{(s)}(a)$. To begin with, we define
$E$-functions $E_{a,s}(z)$ in §4.1 and prove a linear independence result
concerning these functions. Then we prove in §4.2 a formula for
$\Gamma^{(s)}(a)$, namely Eq. (4.4), involving $E_{a,s+1}(-1)$ and the
1-summation of an -function. This enables us to prove Theorem 2 in §4.3 and
Proposition 1 in §4.4.

###  4.1 Linear independence of a family of $E$-functions

To study derivatives of the $\Gamma$ function at rational points, we need the
following lemma. For $s\geq 1$ and $a\in\mathbb{Q}\setminus\mathbb{Z}_{\leq
0}$, we consider the $E$-function
$E_{a,s}(z):=\sum_{n=0}^{\infty}\frac{z^{n}}{n!(n+a)^{s}}$.

######  Lemma 2.

$(i)$ For any $a\in\mathbb{Q}\setminus\mathbb{Z}$ and any $s\geq 1$, the
functions

$1,e^{z},e^{z}E_{a,1}(-z),e^{z}E_{a,2}(-z),\ldots,e^{z}E_{a,s}(-z)$

are linearly independent over $\mathbb{C}(z)$.

$(ii)$ For any $a\in\mathbb{N}^{*}$ and any $s\geq 2$, the functions

$1,e^{z},e^{z}E_{a,2}(-z),\ldots,e^{z}E_{a,s}(-z)$

are linearly independent over $\mathbb{C}(z)$.

######  Remark 2.

Part $(i)$ of the lemma is false if $a\in\mathbb{N}^{*}$ because
$1,e^{z},e^{z}E_{a,1}(-z)$ are $\mathbb{Q}(z)$-linearly dependent in this case
(see the proof of Part $(ii)$ below).

###### Proof.

$(i)$ For simplicity, we set $\psi_{s}(z):=e^{z}E_{a,s}(-z)$. We proceed by
induction on $s\geq 1$. Let us first prove the case $s=1$. The derivative of
$\psi_{1}(z)$ is $(1+(z-a)\psi_{1}(z))/z$. Let us assume the existence of a
relation $\psi_{1}(z)=u(z)e^{z}+v(z)$ with $u,v\in\mathbb{C}(z)$ (a putative
relation $U(z)+V(z)e^{z}+W(z)\psi_{1}(z)=0$ forces $W\neq 0$ because
$e^{z}\notin\mathbb{C}(z)$). Then after differentiation of both sides, we end
up with

$\frac{1+(z-a)\psi_{1}(z)}{z}=\big{(}u(z)+u^{\prime}(z)\big{)}e^{z}+v^{\prime}(z).$

Hence,

$\frac{1+(z-a)\big{(}u(z)e^{z}+v(z)\big{)}}{z}=\big{(}u(z)+u^{\prime}(z)\big{)}e^{z}+v^{\prime}(z).$

Since $e^{z}\notin\mathbb{C}(z)$, the function $v(z)$ is a rational solution
of the differential equation $zv^{\prime}(z)=(z-a)v(z)+1$: $v(z)$ cannot be
identically 0, and it cannot be a polynomial (the degrees do not match on both
sides). It must then have a pole at some point $\omega$, of order $d\geq 1$
say. We must have $\omega=0$ because otherwise the order of the pole at
$\omega$ of $zv^{\prime}(z)$ is $d+1$ while the order of the pole of
$(z-a)v(z)+1$ is at most $d$. Writing $v(z)=\sum_{n\geq-d}v_{n}z^{n}$ with
$v_{-d}\neq 0$ and comparing the term in $z^{-d}$ of $zv^{\prime}(z)$ and
$(z-a)v(z)+1$, we obtain that $d=a$. This forces $a$ to be an integer $\geq
1$, which is excluded. Hence, $1,e^{z},e^{z}E_{a,1}(-z)$ are
$\mathbb{C}(z)$-linearly independent.

Let us now assume that the case $s-1\geq 1$ holds. Let us assume the existence
of a relation over $\mathbb{C}(z)$

$\psi_{s}(z)=v(z)+u_{0}(z)e^{z}+\sum_{j=1}^{s-1}u_{j}(z)\psi_{j}(z).$ (4.1)

(A putative relation $V(z)+U_{0}(z)e^{z}+\sum_{j=1}^{s}U_{j}(z)\psi_{j}(z)=0$
forces $U_{s}\neq 0$ by the induction hypothesis). Differentiating (4.1) and
because
$\psi_{j}^{\prime}(z)=(1-\frac{a}{z})\psi_{j}(z)+\frac{1}{z}\psi_{j-1}(z)$ for
all $j\geq 1$ (where we have let $\psi_{0}(z)=1$), we have

$A(z)\psi_{s}(z)+\frac{1}{z}\psi_{s-1}(z)=v^{\prime}(z)+\big{(}u_{0}(z)+u^{\prime}_{0}(z)\big{)}e^{z}+\sum_{j=1}^{s-1}u^{\prime}_{j}(z)\psi_{j}(z)\\\
+\sum_{j=1}^{s-1}u_{j}(z)\big{(}A(z)\psi_{j}(z)+\frac{1}{z}\psi_{j-1}(z)\big{)},$
(4.2)

where $A(z):=1-a/z$. Substituting the right-hand side of (4.1) for
$\psi_{s}(z)$ on the left-hand side of (4.2), we then deduce that

$v^{\prime}(z)-A(z)v(z)+\big{(}u^{\prime}_{0}(z)+(1-A(z))u_{0}(z)\big{)}e^{z}\\\
+\frac{1}{z}(z-a)u_{1}(z)\psi_{1}(z)+\sum_{j=1}^{s-1}u^{\prime}_{j}(z)\psi_{j}(z)+\frac{1}{z}\sum_{j=1}^{s-1}u_{j}(z)\psi_{j-1}(z)-\frac{1}{z}\psi_{s-1}(z)=0.$

This is a non-trivial $\mathbb{C}(z)$-linear relation between
$1,e^{z},\psi_{1}(z),\psi_{2}(z),\ldots,\psi_{s-1}(z)$ because the coefficient
of $\psi_{s-1}(z)$ is $u_{s-1}^{\prime}(z)-1/z$ and it is not identically 0
because $u_{s-1}^{\prime}(z)$ cannot have a pole of order $1$. But by the
induction hypothesis, we know that such a relation is impossible.

$(ii)$ The proof can be done by induction on $s\geq 2$ similarily. In the case
$s=2$, assume the existence of a relation $\psi_{2}(z)=u(z)e^{z}+v(z)$ with
$u(z),v(z)\in\mathbb{C}(z)$. By differentiation, we obtain

$\Big{(}1-\frac{a}{z}\Big{)}\psi_{2}(z)=-\frac{1}{z}\psi_{1}(z)+\big{(}u(z)+u^{\prime}(z)\big{)}e^{z}+v^{\prime}(z).$

By induction on $a\geq 1$, we have $\psi_{1}(z)=(a-1)!e^{z}/z^{a}+w(z)$ for
some $w(z)\in\mathbb{Q}(z)$. Hence, we have

$\Big{(}1-\frac{a}{z}\Big{)}u(z)=-\Big{(}\frac{(a-1)!}{z^{a+1}}+1\Big{)}u(z)+u^{\prime}(z)$

which is not possible. Let us now assume that the case $s-1\geq 2$ holds, as
well as the existence of a relation over $\mathbb{C}(z)$

$\psi_{s}(z)=v(z)+u_{0}(z)e^{z}+\sum_{j=2}^{s-1}u_{j}(z)\psi_{j}(z).$ (4.3)

We proceed exactly as above by differentiation of both sides of (4.3). Using
the relation
$\psi_{j}^{\prime}(z)=(1-\frac{a}{z})\psi_{j}(z)+\frac{1}{z}\psi_{j-1}(z)$ for
all $j\geq 2$ and the fact that $\psi_{1}(z)=(a-1)!e^{z}/z^{a}+w(z)$, we
obtain a relation
$\widetilde{v}(z)+\widetilde{u}_{0}(z)e^{z}+\sum_{j=2}^{s-1}\widetilde{u}_{j}(z)\psi_{j}(z)=0$
where $\widetilde{u}_{s-1}(z)=u_{s-1}^{\prime}(z)-1/z$ cannot be identically
0. The induction hypothesis rules out the existence of such a relation. ∎

###  4.2 A formula for $\Gamma^{(s)}(a)$

Let $z>0$ and $a\in\mathbb{Q}^{+}$, $a\neq 0$. We have

$\Gamma^{(s)}(a)=\int_{0}^{\infty}t^{a-1}\log(t)^{s}e^{-t}dt=\int_{0}^{z}t^{a-1}\log(t)^{s}e^{-t}dt+\int_{z}^{\infty}t^{a-1}\log(t)^{s}e^{-t}dt.$

On the one hand,

$\displaystyle\int_{0}^{z}t^{a-1}\log(t)^{s}e^{-t}dt$
$\displaystyle=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\int_{0}^{z}t^{a+n-1}\log(t)^{s}dt$
$\displaystyle=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\sum_{k=0}^{s}(-1)^{k}\frac{s!}{(s-k)!}\frac{z^{n+a}\log(z)^{s-k}}{(n+a)^{k+1}}$
$\displaystyle=\sum_{k=0}^{s}\frac{(-1)^{k}s!}{(s-k)!}z^{a}\log(z)^{s-k}E_{a,k+1}(-z);$

recall that $E_{a,j}(z)=\sum_{n=0}^{\infty}\frac{z^{n}}{n!(n+a)^{j}}$. On the
other hand,

$\displaystyle\int_{z}^{\infty}t^{a-1}\log(t)^{s}e^{-t}dt$
$\displaystyle=e^{-z}\int_{0}^{\infty}(t+z)^{a-1}\log(t+z)^{s}e^{-t}dt$
$\displaystyle=z^{a-1}e^{-z}\sum_{k=0}^{s}\binom{s}{k}\log(z)^{s-k}\int_{0}^{\infty}(1+t/z)^{a-1}\log(1+t/z)^{k}e^{-t}dt.$

Now $z>0$ so that

$\mathfrak{f}_{a,k+1;0}(z):=\int_{0}^{\infty}(1+tz)^{a-1}\log(1+tz)^{k}e^{-t}dt=\frac{1}{z}\int_{0}^{\infty}(1+x)^{a-1}\log(1+x)^{k}e^{-x/z}dx$

is the $1$-summation at the origin in the direction 0 of the -function

$\sum_{n=0}^{\infty}n!u_{a,k,n}z^{n},$

where the sequence $(u_{a,k,n})_{n\geq 0}\in\mathbb{Q}^{\mathbb{N}}$ is
defined by the expansion of the $G$-function:

$(1+x)^{a-1}\log(1+x)^{k}=\sum_{n=0}^{\infty}u_{a,k,n}x^{n}.$

Note that if $k=0$ and $a\in\mathbb{N}^{*}$, then $u_{a,k,n}=0$ for any $n\geq
a$, and $\mathfrak{f}_{a,k+1;0}(1/z)$ is a polynomial in $1/z$. Therefore, we
have for any $z>0$:

$\Gamma^{(s)}(a)=\sum_{k=0}^{s}\frac{(-1)^{k}s!}{(s-k)!}z^{a}\log(z)^{s-k}E_{a,k+1}(-z)+z^{a-1}e^{-z}\sum_{k=0}^{s}\binom{s}{k}\log(z)^{s-k}\mathfrak{f}_{a,k+1;0}(1/z).$

In particular, for $z=1$, this relation reads

$\Gamma^{(s)}(a)=(-1)^{s}s!E_{a,s+1}(-1)+e^{-1}\mathfrak{f}_{a,s+1;0}(1).$
(4.4)

Since $\gamma=-\Gamma^{\prime}(1)$ we obtain as a special case the formula

$\gamma=E_{1,2}(-1)-e^{-1}\mathfrak{f}_{1,2;0}(1),$ (4.5)

which is also a special case of Eq. (3.1) proved in [15].

###  4.3 Proof of Theorem 2

Let us assume that $\Gamma^{(s)}(a)\in\overline{\mathbb{Q}}$ for some
$a\in\mathbb{Q}^{+}\setminus\mathbb{N}$ and $s\geq 0$. Then
$e^{z}\Gamma^{(s)}(a)+(-1)^{s+1}s!e^{z}E_{a,s+1}(-z)$ is an $E$-function. The
relation
$e\Gamma^{(s)}(a)+(-1)^{s+1}s!eE_{a,s+1}(-1)=\mathfrak{f}_{a,s+1;0}(1)$ proved
at the end of §4.2 shows that
$\alpha:=e\Gamma^{(s)}(a)+(-1)^{s+1}s!eE_{a,s+1}(-1)\in{\bf E}\cap{\bf D}$.
Hence $\alpha$ is in $\overline{\mathbb{Q}}$ by Conjecture 1 and we have a
non-trivial $\overline{\mathbb{Q}}$-linear relation between $1,e$ and
$eE_{a,s+1}(-1)$: we claim that this is not possible. Indeed, consider the
vector

$Y(z):={}^{t}(1,e^{z},e^{z}E_{a,1}(-z),\ldots,e^{z}E_{a,s+1}(-z)).$

It is solution of a differential system $Y^{\prime}(z)=M(z)Y(z)$ where 0 is
the only pole of $M(z)\in M_{s+3}(\overline{\mathbb{Q}}(z))$ (see the
computations in the proof of Lemma 2 above). Since the components of $Y(z)$
are $\overline{\mathbb{Q}}(z)$-linearly independent by Lemma 2$(i)$, we deduce
from Beukers’ [6, Corollary 1.4] that

$1,\,e,\,eE_{a,1}(-1),\,\ldots,\,eE_{a,s+1}(-1)$

are $\overline{\mathbb{Q}}$-linearly independent, and in particular that $1,e$
and $eE_{a,s+1}(-1)$ are $\overline{\mathbb{Q}}$-linearly independent. This
concludes the proof if $a\in\mathbb{Q}^{+}\setminus\mathbb{N}$.

Let us assume now that $\Gamma^{(s)}(a)\in\overline{\mathbb{Q}}$ for some
$a\in\mathbb{N}^{*}$ and $s\geq 1$. Then
$e^{z}\Gamma^{(s)}(a)+(-1)^{s+1}s!e^{z}E_{a,s+1}(-z)$ is an $E$-function. The
relation
$\Gamma^{(s)}(a)+(-1)^{s+1}s!E_{a,s+1}(-1)=e^{-1}\mathfrak{f}_{a,s+1;0}(1)$
shows that $\alpha:=e\Gamma^{(s)}(a)+(-1)^{s+1}s!eE_{a,s+1}(-1)\in{\bf
E}\cap{\bf D}$. Hence $\alpha$ is in $\overline{\mathbb{Q}}$ by Conjecture 1
and we have a non-trivial $\overline{\mathbb{Q}}$-linear relation between
$1,e$ and $eE_{a,s+1}(-1)$: we claim that this is not possible. Indeed,
consider the vector $Y(z):={}^{t}(1,e^{z},e^{z}E_{a,2}(-z),\ldots,$
$e^{z}E_{a,s+1}(-z))$: it is solution of a differential system
$Y^{\prime}(z)=M(z)Y(z)$ where 0 is the only pole of $M(z)\in
M_{s+2}(\overline{\mathbb{Q}}(z))$. Since the components of $Y(z)$ are
$\overline{\mathbb{Q}}(z)$-linearly independent by Lemma 2$(ii)$, we deduce
again from Beukers’ theorem that

$1,\,e,\,eE_{a,2}(-1),\,\ldots,\,eE_{a,s+1}(-1)$

are $\overline{\mathbb{Q}}$-linearly independent, and in particular that $1,e$
and $eE_{a,s+1}(-1)$ are $\overline{\mathbb{Q}}$-linearly independent. This
concludes the proof of Theorem 2.

###  4.4 Proof of Proposition 1

Recall that Eq. (4.5) proved in §4.2 reads
$eE_{1,2}(-1)-e\gamma=\mathfrak{f}_{1,2;0}(1).$ Assuming that $\gamma\in{\bf
E}$, the left-hand side is in ${\bf E}$ while the right-hand side is in D.
Hence both sides are in $\overline{\mathbb{Q}}$ by Conjecture 1. Note that, by
integration by parts,

$\mathfrak{f}_{1,2;0}(1)=\int_{0}^{\infty}\log(1+t)e^{-t}dt=\int_{0}^{\infty}\frac{e^{-t}}{1+t}dt$

is Gompertz’s constant. Hence, by Corollary 1 (which holds under Conjecture
2), the number $\mathfrak{f}_{1,2;0}(1)$ is not in $\overline{\mathbb{Q}}$.
Consequently, $\gamma\notin{\bf E}$.

Similarly, Eq. (4.4) with $a\in\mathbb{Q}\setminus\mathbb{Z}$ and $s=0$ reads
$e\Gamma(a)-eE_{a,1}(-1)=\mathfrak{f}_{a,1;0}(1).$ Assuming that
$\Gamma(a)\in{\bf E}$, the left-hand side is in ${\bf E}$ while the right-hand
side is in D. Hence both sides are in $\overline{\mathbb{Q}}$ by Conjecture 1.
But by Corollary 1 (which holds under Conjecture 2), the number
$\mathfrak{f}_{a,1;0}(1)=\int_{0}^{\infty}(1+t)^{a-1}e^{-t}dt$ is not in
$\overline{\mathbb{Q}}$. Hence, $\Gamma(a)\notin{\bf E}$.

##  5 Application of Beukers’ method and consequence

In this section we prove Theorem 3 and Corollary 1 stated in the introduction.

###  5.1 Proof of Theorem 3

The proof of Theorem 3 is based on the arguments given in [6], except that
$E$-functions have to be replaced with -functions, and 1-summation in non-
anti-Stokes directions is used for evaluations. Conjecture 2 is used as a
substitute for Theorem A$(i)$.

The main step is the following result, the proof of which is analogous to the
end of the proof of [6, Corollary 2.2].

######  Proposition 4.

Assume that Conjecture 2 holds.

Let $\mathfrak{f}$ be an -function, $\xi\in\overline{\mathbb{Q}}^{*}$ and
$\theta\in(\arg(\xi)-\pi/2,\arg(\xi)+\pi/2)$. Assume that $\theta$ is not
anti-Stokes for $\mathfrak{f}$, and that $\mathfrak{f}_{\theta}(1/\xi)=0$.
Denote by $Ly=0$ a differential equation, of minimal order, satisfied by
$\mathfrak{f}(1/z)$.

Then all solutions of $Ly=0$ are holomorphic and vanish at $\xi$; the
differential operator $L$ has an apparent singularity at $\xi$.

To deduce Theorem 3 from Proposition 4, it is enough to follow [6, §3].

###  5.2 Proof of Corollary 1

Let $s\in\mathbb{Q}\setminus\mathbb{Z}_{\geq 0}$. The -function
$\mathfrak{f}(z):=\sum_{n=0}^{\infty}s(s-1)\ldots(s-n+1)z^{n}$ is solution of
the inhomogeneous differential equation
$z^{2}\mathfrak{f}^{\prime}(z)+(1-sz)\mathfrak{f}(z)-1=0$, which can be
immediately transformed into a differential system satisfied by the vector of
-functions ${}^{t}(1,\mathfrak{f}(z))$. The coefficients of the matrix have
only 0 as pole. Moreover, $\mathfrak{f}(z)$ is a transcendental function
because $s\notin\mathbb{Z}_{\geq 0}$. Hence, by Theorem 3,
$\mathfrak{f}_{0}(1/\alpha)\notin\overline{\mathbb{Q}}$ when
$\alpha\in\overline{\mathbb{Q}}$, $\alpha>0$, because $0$ is not an anti-
Stokes direction of $\mathfrak{f}(z)$. It remains to observe that this
1-sommation is

$\int_{0}^{\infty}(1+tz)^{s}e^{-t}dt.$

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S. Fischler, Université Paris-Saclay, CNRS, Laboratoire de mathématiques
d’Orsay, 91405 Orsay, France.

T. Rivoal, Université Grenoble Alpes, CNRS, Institut Fourier, CS 40700, 38058
Grenoble cedex 9, France.

Keywords: $E$-functions, -functions, $G$-functions, Gamma function, Siegel-
Shidlovskii Theorem.

MSC 2020: 11J91 (Primary), 33B15 (Secondary)