# Third post-Newtonian gravitational radiation from two-body scattering II:
Hereditary Energy radiation

Gihyuk Cho<EMAIL_ADDRESS>Deutsches Elektronen-Synchrotron DESY,
Notkestr. 85, 22607 Hamburg, Germany

###### Abstract

We compute the hereditary part of the third post-Newtonian accurate
gravitational energy radiation from hyperbolic scatterings (and parabolic
scatterings) of non-spinning compact objects. We employ large angular momentum
($j$) expansion, and compute it to the relative $1/j^{11}$ order (so the first
12 terms). For parabolic scattering case, the exact solution is computed. At
the end, the completely collected expression of the energy radiation upto the
third post-Newtonian and from $1/j^{3}$ to $1/j^{15}$ order, is presented
including the instantaneous contribution.

##  I Introduction

Accurate prediction of energy radiation from compact object binaries emitted
via gravitational waves (GW), is one of the crucial inputs to predict dynamics
of the binary system. The behavior of gravitational field at future null
infinity is not just dependent on instantaneous dynamics of the binaries at
the same retarded time, but also dependent on the past history of the system
due to non-linearity of Einstein’s field equation Blanchet (2014). The latter
contribution is called hereditary Blanchet and Damour (1988, 1992); Blanchet
and Schaefer (1993) understood within the framework of post-Newtonian (PN)
theory and multipolar post-Minkwoskian (MPM) expansion Blanchet and Damour
(1986, 1989) and effective field theory approach Galley  _et al._ (2016).
Since more attention has been paid on bound orbits Abbott  _et al._ (2017a, b,
2019, 2021); Venumadhav  _et al._ (2020) in GW/multi-messengers astronomy,
energy radiation as well as angular momentum radiation from bound orbits were
already computed upto the third post-Newtonian (3PN) order, which indicates
$1/c^{6}$ ($c$ is speed of light) correction to leading order term, in Arun
_et al._ (2008a, 2009, b) long ago in both instantaneous/hereditary
contributions. On the other hand, highly accurate description on unbound
orbits is getting more attention today in several complementary methodologies
Cho  _et al._ (2018, 2022); Kälin  _et al._ (2020); Dlapa  _et al._ (2021a,
b); Liu  _et al._ (2021), because of its direct usages such as for detection
in LIGO, LISA and IPTA García-Bellido and Nesseris (2018); Mukherjee  _et al._
(2021); Kocsis  _et al._ (2006); Burke-Spolaor  _et al._ (2019), and its
astronomical applications to gravitational captures Bae  _et al._ (2020), and
also the fact that analytic solutions describing dynamics of unbound orbits
have comprehensive information on both bound and unbound orbits Kälin and
Porto (2020a, b); Cho  _et al._ (2021).
In this work, we compute the hereditary contribution of energy radiation
($\Delta{\cal E}$) during hyperbolic and parabolic scattering of non-spinning
compact objects upto 3PN order. This is a sequel of Ref. Cho  _et al._ (2022),
where 3PN instantaneous contribution of $\Delta{\cal E}$ and angular momentum
radiation ($\Delta{\cal J}$) were computed. By Ref. Cho  _et al._ (2022) and
this work, we complete 3PN energy radiation. Historically, the leading
(Newtonian) order energy radiation was computed in Hansen (1972), and the
extension to 1PN order was made in Blanchet and Schäfer (1989). And post-
Minkowskian (PM) result was computed at the leading $G^{3}$ order in Herrmann
_et al._ (2021) ($G$ is Newton’s constant). Recently, the hereditary
contribution of both $\Delta{\cal E}$ and $\Delta{\cal J}$ was tackled in Bini
and Geralico (2021) partially upto $1/j^{7}$ order, at the leading order of
each tail, tail-of-tail, tail-squared pieces. In independent way, we compute
all pieces of the hereditary contribution upto 3PN order including next-
leading order of tail piece, and provide upto $1/j^{15}$ or $1/j^{16}$ order
in large-$j$ (or, large eccentricity) expansion. Since taking large angular
momentum approximation after PN approximation sequentially is equivalent to
small energy approximation after PM expansion, this is corresponding to
compute upto $G^{15}$ (for tail), and $G^{16}$ (for tail-of-tail, tail-
squared) orders. The reason that the aimed $G$ order is different is that our
purpose is not only providing a truncated solution in a certain $G$ order, but
provide a decent approximation with 12 terms for practical usages. Since tail
piece starts at $G^{4}$ order, while tail-of-tail and tail-squared pieces
start at $G^{5}$ order, they could end at the different orders. Also, for this
reason, we work with two variables, angular momentum and eccentricity to
characterize hyperbolic orbits. We will see that this choice exhibits
remarkable convergence in Sec.VI.

This paper is organized as what follows. In Sec.II, we explain all inputs
needed for entire computation. Next, we do the main computation. The three
different pieces are solved via several strategies adopted for each pieces,
leading order of tail in Sec.III, leading order of tail-of-tail, tail-squared
in Sec.IV and 1PN correction of tail in Sec.V. In Sec.VI, we provide exact
value of energy radiation from parabolic orbits, and it will be compared to
hyperbolic energy radiation. In Appendix.A, some required integrals are
computed exactly. In Appendix.C, we collect all contributions upto 3PN and
from $G^{3}$ to $G^{15}$ including the instantaneous part. In the ancillary
file accompanied with this paper, every results are provided in Wolfram
language.

##  II Formalism & Notations

We are computing total energy radiation from scatterings of compact binaries
of which each components of which masses are $m_{1}$, $m_{2}$, and their
positions are $x_{1}^{i}$, $x_{2}^{i}$ in an almost Minkowskian coordinate.
Since total energy radiation $\Delta\mathcal{E}$ is composed of several pieces
schematically,

$\displaystyle\Delta\mathcal{E}=\Delta\mathcal{E}_{\text{inst}}+\Delta\mathcal{E}_{\text{hered}}\,,$
(1)

where $\Delta\mathcal{E}_{\text{hered}}$, of our interest, is consist with
sub-pieces,

$\displaystyle\Delta\mathcal{E}_{\text{hered}}=\Delta\mathcal{E}_{\text{tail}}+\Delta\mathcal{E}_{(\text{tail})^{2}}+\Delta\mathcal{E}_{\text{tail(tail)}}\,.$
(2)

Since $\Delta\mathcal{E}_{\text{tail}}$ starts at 1.5PN order, we need the
leading term and its 1PN correction (2.5PN), while only leading contribution
of
$\Delta\mathcal{E}_{(\text{tail})^{2}}+\Delta\mathcal{E}_{\text{tail(tail)}}$
is required, because they start at 3PN order. To do this, we need 1PN
quadrupole mass moment $I_{ij}$, Newtonian octupolar mass moment $I_{ijk}$ and
Newtonian quadrupole current moment $J_{ij}$ in terms of relative position
$x^{i}=x_{1}^{i}-x_{2}^{i}$ of binaries in the center of mass frame, which can
be found in Arun  _et al._ (2008a), and we present them here for convenience,

$\displaystyle I_{ij}$
$\displaystyle=M\,\nu\,\Big{(}x_{i}\,x_{j}\Big{)}_{\text{STF}}\,\Bigg{[}1+v^{2}\,\Big{(}\frac{29}{42}-\frac{29}{14}\nu\Big{)}+\frac{G\,M}{R}\Big{(}-\frac{5}{7}+\frac{8}{7}\nu\Big{)}\Bigg{]}$
$\displaystyle+\frac{M\,\nu}{c^{2}}\,\Big{(}v_{i}\,v_{j}\Big{)}_{\text{STF}}\,R^{2}\,\Big{(}\frac{11}{21}-\frac{11}{7}\,\nu\Big{)}+\frac{M\,\nu}{c^{2}}\,\Big{(}x_{i}\,v_{j}\Big{)}_{\text{STF}}\,\frac{d(R^{2})}{d\hat{t}}\,\Big{(}-\frac{2}{7}+\frac{6}{7}\,\nu\Big{)}\,,$
(3a) $\displaystyle I_{ijk}$
$\displaystyle=-M\,\nu\,\sqrt{1-4\,\nu}\,\Big{(}x_{i}\,x_{j}\,x_{k}\Big{)}_{\text{STF}}\,,$
(3b) $\displaystyle J_{ij}$
$\displaystyle=M\,\nu\,\sqrt{1-4\,\nu}\,\,\Big{(}x^{i}(x\times
v)^{j}\Big{)}_{\text{STF}}\,.$ (3c)

Therein, the followings are defined : $v^{i}=\frac{dx^{i}}{d\hat{t}}$ where
$\hat{t}$ stands for the physical time, total mass $M=m_{1}+m_{2}$, symmetric
mass ratio $\nu=\frac{m_{1}\,m_{2}}{M^{2}}$, $R=\sqrt{x^{i}x^{j}\delta_{ij}}$,
and $()_{\text{STF}}$ means the symmetric and trace-free (STF) part. And
$x^{i}$ in terms of physical time $\hat{t}$ upto 3PN can be found in Ref. Cho
_et al._ (2018) (we need only 1PN), and is written here with introducing
reduced variables such as $r:=\frac{R}{G\,M}$ and
$t:=\frac{t_{\text{phys}}}{G\,M}$. Note that $r$ has a dimension of
$\frac{\text{time}^{2}}{\text{length}^{2}}$ while $t$ has
$\frac{\text{time}^{3}}{\text{length}^{3}}$. And we define the azimuthal angle
as $\phi:=\arctan(\frac{x^{2}}{x^{1}})$ assuming that $x^{3}=0$. The analytic
expressions for these dynamical variables are given in time implicitly via
eccentric anomaly $u$ ($-\infty\leq u\leq\infty$) characterized by $e_{t}$
(time eccentricity) in 1PN relation,

$\displaystyle r$ $\displaystyle=j^{2}\,\frac{e_{t}\,\cosh
u-1}{e_{t}^{2}-1}+\frac{1}{c^{2}}\,\frac{8+4\,e_{t}^{2}\,(-3+\nu)-e_{t}\,\big{(}e_{t}^{2}\,(-4+\nu)+3\,\nu\big{)}\,\cosh
u}{2\,(e_{t}^{2}-1)}\,,$ (4a) $\displaystyle\phi$
$\displaystyle=2\arctan\left(\sqrt{\frac{e_{t}+1}{e_{t}-1}}\tanh\left(\frac{u}{2}\right)\right)+\frac{1}{c^{2}\,j^{2}}\left(6\arctan\left(\sqrt{\frac{e_{t}+1}{e_{t}-1}}\tanh\left(\frac{u}{2}\right)\right)-\frac{e_{t}\sqrt{e_{t}^{2}-1}\,(-4+\nu)\sinh(u)}{(e_{t}\,\cosh(u)-1)}\right)\,,$
(4b) $\displaystyle t$ $\displaystyle=\frac{1}{n}(e_{t}\,\sinh u-u)\,,$ (4c)

where

$\displaystyle
n=\frac{(e_{t}^{2}-1)^{3/2}}{j^{3}}+\frac{1}{c^{2}\,j^{5}}\,\frac{(e_{t}^{2}-1)^{3/2}\,(3+\nu+e_{t}^{2}\,(-9+5\,\nu))}{2}\,,$
(5)

and $e_{t}$ is dimensionless and given in reduced energy
$E=\frac{\mathcal{E}}{M\,\nu}$, reduced angular momentum
$j=\frac{\mathcal{J}}{G\,M^{2}\,\nu}$ as,

$\displaystyle
e_{t}^{2}=1+2\,E\,j^{2}+\frac{E}{2\,c^{2}}\,\Big{(}8-8\,\nu+(17-7\,\nu)(2\,E\,j^{2})\Big{)}\,.$
(6)

Note that we will adopt $e_{t}$ and $j$ to characterize hyperbolic orbits,
because $e_{t}$ gives the simplest structure in $t$ which makes the following
computations easier, and also large-$e_{t}$ expansion (which is equivalent to
large-$j$ expansion) with $j$ exhibits great convergence rate even at
$e_{t}=1$ as reported in Cho  _et al._ (2022).

##  III Leading order Tail Contribution

In this section, we are going to get the analytic expressions of the Newtonian
contribution of each tail contributions of energy radiation. The computation
done in this section is influenced by Bini and Damour (2017); Bini  _et al._
(2020, 2021). The tail contribution up to 3PN order, the explicit expressions
Arun  _et al._ (2008b) are given as

$\displaystyle\Delta\mathcal{E}_{\text{tail}}$
$\displaystyle=\frac{1}{c^{8}}\,\frac{4\,G^{2}\,\mathcal{M}}{5}\,\frac{1}{G^{6}\,M^{6}}\int^{\infty}_{-\infty}\,dt\,\int^{\infty}_{0}\,d\tau\,I^{(3)}_{ij}(t)\,I^{(5)}_{ij}(t-\tau)\,\Bigg{[}\,\log\Big{(}\frac{c\,\tau\,G\,M}{2\,r_{0}}\Big{)}+\frac{11}{12}\Bigg{]}$
$\displaystyle+\frac{1}{c^{10}}\frac{4\,G^{2}\,\mathcal{M}}{189}\,\frac{1}{G^{8}\,M^{8}}\int^{\infty}_{-\infty}\,dt\,\int^{\infty}_{0}\,d\tau\,I^{(4)}_{ijk}(t)\,I^{(6)}_{ijk}(t-\tau)\,\Bigg{[}\,\log\Big{(}\frac{c\,\tau\,G\,M}{2\,r_{0}}\Big{)}+\frac{97}{60}\Bigg{]}$
$\displaystyle+\frac{1}{c^{10}}\frac{64\,G^{2}\,\mathcal{M}}{45}\,\frac{1}{G^{6}\,M^{6}}\int^{\infty}_{-\infty}\,dt\,\int^{\infty}_{0}\,d\tau\,J^{(3)}_{ij}(t)\,J^{(5)}_{ij}(t-\tau)\,\Bigg{[}\,\log\Big{(}\frac{c\,\tau\,G\,M}{2\,r_{0}}\Big{)}+\frac{7}{6}\Bigg{]}\,,$
(7)

where $\mathcal{M}$ is ADM mass

$\displaystyle\mathcal{M}=M\,\Bigg{(}1+\frac{\nu}{c^{2}}\,\frac{e_{t}^{2}-1}{2\,j^{2}}\Bigg{)}\,.$
(8)

Note that the extra $G\,M$s in the above expressions, which cannot be found in
Arun  _et al._ (2008b) appear because of using $t$ and $\tau$ as the reduced
time variable. In favor of computation, we decompose
$\Delta\mathcal{E}_{\text{tail}}$ into several pieces in Fourier domain such
as

$\displaystyle\Delta\mathcal{E}_{(0)}+\Delta\mathcal{E}_{(1)}=\frac{2}{5}\frac{n^{6}\,\mathcal{M}}{c^{8}\,G^{4}\,M^{6}\,e_{t}^{6}}\int^{\infty}_{0}dp\,|\hat{I}_{ij}(p)|^{2}\,p^{7}\,,$
(9a)
$\displaystyle\Delta\mathcal{E}_{(2)}=\frac{2}{189}\frac{n^{8}\,\mathcal{M}}{c^{10}\,G^{6}\,M^{8}\,e_{t}^{8}}\,\int^{\infty}_{0}\,dp\,|\hat{I}_{ijk}(p)|^{2}\,p^{9}\,,$
(9b)
$\displaystyle\Delta\mathcal{E}_{(3)}=\frac{32}{45}\,\frac{n^{6}\,\mathcal{M}}{c^{10}\,G^{4}\,M^{6}\,e_{t}^{6}}\,\int^{\infty}_{0}\,dp\,|\hat{J}_{ij}(p)|^{2}\,p^{7}\,,$
(9c)

where $\Delta\mathcal{E}_{(0)}$, $\Delta\mathcal{E}_{(2)}$ and
$\Delta\mathcal{E}_{(3)}$ denote the leading(Newtonian) order contributions of
the right hand sides respectively, whereas $\Delta\mathcal{E}_{(1)}$ refers to
1PN correction of the right hand side. Therein, we have defined the Fourier
transform as

$\displaystyle\hat{K}(p)$
$\displaystyle:=\int^{+\infty}_{-\infty}d\bar{l}\,\text{e}^{i\,p\,\bar{l}}K(\bar{l})\,,$
(10a) $\displaystyle K(t)$
$\displaystyle=\frac{1}{2\,\pi}\int^{+\infty}_{-\infty}dp\,\text{e}^{-i\,p\,\bar{l}}\hat{K}(p)\,,$
(10b)

where the reduced mean anomaly is defined as $\bar{l}:=\sinh
u-\frac{u}{e_{t}}=\frac{n}{e_{t}}\,(t-t_{0})$ in the preparation of large
$e_{t}$ expansion, and $p$ and $l$ are both dimensionless for the convenience
of forthcoming computation. Also, we should mention that we use the following
integration reported in Arun  _et al._ (2008b) (with $\gamma$ being Euler’s
constant),

$\displaystyle\int^{+\infty}_{0}dt\,\text{e}^{i\,\sigma\,t}\log(t)=-\frac{1}{\sigma}\,\Bigg{[}\frac{\pi}{2}\,\text{sign}(\sigma)+i\,\Big{(}\log\big{(}|\sigma|\big{)}+\gamma\Big{)}\Bigg{]}$
(11)

in the middle of obtaining the Fourier domain expressions in Eqs. (9).
Among them, we compute three of them $\Delta\mathcal{E}_{(0)}$,
$\Delta\mathcal{E}_{(2)}$ and $\Delta\mathcal{E}_{(3)}$ here, because they
have analytic expressions in Fourier domain, whereas it is not the case for
$\Delta\mathcal{E}_{(1)}$. We will come back to $\Delta\mathcal{E}_{(1)}$ in
Sec. V.

Using Eqs. (4), we can write all components of multipole moments in terms of
the eccentric anomaly $u$ (particularly the form of $\text{e}^{ku}$ with
integers $k$). By considering $d\bar{l}=du(\cosh u-\frac{1}{e_{t}})$, every
terms encountered in the Fourier transformation could be written in form of
the typical integral,

$\displaystyle I_{t}:=\int^{+\infty}_{-\infty}du\,\text{e}^{i\,p\,\sinh
u+k\,u-i\,\frac{p}{e_{t}}\,u}\,.$ (12)

This is nothing but one of integral representations of Hankel function of the
first kind $H^{(1)}_{a}(x)$ (we will call it just the Hankel function
henceforth),

$\displaystyle
H^{(1)}_{a}(x)=\frac{1}{i\,\,\pi}\int^{+\infty}_{-\infty}\,\text{e}^{x\sinh
u-au}\,du\,.$ (13)

Note that the integral on the right hand side is only convergent when the
order is in $-1<\text{Re}(a)<1$, which is not our case ($k$ could be bigger
than 1 in $I_{t}$). It implies that the typical integral $I_{t}$ is actually
divergent in the usual integral sense. But the Hankel function is even
analytic outside of $-1<\text{Re}(a)<1$, hence, we will use the Hankel
function as a regularized, hence finite value of the typical divergent
integral by the argument of analytic continuation resulting in

$\displaystyle I_{t}=i\,\pi\,H^{(1)}_{i\,\frac{p}{e_{t}}-k}(i\,p).$ (14)

After successive uses of the relations

$\displaystyle
H^{(1)}_{a-1}(x)+H^{(1)}_{a+1}(x)=\frac{2a}{x}H^{(1)}_{a}(x)\,,$ (15)

to unify all the orders appearing in the integrands to either $0$ or $1$, we
obtain the intermediate results of $|\hat{I}_{ij}(p)|^{2}$,
$|\hat{I}_{ijk}(p)|^{2}$ and $|\hat{J}_{ij}(p)|^{2}$ at the leading order. As
pointed out in Bini and Damour (2017), the large-$e_{t}$ expansion drops the
$p$ dependence in the order of the Hankel function, which makes the
computation much simpler. After taking the large-$e_{t}$ expansion, every
terms can be dealt by the following master integral (similar expression is
reported in Eq.(7.24) of Bini  _et al._ (2021)),

$\displaystyle\int^{+\infty}_{-\infty}\,dp\,p^{n}H^{(1)}_{a}(i\,p)\,H^{(1)}_{b}(i\,p)=$
$\displaystyle\frac{2^{n}\,\text{e}^{-i\frac{(1+a+b)\,\pi}{2}}}{i\,\,\pi\,\Gamma(1+n)}\,\Gamma\Big{(}\frac{1-a-b+n}{2}\Big{)}\,\Gamma\Big{(}\frac{1+a-b+n}{2}\Big{)}$
$\displaystyle\times\Gamma\Big{(}\frac{1-a+b+n}{2}\Big{)}\,\Gamma\Big{(}\frac{1+a+b+n}{2}\Big{)}\,,$
(16)

Now that all the computation is reduced down to the straightforward but
extensive machinery work, we only present the final results with the first 12
terms in the large $e_{t}$ expansion $\Delta\mathcal{E}^{(l)}_{(0)}$ in
Appendix.B. (We use the superscript $(l)$ to denote that this is not exact one
but expressed in large-$e_{t}$ (or, large-$j$) expansion.) We have found that
the first three terms are coincident to Eq.(4.14) in Bini and Geralico (2021).
By the exactly same procedure, we also get the first 12 terms of
$\Delta\mathcal{E}_{(2)}$ and $\Delta\mathcal{E}_{(3)}$ presented in
Appendix.B.

##  IV Leading Quadratic tail contribution

In this section, we are going to obtain analytic expressions of
$(\text{tail})^{2}$ and $\text{tail}(\text{tail})$ contributions. In terms of
mulitpole moments, the leading order terms of each pieces Arun  _et al._
(2008b) are

$\displaystyle\Delta\mathcal{E}_{(\text{tail})^{2}}$
$\displaystyle=\frac{4\,G^{3}\,\mathcal{M}^{2}}{c^{11}\,5}\,\frac{1}{G^{7}\,M^{7}}\int^{+\infty}_{-\infty}\,dt\,\Bigg{(}\int^{+\infty}_{0}\,d\tau\,I^{(5)}_{ij}(t-\tau)\Bigg{[}\log\Big{(}\frac{c\,\tau\,G\,M}{2\,r_{0}}\Big{)}+\frac{11}{12}\Bigg{]}\Bigg{)}^{2}\,,$
(17a) $\displaystyle\Delta\mathcal{E}_{\text{tail}(\text{tail})}$
$\displaystyle=\frac{4\,G^{3}\,\mathcal{M}^{2}}{c^{11}\,5}\,\frac{1}{G^{7}\,M^{7}}\int^{+\infty}_{-\infty}\,dt\,I^{(3)}_{ij}(t)\int^{+\infty}_{0}\,d\tau\,I^{(6)}_{ij}(t-\tau)\,\Bigg{[}\log^{2}\Big{(}\frac{c\,\tau\,G\,M}{2\,r_{0}}\Big{)}+\frac{57}{70}\,\log\Big{(}\frac{c\,\tau\,G\,M}{2\,r_{0}}\Big{)}+\frac{124627}{44100}\Bigg{]}\,.$
(17b)

Similarly, they are rearranged compactly in Fourier domain into

$\displaystyle\Delta\mathcal{E}_{(4)}$
$\displaystyle=\frac{\mathcal{M}^{2}}{c^{11}\,G^{4}\,M^{7}}\,\frac{2}{5\,\pi}\,\frac{n^{7}}{e_{t}^{7}}\,\Bigg{[}\frac{2\,\pi^{2}}{3}-\frac{214}{105}\Bigg{(}\,\log\Big{(}\frac{2\,n\,r_{0}}{c\,e_{t}\,G\,M}\Big{)}+\gamma_{E}\Bigg{)}-\frac{116761}{29400}\Bigg{]}\,\Big{(}\int^{\infty}_{0}dp\,|\hat{I}_{ij}|^{2}\,p^{8}\Big{)}\,,$
(18a) $\displaystyle\Delta\mathcal{E}_{(5)}$
$\displaystyle=-\frac{\mathcal{M}^{2}}{c^{11}\,G^{4}\,M^{7}}\,\frac{428}{525\,\pi}\,\frac{n^{7}}{e_{t}^{7}}\,\int^{\infty}_{0}dp\,|\hat{I}_{ij}|^{2}\,p^{8}\,\log
p\,,$ (18b)

where we use the relations Arun  _et al._ (2008b), Eq.(11) and

$\displaystyle\int^{+\infty}_{0}dt\,\text{e}^{i\,\sigma\,t}\log^{2}(t)=\frac{i}{\sigma}\,\Bigg{\\{}\frac{\pi^{2}}{6}-\Bigg{[}\frac{\pi}{2}\,\text{sign}(\sigma)+i\,\Big{(}\log\big{(}|\sigma|\big{)}+\gamma\Big{)}\Bigg{]}^{2}\Bigg{\\}}\,.$
(19)

Thanks to even power of $p$ in the integrand, $\Delta\mathcal{E}_{(4)}$ can be
integrated in an exact way. Using the elementary integrals $E^{i}(n)$ defined
in Appendix.A,

$\displaystyle\int^{\infty}_{0}dp\,|\hat{I}_{ij}|^{2}\,p^{8}=8\,\pi^{2}\,e_{t}^{3}\,j^{8}\,G^{4}\,M^{6}\,\nu^{2}$
$\displaystyle\times\Bigg{[}\frac{\left(3\,e_{t}^{2}-4\right)\,e_{t}\,E^{y}(5)}{\left(e_{t}^{2}-1\right)^{3}}+\frac{e_{t}^{2}\,E^{z}(4)}{\left(e_{t}^{2}-1\right)^{3}}+\frac{e_{t}^{2}\,E^{z}(6)}{\left(e_{t}^{2}-1\right)^{2}}-\frac{E^{x}(6)}{e_{t}^{2}-1}-\frac{\left(e_{t}^{4}-3\,e_{t}^{2}+3\right)\,E^{x}(4)}{3\,\left(e_{t}^{2}-1\right)^{4}}\Bigg{]}\,,$
(20)

where the exact solutions of each elementary integrals are listed in Eqs.(51).
Gathering all pieces, we obtain the exact expression ($\mathcal{M}=M$ at this
order) displayed in Eq.(B) of Appendix.B. We have checked that the $r_{0}$
dependence in $\Delta\mathcal{E}_{(4)}$ is canceled with $r_{0}$ in Eq.(31.d)
of Cho  _et al._ (2022), so the final result does not depend on $r_{0}$.
In the case of $\Delta\mathcal{E}_{(5)}$, since there is no closed form
solution, we use the same strategy of computing the Newtonian tails explained
in Sec. III. The picky $\,\log p$ can be dealt by the derivative with respect
to $n$ in the master integral Eq.(III), and hence is reduced down to the
machinery work as well. Likewise, we display the first 12 terms of the
large-$e_{t}$ expansion of $\Delta\mathcal{E}_{(5)}$ in Appendix.B. And the
agreement is found in the first three terms with Eq.(4.14) in Ref.Bini and
Geralico (2021).111Only the second arxiv version of Bini and Geralico (2021)
for now.

##  V 1PN tail contribution

Now, we need to tackle down the most tricky computation of
$\Delta\mathcal{E}_{(1)}$ in this paper. Since we could not obtain Fourier
transform of the quadrupole moment at the 1PN order in general, time domain
integration is attempted. Contrary to the way we have computed, we could not
find the systematic and simple automatic way of integration i.e. master
integral method, so the integrations should be treated in a rather heuristic
way. For the first 4 terms, we could find exact values of the coefficients by
symbolic integration, but we failed for the other 8 terms. Instead, numerical
integration and PSLQ algorithm Bailey and Ferguso (1989) are adopted to find
analytic form of the coefficients. Since we have encountered more than 10
thousands terms, we will explain the process of computation roughly.
By putting all dynamical variables Eq.(4) in the quadrupole moment Eq.(3a), we
construct the integral over $u$ and $v$ which are given via the relations to
the time variables

$\displaystyle t$ $\displaystyle=\frac{1}{n}\big{(}e_{t}\sinh u-u\big{)}\,,$
(21a) $\displaystyle t-\tau$ $\displaystyle=\frac{1}{n}\big{(}e_{t}\sinh
v-v\big{)}\,.$ (21b)

For the efficient computation, we subsequently choose two dynamical parameters
$(\kappa,\,q)$ in the integrand,

$\displaystyle\kappa$ $\displaystyle=\text{e}^{u}\,,$ (22a) $\displaystyle q$
$\displaystyle=\frac{\text{e}^{v}}{\kappa}\,.$ (22b)

Schematically, the tail radiation in time domain has the following form

$\displaystyle\Delta\mathcal{E}$
$\displaystyle\sim\int^{+\infty}_{-\infty}dt\int^{t}_{-\infty}d(t-\tau)\,\mathcal{I}_{1}(t,t-\tau)\,,$
(23a)
$\displaystyle\sim\int^{+\infty}_{-\infty}du\int^{u}_{-\infty}dv\,\mathcal{I}_{2}(u,v)\,,$
(23b)
$\displaystyle\sim\int^{+\infty}_{0}d\kappa\int^{1}_{0}dq\,\mathcal{I}_{3}(\kappa,q)\,,$
(23c)

where $\mathcal{I}_{1,2,3}$ represent some integrands. One can easily see that
the choice of $q$ makes two integrals independent. All terms we encounter is
classified into four categories,

$\displaystyle I_{1}$
$\displaystyle=\frac{(q\,\kappa^{2}+1)^{n}\,\kappa^{l}}{(q-1)^{a}\,(q^{2}\,\kappa^{2}+1)^{b}}\,,$
(24a) $\displaystyle I_{2}$
$\displaystyle=\frac{(q\,\kappa^{2}+1)^{n}\,\,\kappa^{l}\,\arctan(\frac{\kappa-1}{\kappa+1})}{(q-1)^{a}\,(q^{2}\,\kappa^{2}+1)^{b}}\,,$
(24b) $\displaystyle I_{3}$
$\displaystyle=\frac{(q\,\kappa^{2}+1)^{n}\,\,\kappa^{l}\,\log^{m}(q)}{(q-1)^{a}\,(q^{2}\,\kappa^{2}+1)^{c}}\,,$
(24c) $\displaystyle I_{4}$
$\displaystyle=\frac{(q\,\kappa^{2}+1)^{n}\,\,\kappa^{l}\,\arctan(\frac{\kappa-1}{\kappa+1})\,\log^{m}(q)}{(q-1)^{a}\,(q^{2}\,\kappa^{2}+1)^{b}}\,,$
(24d)

where $m,l,a,b,c$ denote some positive integers whereas $n$ is an integer. To
solve them, we have found that

$\displaystyle\int^{1}_{0}\,dq\,\frac{\log^{n}(q)\,q^{m}}{(q+a)(q+b)(q+c)(q+d)}$
(25) $\displaystyle=$
$\displaystyle(-1)^{n+m+1}\,n!\,\Bigg{[}\frac{\,a^{m}\,\text{Li}_{n+1}(\frac{-1}{a})}{(a-b)(a-c)(a-d)}+\frac{b^{m}\,\text{Li}_{n+1}(\frac{-1}{b})}{(b-a)(b-c)(b-d)}+\frac{c^{m}\,\text{Li}_{n+1}(\frac{-1}{c})}{(c-a)(c-b)(c-d)}+\frac{d^{m}\,\text{Li}_{n+1}(\frac{-1}{d})}{(d-a)(d-b)(d-c)}\Bigg{]}\,,$

where $n,m$ are positive integer with $m\leq 3$, and $\text{Li}_{n}$ is poly-
logarithm. After taking successive derivatives with respect to $a,b,c,d$, we
manage to integrate all types of integrand over $q$. In the cases of $I_{1}$,
$I_{2}$ and $I_{3}$, further integration over $\kappa$ is also possible
analytically (hence, every coefficients of $\nu^{3}$ term are determined
exactly). However, in the case of $I_{4}$, we encounter multi-logarithm terms
such as

$\displaystyle\int\,dq\,I_{4}\sim g(Q)\,\arctan(Q)\,\text{Li}_{n}(f(Q))\,,$
(26)

where $f(Q),g(Q)$ are some rational function of polynomial of $Q$ with
$Q=\frac{\kappa-1}{\kappa+1}$. Given that $\arctan$ is also a logarithmic
function, we need to do integration over two or more logarithm. When $n=2$
(i.e. dilogarithm), we could get rid of $\text{Li}_{2}$ by integration-by-
part, but when $n\geq 3$, because the derivative of $\text{Li}_{n}$ is still a
poly-logarithm, we could not find an analytic way of integration. Instead, for
remaining terms, we integrate $I_{4}$ type integrals ($q$ is already
integrated), numerically with $600\sim 1000$ digits, and attempt to find their
analytic forms using PSLQ algorithm. Since PSLQ algorithm is used for guessing
a relation between integers, we need to make a list of candidates. For even
orders $e_{t}^{n}$ ($n=8,6,4\cdots$), the candidate list is consist with
combinations of even powers of $\pi$, and $\zeta(m)$ with odd integer $m$,
such as

$\displaystyle\\{1,\pi^{2},\pi^{4},\cdots,\zeta(3),\pi^{2}\zeta(3),\cdots\zeta(5),\pi^{2}\zeta(5),\cdots\\}\,,$
(27)

and odd powers of $\pi$ for odd powers of $e_{t}$,

$\displaystyle\\{\pi,\pi^{3},\pi^{5}\cdots\\}\,.$ (28)

The higher order of $e_{t}$ we take, the longer candidate lists are required.
We first guess analytic forms of numerical coefficients using first $400\sim
800$ digits, and check if the number of digits used in guessed integers is
much smaller than $400\sim 800$. For example, if from this 40 digits number,

$46.50941502044973026321447260065209280334\,,$

PSLQ guesses these two long integers (in each numerator and denominator) in
totally 55 digits,

$\frac{4320335101191349704192837044}{291827644122831121457024899}\pi\,,$

we regard it as a failure, because it just stores 40 digits with 55 digits.
But if PSLQ find two digits such as $\frac{3}{2}\pi^{3}$, we regard it
trustworthy. After meeting this standard, we also check if they can predict
the next 200 digits, and if so, we confirm them. Using the coefficients
determined in this way, we compute the first 12 terms of
$\Delta\mathcal{E}_{(1)}$ which can be found in Appendix.B.

##  VI Radiation from Parabolic orbits

In this section, we are going to compute hereditary part of energy radiation
$\Delta\mathcal{E}^{(p)}_{\text{hered}}$ from parabolic orbits and this could
be understood as the limit of $e_{t}\to 1$ (or, $E\to 0$) in the hyperbolic
radiation. As pointed out in Cho  _et al._ (2022), the choice of the
parametrization $(e_{t},h)$ (or $(e,h)$) of orbits allows us quickly
converging large eccentricity expansions even approximating well at $e_{t}=1$.
We check if this works even in the hereditary contribution as well as the
validity of our hyperbolic computations.
The parabolic version of keplerian parametrization can be derived by changing
$u$ to $w$ via $u=(e_{t}^{2}-1)^{1/2}\,w$ in Eqs.(4). After taking the limit
of $e_{t}\to 1$, one will get

$\displaystyle r$
$\displaystyle=\frac{j^{2}w^{2}}{2}+\frac{j^{2}}{2}+\frac{1}{c^{2}}\,\left(\frac{\nu-6}{2}+(1-\nu)w^{2}\right)\,,$
(29a) $\displaystyle t$
$\displaystyle=\frac{j^{3}w^{3}}{6}+\frac{j^{3}w}{2}+\frac{j}{c^{2}}\,\left(-\frac{1}{2}(\nu-1)w^{3}-\frac{3}{2}(\nu-1)w\right)\,,$
(29b) $\displaystyle\phi$
$\displaystyle=2\arctan(w)+\frac{2\left(3\,w^{2}\,\arctan(w)-\nu\,w+4\,w+3\,\arctan(w)\right)}{c^{2}\,j^{2}\,\left(1+w^{2}\right)}\,.$
(29c)

By these, it is possible to construct every mulipole moments in terms of $w$.
In the computation of $\Delta\mathcal{E}^{(p)}_{(0)}$ and
$\Delta\mathcal{E}^{(p)}_{(1)}$, the most of the integrations can be done
straightforwardly except

$\displaystyle
I_{p}:=\int^{\infty}_{-\infty}dw\,\frac{w\left(w^{6}+6w^{4}-18w^{2}+4\right)\arctan(w)}{\left(w^{2}+1\right)^{7}\left(w^{2}+4\right)^{7/2}}\,,$
(30)

which arises in 1PN order of quadrupole tail part. After inserting the
integral representation of

$\arctan(w)=\int^{1}_{0}{dk}\frac{w}{1+w^{2}\,k^{2}}\,,$

the successive integrations over $w$ and $k$ could yield the analytic result.
By simplifying the result with the help of the second order of Clausen
function, we managed to obtain

$\displaystyle
I_{p}=\frac{385\,\text{Cl}_{2}\left(\frac{\pi}{3}\right)}{1024\,\sqrt{3}}-\frac{5221}{115200}-\frac{501151\,\pi}{2799360\,\sqrt{3}}+\frac{77\,\pi\,\log(3)}{1024\,\sqrt{3}}\,,$
(31)

where

$\displaystyle\text{Cl}_{2}\left(z\right):=-\int^{z}_{0}\,dx\,\log\Big{|}2\sin\frac{x}{2}\Big{|}.$
(32)

Specially, $\text{Cl}_{2}\left(\frac{\pi}{3}\right)$ is the maximum value of
$\text{Cl}_{2}\left(z\right)$ when $z$ is real. As a result, we obtain the
quadrupolar tail piece of energy radiation from parabolic orbits
$\Delta\mathcal{E}^{(p)}$ ,

$\displaystyle\Delta\mathcal{E}^{(p)}_{(0,1)}=\frac{M\,\nu^{2}\,\pi}{c^{8}\,j^{10}}\bBigg@{3}[\frac{169984}{45\,\sqrt{3}}+\frac{1}{c^{2}\,j^{2}}\left(27720\,\text{Cl}_{2}\left(\frac{\pi}{3}\right)+\frac{7949344\,\pi}{1215}-\frac{41631152}{525\,\sqrt{3}}+5544\,\pi\,\log(3)+\frac{2818048\,\nu}{63\sqrt{3}}\right)\bBigg@{3}]\,$

In the case of contributions $\Delta\mathcal{E}^{(p)}_{(2)}$,
$\Delta\mathcal{E}^{(p)}_{(3)}$, all computation is trivial. And for
$\Delta\mathcal{E}^{(p)}_{(4)}$, it is easily obtained by putting 1 in $e_{t}$
in Eq.(B) because it is exact,

$\displaystyle\Delta\mathcal{E}^{(p)}_{(2)}$
$\displaystyle=\frac{M\,\nu^{2}\,(1-4\,\nu)\,\pi}{c^{10}\,j^{12}}\frac{1448960}{63\,\sqrt{3}}\,,$
(33a) $\displaystyle\Delta\mathcal{E}^{(p)}_{(3)}$
$\displaystyle=\frac{M\,\nu^{2}\,(1-4\,\nu)\,\pi}{c^{10}\,j^{12}}\,\frac{4096}{9\,\sqrt{3}}\,,$
(33b) $\displaystyle\Delta\mathcal{E}^{(p)}_{(4)}$
$\displaystyle=\frac{M\,\nu^{2}\,\pi}{c^{11}\,j^{13}}\,\Bigg{[}-\frac{161249\,\gamma}{15}-\frac{161249}{15}\,\log\left(\frac{2\,r_{0}\,n}{e_{t}\,G\,M\,c}\right)$
$\displaystyle+\frac{10549\,\pi^{2}}{3}-\frac{175958827}{8400}\Bigg{]}\,,$
(33c)

where we keep $\log(n)$, because it is divergent at $e_{t}=1$, which will be
canceled by $\Delta\mathcal{E}_{(5)}$. The fifth piece
$\Delta\mathcal{E}^{(p)}_{(5)}$ is left. After performing inverse Fourier
transform, $\Delta\mathcal{E}_{(5)}$ is re-expressed in time domain as

$\displaystyle\Delta\mathcal{E}_{(5)}=\frac{428}{525}\Bigg{[}\Big{(}\gamma+\log\left(\frac{n}{c^{3}\,e_{t}}\right)\Big{)}\,\int^{+\infty}_{-\infty}dt\,\Big{(}I_{ij}^{(4)}(t)\Big{)}^{2}\,$
$\displaystyle+\int^{+\infty}_{-\infty}dt\,\int^{+\infty}_{t}d\tau\,I_{ij}^{(5)}(t+\tau)I_{ij}^{(4)}(\tau)\log(\tau\,c^{3})\Bigg{]}\,.$
(34)

As previously computed, almost all term can be dealt by elementary integration
methods, but $\arctan$ term should be replaced by its integral representation
again. One could get

$\displaystyle\Delta\mathcal{E}^{(p)}_{(5)}=\frac{M\,\nu^{2}\,\pi}{c^{11}\,j^{13}}\Bigg{[}\frac{161249\,\gamma}{15}+\frac{161249}{15}\,\log\left(\frac{n}{c^{3}\,e_{t}}\right)$
(35)
$\displaystyle-\frac{90210202}{1575}+\frac{161249}{5}\,\log(2\,c\,j)+\frac{161249}{30}\,\log(3)\Bigg{]}\,.$

Indeed, the $\log(n)$ term in $\Delta\mathcal{E}^{(p)}_{(4)}$ and
$\Delta\mathcal{E}^{(p)}_{(5)}$ are canceled. Contrary to
$\Delta\mathcal{E}^{(p)}_{(5)}$ having $\log(j)$ dependence,
$\Delta\mathcal{E}^{(l)}_{(5)}$ does not have it. This missing $\log(j)$ comes
in from $\log(n)=\log\left(\frac{(e_{t}^{2}-1)^{3/2}}{j^{3}}\right)$ in
$\Delta\mathcal{E}_{(4)}$ (Eq.(B)), which also indicates that
$\Delta\mathcal{E}_{(4)}$ and $\Delta\mathcal{E}_{(5)}$ should not be treated
independently. Finally, we compare the exact parabolic results
$\Delta\mathcal{E}^{(p)}$ and the hyperbolic results in large-$e_{t}$
expansion $\Delta\mathcal{E}^{(l)}$ at $e_{t}=1$ in Table.1. One could find
that they have very close values, which shows the robustness of our
computations as well as the efficiency of large-$e_{t}$ expansion.

| $\Delta\mathcal{E}^{(p)}$ | $\Delta\mathcal{E}^{(l)}|_{e_{t}=1}$
---|---|---
$\Delta\mathcal{E}_{(0)}$ | 107.0544 | 107.0545
$\Delta\mathcal{E}_{(1)}$ | 1240.453 | 1240.384
$\Delta\mathcal{E}_{(2)}+\Delta\mathcal{E}_{(3)}$ | 332.3571 | 332.3535
$\Delta\mathcal{E}_{(4)}+\Delta\mathcal{E}_{(5)}$ | -1114.843 | -1114.842
Table 1: 7 digits comparisons. $G=M=c=j=r_{0}=1$ and $\nu=1/8$ are chosen

For the completeness, we present the completed result of
$\Delta\mathcal{E}^{(p)}=\Delta\mathcal{E}^{(p)}_{\text{inst}}+\Delta\mathcal{E}^{(p)}_{\text{hered}}$
from parabolic orbits upto 3PN combined with the instantaneous contribution
(Eq.(43) of Cho  _et al._ (2022)),

$\displaystyle\Delta\mathcal{E}^{(p)}$
$\displaystyle=\frac{M\,\nu^{2}\,\pi}{c^{5}\,j^{7}}\,\bBigg@{3}\\{\frac{170}{3}+\frac{1}{c^{2}\,j^{2}}\,\Bigg{(}\frac{13447}{20}-\frac{1127\,\nu}{3}\Bigg{)}+\frac{1}{c^{3}\,j^{3}}\,\frac{169984}{45\sqrt{3}}+\frac{1}{c^{4}\,j^{4}}\,\Bigg{(}\frac{5839651}{1008}-\frac{258051\,\nu}{40}+\frac{5481\,\nu^{2}}{4}\Bigg{)}$
$\displaystyle+\frac{1}{c^{5}\,j^{5}}\,\Bigg{(}27720\,\text{Cl}_{2}\left(\frac{\,\pi}{3}\right)-\frac{29317552}{525\sqrt{3}}+\frac{7949344\,\pi}{1215}+5544\,\pi\log(3)-\frac{3092480\,\nu}{63\sqrt{3}}\Bigg{)}$
$\displaystyle+\frac{1}{c^{6}\,j^{6}}\,\Bigg{[}\frac{161249\log(2\,\sqrt{3}\,c\,j)}{15}+\frac{10549\,\pi^{2}}{3}+\frac{16880186749}{241920}+\left(-\frac{3972009943}{30240}+\frac{208813\,\pi^{2}}{160}\right)\,\nu$
$\displaystyle\quad+\frac{1157409\,\nu^{2}}{32}-\frac{29645\,\nu^{3}}{8}\Bigg{]}\bBigg@{3}\\}+\mathcal{O}\Bigg{(}\frac{1}{c^{12}}\Bigg{)}.$
(41)

##  VII Discussion & Future work

We have computed the hereditary part of energy radiation from hyperbolic
scatterings as well as parabolic scatterings. For hyperbolic computation, we
rely on large-$e_{t}$ (or, large-$j$) expansion while the parabolic result is
exact. As expected, the hyperbolic result in large-$e_{t}$ is close to the
value of the parabolic one numerically. It is also interesting to point out
that the irrational values appear in the parabolic case is $\pi$, $\log(2)$,
$\log(3)$, $\sqrt{3}$ and $\text{Cl}_{2}(\frac{\pi}{3})$ whereas $\pi$,
$\log(2)$, $\gamma$ and $\zeta(2n+1)$ in the hyperbolic case. In order to
complete 3PN radiation reaction to scattering angle, impact parameter, we need
to compute angular momentum radiation. This is going to be done in a
subsequent paper using the same techniques elaborated in this paper except the
memory effect which will probably require another treatment. We also expect
that the practical knowledge on the integrations gained from this work, might
be helpful to extend our knowledge specially in non-local in time conservative
dynamics, which has the similar structure with hereditary radiation.

## Acknowledgments

We thank Yannick Boetzel for the setup of the formalism at the beginning
phase, Rafael A. Porto for discussion and Hyung-Mok Lee for technical support.
G.C is supported by the ERC Consolidator Grant “Precision Gravity: From the
LHC to LISA,” provided by the European Research Council (ERC) under the
European Union’s H2020 research and innovation programme (grant No. 817791).

##  Appendix A Exact integrals

The aim of this section is to obtain the following element integrals,

$\displaystyle E^{x}(n)$
$\displaystyle:=\int^{\infty}_{0}dq\,q^{n}\,(H^{(1)}_{i\,q}(i\,q\,e_{t}))^{2}\,,$
(42a) $\displaystyle E^{y}(n)$
$\displaystyle:=i\,\int^{\infty}_{0}dq\,q^{n}\,H_{i\,q}(i\,q\,e_{t})\,\dot{H}^{(1)}_{i\,q}(i\,q\,e_{t})\,,$
(42b) $\displaystyle E^{z}(n)$
$\displaystyle:=\int^{\infty}_{0}dq\,q^{n}\,(\dot{H}^{(1)}_{i\,q}(i\,q\,e_{t}))^{2}\,,$
(42c)

where $\dot{H}^{(1)}_{a}(x)=\frac{\partial}{\partial x}H^{(1)}_{a}(x)$.
Inspired by Appendix of Ref.Peters and Mathews (1963), we start at the
following relation between $(l,u,e_{t})$,

$\displaystyle l=e_{t}\,\sinh u-u\,.$ (43)

This gives values of $l$ from $e_{t}$ and $u$ explicitly. What we want to do
is to obtain an explicit expression $u=U(l,e_{t})$. First, we seek Fourier
transform $K(q)$ of $e\,\sinh u$,

$\displaystyle e_{t}\,\sinh
u:=\int^{+\infty}_{-\infty}\,K(q)\,\text{e}^{-i\,q\,l}\,dq\,,$ (44)

or

$\displaystyle K(q)$ $\displaystyle=\frac{e_{t}}{2\pi}\,\int\,\sinh
u\,\text{e}^{i\,q\,l}dl\,,$
$\displaystyle=-\frac{H^{(1)}_{i\,q}(i\,q\,e_{t})}{2\,q}\,.$ (45)

So,

$\displaystyle
U(l,e_{t})=-l-\int^{+\infty}_{-\infty}\,\frac{H^{(1)}_{i\,q}(i\,q\,e_{t})}{2\,q}\,\text{e}^{-i\,q\,l}\,dq\,.$
(46)

Now, we have two definitions for $U$, the implicit one Eq.(43), and the
explicit one Eq.(46). If one takes an operator
$\nabla_{e_{t}}:=\partial^{2}_{e_{t}}+\frac{1}{{e_{t}}}\partial_{e_{t}}$ on
the definitions, two seemingly different (but actually the same) expressions
are obtained,

$\displaystyle\nabla_{e_{t}}\,U$
$\displaystyle=\frac{\left(e_{t}^{2}-1\right)\sinh(u)}{e_{t}\,(e_{t}\cosh(u)-1)^{3}}\quad(\text{from
the implicit side})\,,$ (47)
$\displaystyle=\frac{(e_{t}^{2}-1)}{2\,e_{t}^{2}}\int^{+\infty}_{-\infty}\,q\,H^{(1)}_{i\,q}(i\,q\,e_{t})\,\text{e}^{-i\,q\,l}\,dq\quad(\text{from
the explicit side})\,,$ (48)

in which we apply

$\dot{H}^{(1)}_{a}(x)=\frac{a\,H_{a}^{(1)}(x)}{x}-H_{a+1}^{(1)}(x)\,.$

If one repeats this several times, since

$\displaystyle\big{(}\nabla_{e_{t}}\big{)}^{n}\,U\sim\int^{+\infty}_{-\infty}\,q^{2n-1}\,H^{(1)}_{i\,q}(i\,q\,e_{t})\,\text{e}^{-i\,q\,l}\,dq\,,$
(49)

from the explicit side, we multiply it with its complex conjugate ($*$) to get
the desirable form,

$\displaystyle\Big{[}\big{(}\nabla_{e_{t}}\big{)}^{n}U\Big{]}\Big{[}\big{(}\nabla_{e_{t}}\big{)}^{m}\,U\Big{]}^{*}\sim\int\int
dq\,dk\,q^{2n-1}\,k^{2m-1}\,H^{(1)}_{iq}(iqe_{t})\,H^{(1)*}_{ik}(ike_{t})\text{e}^{-i(q-k)l}\,.$
(50)

On the other hand, one will get elementary expressions from the implicit side.
After integrating over $-\infty<l<\infty$ on the both sides and considering
$H^{(1)}_{iq}(iqe_{t})$ is pure imaginary number, one can get exact values of
$E^{x}$ type integrals. In order to get $E^{y}$ and $E^{z}$ types, taking
derivatives with respect to $e_{t}$ is sufficient. As a result, we list the
required integrals below :

$\displaystyle E^{x}(2)$
$\displaystyle=\frac{13\,e_{t}^{2}+2}{12\,\pi\left(\,e_{t}^{2}-1\right)^{3}}+\frac{\left(\,e_{t}^{2}+4\right)\,e_{t}^{2}\arccos\left(-\frac{1}{\,e_{t}}\right)}{4\,\pi\left(\,e_{t}^{2}-1\right)^{7/2}}\,,$
(51a) $\displaystyle E^{x}(4)$
$\displaystyle=-\frac{3691\,e_{t}^{6}+11082\,e_{t}^{4}+2568\,e_{t}^{2}-16}{960\,\pi\left(\,e_{t}^{2}-1\right)^{6}}-\frac{\left(27\,e_{t}^{6}+472\,e_{t}^{4}+592\,e_{t}^{2}+64\right)\,e_{t}^{2}\arccos\left(-\frac{1}{\,e_{t}}\right)}{64\,\pi\left(\,e_{t}^{2}-1\right)^{13/2}}\,,$
(51b) $\displaystyle E^{x}(6)$
$\displaystyle=-\frac{4954041\,e_{t}^{10}+45033746\,e_{t}^{8}+65383216\,e_{t}^{6}+18070896\,e_{t}^{4}+565696\,e_{t}^{2}+1280}{161280\,\pi\left(\,e_{t}^{2}-1\right)^{9}}$
$\displaystyle-\frac{\left(1125\,e_{t}^{10}+45820\,e_{t}^{8}+189040\,e_{t}^{6}+161152\,e_{t}^{4}+27776\,e_{t}^{2}+512\right)\,e_{t}^{2}\arccos\left(-\frac{1}{\,e_{t}}\right)}{512\,\pi\left(\,e_{t}^{2}-1\right)^{19/2}}\,,$
(51c) $\displaystyle E^{y}(3)$
$\displaystyle=\frac{5\,e_{t}\left(11\,e_{t}^{2}+10\right)}{24\,\pi\left(\,e_{t}^{2}-1\right)^{4}}+\frac{\,e_{t}\left(3\,e_{t}^{4}+24\,e_{t}^{2}+8\right)\arccos\left(-\frac{1}{\,e_{t}}\right)}{8\,\pi\left(\,e_{t}^{2}-1\right)^{9/2}}\,,$
(51d) $\displaystyle E^{y}(5)$
$\displaystyle=\frac{\,e_{t}\left(7517\,e_{t}^{6}+39294\,e_{t}^{4}+26296\,e_{t}^{2}+1968\right)}{640\,\pi\left(\,e_{t}^{2}-1\right)^{7}}+\frac{\,e_{t}\left(135\,e_{t}^{8}+3520\,e_{t}^{6}+8160\,e_{t}^{4}+3072\,e_{t}^{2}+128\right)\arccos\left(-\frac{1}{\,e_{t}}\right)}{128\,\pi\left(\,e_{t}^{2}-1\right)^{15/2}}\,,$
(51e) $\displaystyle E^{z}(4)$
$\displaystyle=\frac{5469\,e_{t}^{6}+12598\,e_{t}^{4}+2392\,e_{t}^{2}+16}{960\,\pi\,e_{t}^{2}\left(\,e_{t}^{2}-1\right)^{5}}+\frac{\left(45\,e_{t}^{6}+632\,e_{t}^{4}+624\,e_{t}^{2}+64\right)\arccos\left(-\frac{1}{\,e_{t}}\right)}{64\,\pi\left(\,e_{t}^{2}-1\right)^{11/2}}\,,$
(51f) $\displaystyle E^{z}(6)$
$\displaystyle=\frac{6581763\,e_{t}^{10}+53772430\,e_{t}^{8}+70576832\,e_{t}^{6}+18257424\,e_{t}^{4}+587456\,e_{t}^{2}-1280}{161280\,\pi\,e_{t}^{2}\left(\,e_{t}^{2}-1\right)^{8}}$
$\displaystyle+\frac{\left(1575\,e_{t}^{10}+58140\,e_{t}^{8}+217360\,e_{t}^{6}+169856\,e_{t}^{4}+28032\,e_{t}^{2}+512\right)\arccos\left(-\frac{1}{\,e_{t}}\right)}{512\,\pi\left(\,e_{t}^{2}-1\right)^{17/2}}\,.$
(51g)

Note that for $E^{x}(n)$ and $E^{z}(n)$, $n$ should be even, and odd for
$E^{y}(n)$. Otherwise, this strategy does not work.

##  Appendix B List of Partial results

$\displaystyle\Delta\mathcal{E}^{(l)}_{(0)}$
$\displaystyle=\frac{M\,\nu^{2}}{c^{8}\,j^{10}}\,\bBigg@{3}[\,\frac{3136\,e_{t}^{6}}{45}+\frac{297\,\pi^{3}\,e_{t}^{5}}{20}+\left(\frac{88576\,\pi^{2}}{675}-\frac{64}{45}\right)\,e_{t}^{4}$
$\displaystyle+\left(\frac{11741\,\pi^{3}}{24}-\frac{2755\,\pi^{5}}{64}\right)e_{t}^{3}+\left(-\frac{27776}{225}+\frac{844288\,\pi^{2}}{675}-\frac{280576\,\pi^{4}}{2625}\right)e_{t}^{2}$
$\displaystyle+\left(\frac{202289\,\pi^{3}}{96}-\frac{3419707\,\pi^{5}}{1152}+\frac{17885\,\pi^{7}}{64}\right)e_{t}+\left(\frac{55936}{1575}+\frac{421888\,\pi^{2}}{135}-\frac{509827072\,\pi^{4}}{212625}+\frac{104726528\,\pi^{6}}{496125}\right)$
$\displaystyle+\frac{1}{e_{t}}\Bigg{(}\frac{3553649\,\pi^{3}}{960}-\frac{154014913\,\pi^{5}}{4608}+\frac{102671023\,\pi^{7}}{3200}-\frac{47703411\,\pi^{9}}{16384}\Bigg{)}$
$\displaystyle+\frac{1}{e_{t}^{2}}\Bigg{(}\frac{1856}{189}+\frac{4238336\,\pi^{2}}{945}-\frac{1355776\,\pi^{4}}{81}+\frac{7286226944\,\pi^{6}}{893025}-\frac{719847424\,\pi^{8}}{1091475}\Bigg{)}$
$\displaystyle+\frac{1}{e_{t}^{3}}\Bigg{(}\frac{36730841\,\pi^{3}}{7680}-\frac{7602883699\,\pi^{5}}{46080}+\frac{24445251593\,\pi^{7}}{38400}-\frac{406754528303\,\pi^{9}}{819200}+\frac{2879946531\,\pi^{11}}{65536}\Bigg{)}$
$\displaystyle+\frac{1}{e_{t}^{4}}\Bigg{(}\frac{5312}{1485}+\frac{5176832\,\pi^{2}}{945}-\frac{905998336\,\pi^{4}}{14175}+\frac{12913147904\,\pi^{6}}{127575}-\frac{209391976448\,\pi^{8}}{5457375}+\frac{2021654528\,\pi^{10}}{693693}\Bigg{)}$
$\displaystyle+\frac{1}{e_{t}^{5}}\Bigg{(}\frac{86668951\,\pi^{3}}{15360}-\frac{191346755029\,\pi^{5}}{368640}+\frac{426736282133\,\pi^{7}}{76800}$
$\displaystyle\quad-\frac{440356873312201\,\pi^{9}}{29491200}+\frac{2142139802884279\,\pi^{11}}{206438400}-\frac{4738568225391\,\pi^{13}}{5242880}\Bigg{)}+\mathcal{O}(1/e_{t}^{6})\Bigg{]}\,,$
(52) $\displaystyle\Delta\mathcal{E}^{(l)}_{(1)}$
$\displaystyle=\frac{M\,\nu^{2}}{c^{10}\,j^{12}}\,\bBigg@{3}\\{e_{t}^{8}\,\Bigg{(}-\frac{288256}{315}+\frac{185824\,\,\nu}{315}\Bigg{)}+e_{t}^{7}\,\Bigg{(}\frac{9216\,\pi}{35}-\frac{110367\,\pi^{3}}{560}+\frac{873\,\pi^{3}\,\nu}{7}\Bigg{)}$
$\displaystyle+e_{t}^{6}\,\Bigg{[}\frac{467596}{315}-\frac{6344704\,\pi^{2}}{4725}+\frac{2898\,\zeta(3)}{5}+\left(\frac{81856}{315}+\frac{1715456\,\pi^{2}}{1575}\right)\,\nu\Bigg{]}$
$\displaystyle+e_{t}^{5}\,\Bigg{[}\frac{2816\,\pi}{225}-\frac{216981733\,\pi^{3}}{36960}+\frac{514505\,\pi^{5}}{896}+\left(\frac{682223\,\pi^{3}}{168}-\frac{315785\,\pi^{5}}{896}\right)\,\nu\Bigg{]}$
$\displaystyle+e_{t}^{4}\,\Bigg{[}\frac{3920593}{3150}-\frac{22037087\,\pi^{2}}{3780}+\frac{177987584\,\pi^{4}}{202125}+\frac{54936\,\zeta(3)}{5}-\frac{8883\,\pi^{2}\,\zeta(3)}{8}-\frac{118017\,\zeta(5)}{8}$
$\displaystyle\quad+\left(-\frac{1670624}{1575}+\frac{49979392\,\pi^{2}}{4725}-\frac{5280768\,\pi^{4}}{6125}\right)\,\nu\Bigg{]}$
$\displaystyle+e_{t}^{3}\,\Bigg{[}-\frac{79616\,\pi}{135}-\frac{202204565099\,\pi^{3}}{11088000}+\frac{2230868679727\,\pi^{5}}{57657600}-\frac{7622335\,\pi^{7}}{2048}$
$\displaystyle\quad+\left(\frac{6238315\,\pi^{3}}{336}-\frac{42527319\,\pi^{5}}{1792}+\frac{2272095\,\pi^{7}}{1024}\right)\,\nu\Bigg{]}$
$\displaystyle+e_{t}^{2}\,\Bigg{[}-\frac{5903108}{2205}+\frac{2185941973\,\pi^{2}}{226800}+\frac{5354923998451\,\pi^{4}}{523908000}-\frac{11762012782592\,\pi^{6}}{8442559125}+\frac{76944\,\zeta(3)}{5}$
$\displaystyle\quad-\frac{443177\,\pi^{2}\,\zeta(3)}{8}+\frac{174195\,\pi^{4}\,\zeta(3)}{64}-\frac{5887923\,\zeta(5)}{8}+\frac{3857175\,\pi^{2}\,\zeta(5)}{64}+\frac{47405925\,\zeta(7)}{64}$
$\displaystyle\quad+\left(-\frac{1834496}{11025}+\frac{137875712\,\pi^{2}}{4725}-\frac{28483790848\,\pi^{4}}{1488375}+\frac{1145110528\,\pi^{6}}{694575}\right)\,\nu\Bigg{]}$
$\displaystyle+e_{t}\,\Bigg{[}\frac{12832\,\pi}{45}-\frac{13673548523\,\pi^{3}}{1056000}+\frac{2331631085671213\,\pi^{5}}{5765760000}-\frac{181467504528103\,\pi^{7}}{429977600}+\frac{1265629617\,\pi^{9}}{32768}$
$\displaystyle\quad+\left(\frac{246285677\,\pi^{3}}{6720}-\frac{17308193471\,\pi^{5}}{64512}+\frac{12669225577\,\pi^{7}}{51200}-\frac{732535083\,\pi^{9}}{32768}\right)\,\nu\Bigg{]}$
$\displaystyle+\Bigg{[}\frac{1423722529}{2116800}+\frac{735046449181\,\pi^{2}}{19051200}-\frac{14545675750451\,\pi^{4}}{41912640000}-\frac{1052053473232830449\,\pi^{6}}{38903312448000}$
$\displaystyle\quad+\frac{9956950540288\,\pi^{8}}{2916520425}+\frac{10752\,\zeta(3)}{5}-\frac{1682177\,\pi^{2}\,\zeta(3)}{4}+\frac{39521979\,\pi^{4}\,\zeta(3)}{160}-\frac{9183825\,\pi^{6}\,\zeta(3)}{1024}$
$\displaystyle\quad-\frac{22348923\,\zeta(5)}{4}+\frac{175025907\,\pi^{2}\,\zeta(5)}{32}-\frac{512457435\,\pi^{4}\,\zeta(5)}{2048}+\frac{2151124857\,\zeta(7)}{32}-\frac{10497111975\,\pi^{2}\,\zeta(7)}{2048}$
$\displaystyle\quad-\frac{126709233525\,\zeta(9)}{2048}+\Bigg{(}\frac{804128}{3675}+\frac{1533979136\,\pi^{2}}{33075}-\frac{202340205568\,\pi^{4}}{1488375}+\frac{26006683648\,\pi^{6}}{416745}$
$\displaystyle\quad-\frac{12671254528\,\pi^{8}}{2546775}\Bigg{)}\,\nu\Bigg{]}$
$\displaystyle+\frac{1}{e_{t}}\,\Bigg{[}\frac{8\,\pi}{3}-\frac{13799463626119\,\pi^{3}}{14370048000}+\frac{108826027744604749\,\pi^{5}}{62270208000}-\frac{21125048445944211030463\,\pi^{7}}{2628134885376000}$
$\displaystyle\quad+\frac{5848151327145833\,\pi^{9}}{896860160}-\frac{152039294253\,\pi^{11}}{262144}+\Bigg{(}\frac{341022193\,\pi^{3}}{6720}-\frac{878042967227\,\pi^{5}}{645120}$
$\displaystyle\quad+\frac{2979531597611\,\pi^{7}}{614400}-\frac{42327561853999\,\pi^{9}}{11468800}+\frac{21336464457\,\pi^{11}}{65536}\Bigg{)}\,\nu\Bigg{]}$
$\displaystyle+\frac{1}{e_{t}^{2}}\,\Bigg{[}\frac{334356791}{3326400}+\frac{199332473447\,\pi^{2}}{3386880}-\frac{373142823899\,\pi^{4}}{1587600}+\frac{3907498382463157\,\pi^{6}}{424569600000}$
$\displaystyle\quad+\frac{584827839010454989447\,\pi^{8}}{6570337213440000}-\frac{6152058407223296\,\pi^{10}}{541111756185}-\frac{27597339\,\pi^{2}\,\zeta(3)}{20}+\frac{1213492203\,\pi^{4}\,\zeta(3)}{320}$
$\displaystyle\quad-\frac{6498156357\,\pi^{6}\,\zeta(3)}{5120}+\frac{3215354373\,\pi^{8}\,\zeta(3)}{81920}-\frac{366650361\,\zeta(5)}{20}+\frac{5374036899\,\pi^{2}\,\zeta(5)}{64}$
$\displaystyle\quad-\frac{1812985623603\,\pi^{4}\,\zeta(5)}{51200}+\frac{10171018935\,\pi^{6}\,\zeta(5)}{8192}+\frac{66048647049\,\zeta(7)}{64}-\frac{7427392716051\,\pi^{2}\,\zeta(7)}{10240}$
$\displaystyle\quad+\frac{525021435477\,\pi^{4}\,\zeta(7)}{16384}-\frac{89655063257529\,\zeta(9)}{10240}+\frac{10562439115305\,\pi^{2}\,\zeta(9)}{16384}$
$\displaystyle\quad+\frac{126935300600955\,\zeta(11)}{16384}+\Bigg{(}\frac{704192}{10395}+\frac{78030592\,\pi^{2}}{1323}-\frac{53212499968\,\pi^{4}}{99225}+\frac{137693298688\,\pi^{6}}{178605}$
$\displaystyle\quad-\frac{10766761459712\,\pi^{8}}{38201625}+\frac{308738523136\,\pi^{10}}{14567553}\Bigg{)}\,\nu\Bigg{]}$
$\displaystyle+\frac{1}{e_{t}^{3}}\,\Bigg{[}\frac{1054\,\pi}{135}+\frac{212624533804289\,\pi^{3}}{36951552000}+\frac{11574755701298686367\,\pi^{5}}{2490808320000}-\frac{25790664698470750454305469\,\pi^{7}}{394220232806400000}$
$\displaystyle\quad+\frac{2025121269359422584900830537\,\pi^{9}}{10638690016002048000}-\frac{5826127395266536829411\,\pi^{11}}{42941664460800}+\frac{124362967035879\,\pi^{13}}{10485760}$
$\displaystyle\quad+\Bigg{(}\frac{6592502233\,\pi^{3}}{107520}-\frac{22857379501687\,\pi^{5}}{5160960}+\frac{52153564678267\,\pi^{7}}{1228800}-\frac{14962473720996481\,\pi^{9}}{137625600}$
$\displaystyle\quad+\frac{1787772436877977\,\pi^{11}}{24084480}-\frac{13523632850871\,\pi^{13}}{2097152}\Bigg{)}\,\nu\Bigg{]}+\mathcal{O}(1/e_{t}^{4})\bBigg@{3}\\}\,.$
(53)
$\displaystyle\Delta\mathcal{E}_{(2)}^{(l)}=\frac{M\,\nu^{2}\,(1-4\nu)}{c^{10}\,j^{12}}\Bigg{[}\frac{81856\,e_{t}^{8}}{945}+\frac{4053\,\pi^{3}\,e_{t}^{7}}{160}+\left(\frac{50560}{189}+\frac{275456\,\,\pi^{2}}{945}\right)e_{t}^{6}$
$\displaystyle+\left(\frac{28654757\,\pi^{3}}{20160}-\frac{1292765\,\pi^{5}}{10752}\right)e_{t}^{5}+\left(-\frac{192064}{525}+\frac{4470784\,\pi^{2}}{945}-\frac{20166656\,\pi^{4}}{55125}\right)e_{t}^{4}$
$\displaystyle+\left(\frac{163927481\,\pi^{3}}{16128}-\frac{795673307\,\pi^{5}}{64512}+\frac{2352595\,\pi^{7}}{2048}\right)e_{t}^{3}$
$\displaystyle+\left(-\frac{167936}{1225}+\frac{261379072\,\pi^{2}}{14175}-\frac{53227319296\,\pi^{4}}{4465125}+\frac{10682630144\,\pi^{6}}{10418625}\right)e_{t}^{2}$
$\displaystyle+\left(\frac{270990515\,\pi^{3}}{10752}-\frac{16781634175\,\pi^{5}}{86016}+\frac{3732414369\,\pi^{7}}{20480}-\frac{2161208385\,\pi^{9}}{131072}\right)e_{t}$
$\displaystyle+\left(\frac{3086144}{33075}+\frac{471584768\,\pi^{2}}{14175}-\frac{71593908224\,\pi^{4}}{637875}+\frac{1672914731008\,\pi^{6}}{31255875}-\frac{4699193344\,\pi^{8}}{1091475}\right)$
$\displaystyle+\frac{1}{e_{t}}\Bigg{(}\frac{9587828293\,\pi^{3}}{258048}-\frac{3207169211179\,\pi^{5}}{2580480}+\frac{701408449909\,\pi^{7}}{147456}-\frac{509401541368309\,\pi^{9}}{137625600}$
$\displaystyle+\frac{171695021883\,\pi^{11}}{524288}\Bigg{)}+\frac{1}{e_{t}^{2}}\Bigg{(}\frac{255104}{10395}+\frac{288994304\,\pi^{2}}{6615}-\frac{52566433792\,\pi^{4}}{99225}+\frac{758425714688\,\pi^{6}}{893025}$
$\displaystyle-\frac{12350325784576\,\pi^{8}}{38201625}+\frac{1074396135424\,\pi^{10}}{43702659}\Bigg{)}+\frac{1}{e_{t}^{3}}\Bigg{(}\frac{118105384883\,\pi^{3}}{2580480}-\frac{96013915997417\,\pi^{5}}{20643840}$
$\displaystyle+\frac{380628106184173\,\pi^{7}}{7372800}-\frac{232396062526713293\,\pi^{9}}{1651507200}+\frac{1136196879582428477\,\pi^{11}}{11560550400}$
$\displaystyle-\frac{359312576835069\,\pi^{13}}{41943040}\Bigg{)}+\mathcal{O}(1/e_{t}^{4})\Bigg{]}\,,$
(54)
$\displaystyle\Delta\mathcal{E}_{(3)}^{(l)}=\frac{M\,\nu^{2}\,(1-4\nu)}{c^{10}\,j^{12}}\Bigg{[}\,\frac{512\,e_{t}^{8}}{135}+\frac{15\,\,\pi^{3}\,e_{t}^{7}}{16}+\left(\frac{1024}{135}+\frac{2048\,\pi^{2}}{225}\right)e_{t}^{6}+\left(\frac{10801\,\pi^{3}}{288}-\frac{1225\,\pi^{5}}{384}\right)e_{t}^{5}$
$\displaystyle+\left(-\frac{1024}{75}+\frac{14336\,\pi^{2}}{135}-\frac{65536\,\pi^{4}}{7875}\right)e_{t}^{4}+\left(\frac{226961\,\pi^{3}}{1152}-\frac{186277\,\pi^{5}}{768}+\frac{46305\,\pi^{7}}{2048}\right)e_{t}^{3}$
$\displaystyle+\left(-\frac{1024}{525}+\frac{641024\,\pi^{2}}{2025}-\frac{130777088\,\pi^{4}}{637875}+\frac{1048576\,\pi^{6}}{59535}\right)e_{t}^{2}$
$\displaystyle+\left(\frac{1521919\,\pi^{3}}{3840}-\frac{9146851\,\pi^{5}}{3072}+\frac{282281979\,\pi^{7}}{102400}-\frac{8164233\,\pi^{9}}{32768}\right)e_{t}$
$\displaystyle+\left(\frac{4096}{1575}+\frac{7026688\,\pi^{2}}{14175}-\frac{985563136\,\pi^{4}}{637875}+\frac{214433792\,\pi^{6}}{297675}-\frac{4194304\,\pi^{8}}{72765}\right)$
$\displaystyle+\frac{1}{e_{t}}(\frac{9909613\,\pi^{3}}{18432}-\frac{1446169151\,\pi^{5}}{92160}+\frac{42478149259\,\pi^{7}}{737280}-\frac{217594014011\,\pi^{9}}{4915200}+\frac{1025047023\,\pi^{11}}{262144})$
$\displaystyle+\frac{1}{e_{t}^{2}}\Bigg{(}\frac{1024}{1485}+\frac{83968\,\pi^{2}}{135}-\frac{2523136\,\pi^{4}}{405}+\frac{1194852352\,\pi^{6}}{127575}-\frac{2716860416\,\pi^{8}}{779625}+\frac{33554432\,\pi^{10}}{127413}\Bigg{)}$
$\displaystyle+\frac{1}{e_{t}^{3}}\Bigg{(}\frac{118881539\,\pi^{3}}{184320}-\frac{38059061669\,\pi^{5}}{737280}+\frac{3865364478731\,\pi^{7}}{7372800}$
$\displaystyle\quad-\frac{27120739832417\,\pi^{9}}{19660800}+\frac{10474810693061\,\pi^{11}}{11010048}-\frac{173590231815\,\pi^{13}}{2097152}\Bigg{)}+\mathcal{O}(1/e_{t}^{4})\Bigg{]}\,,$
(55) $\displaystyle\Delta\mathcal{E}_{(4)}$
$\displaystyle=\frac{M\,\nu^{2}}{c^{11}\,j^{13}}\Bigg{[}\left(\frac{163787\,e_{t}^{6}}{450}+\frac{180379\,e_{t}^{4}}{75}+\frac{166996\,e_{t}^{2}}{75}+\frac{62744}{225}\right)\sqrt{e_{t}^{2}-1}$
$\displaystyle+\left(\frac{297\,e_{t}^{8}}{10}+\frac{14008\,e_{t}^{6}}{15}+\frac{41368\,e_{t}^{4}}{15}+\frac{4352\,e_{t}^{2}}{3}+\frac{512}{5}\right)\arccos\left(-\frac{1}{e_{t}}\right)\Bigg{]}$
$\displaystyle\times\Bigg{[}\frac{2\,\pi^{2}}{3}-\frac{214}{105}\,\left(\,\log\left(\frac{2\,n\,r_{0}}{G\,M\,c\,e_{t}}\right)+\gamma\right)-\frac{116761}{29400}\Bigg{]}\,,$
(56)
$\displaystyle\Delta\mathcal{E}^{(l)}_{(5)}=\frac{M\,\nu^{2}}{c^{11}\,j^{13}}\,\Bigg{[}\,e_{t}^{8}\left(\frac{10593\,\gamma\,\pi}{350}-107\,\pi+\frac{31779}{350}\,\pi\,\log(2)\right)+\frac{54784\,e_{t}^{7}}{354375}\Bigg{(}5190\,\gamma-5047-5190\,\log(2)\Bigg{)}$
$\displaystyle+\frac{107\,e_{t}^{6}}{151200}\Bigg{(}-2603475\,\pi\,\zeta(3)+1344768\,\gamma\,\pi-2290582\,\pi+4034304\,\pi\,\log(2)\Bigg{)}$
$\displaystyle+\frac{54784\,e_{t}^{5}}{2480625}\Bigg{(}-443880\,\zeta(3)+291795\,\gamma+45734-291795\,\log(2)\Bigg{)}$
$\displaystyle+\frac{107\,e_{t}^{4}}{302400}\Bigg{(}-351258810\,\pi\,\zeta(3)+374243625\,\pi\,\zeta(5)+7942656\,\gamma\,\pi-6335000\,\pi+23827968\,\pi\,\log(2)\Bigg{)}$
$\displaystyle+\frac{6848\,e_{t}^{3}}{7441875}\Bigg{(}-223001760\,\zeta(3)+220907520\,\zeta(5)+8625330\,\gamma+8763361-8625330\,\log(2)\Bigg{)}$
$\displaystyle+\frac{107\,e_{t}^{2}}{193536000}\Bigg{(}-2243993740800\,\pi\,\zeta(3)+27140132920320\,\pi\,\zeta(5)-25257903843375\,\pi\,\zeta(7)$
$\displaystyle\quad\quad+2673868800\,\gamma\,\pi+186045664\,\pi+8021606400\,\pi\,\log(2)\Bigg{)}$
$\displaystyle+\frac{6848\,e_{t}}{573024375}\Bigg{(}-101159732520\,\zeta(3)+631953315840\,\zeta(5)-531449856000\,\zeta(7)$
$\displaystyle\quad\quad+140502285\,\gamma+277115017-140502285\,\log(2)\Bigg{)}$
$\displaystyle+\Bigg{(}-\frac{17869266109\,\pi\,\zeta(3)}{3600}+\frac{1338717347453\,\pi\,\zeta(5)}{4800}-\frac{483420808341489\,\pi\,\zeta(7)}{204800}+\frac{137148534994779\,\pi\,\zeta(9)}{65536}$
$\displaystyle\quad\quad+\frac{54784\,\gamma\,\pi}{525}+\frac{2277666521\,\pi}{16128000}+\frac{54784}{175}\,\pi\,\log(2)\Bigg{)}$
$\displaystyle+\frac{428}{7449316875\,e_{t}}\Bigg{(}-63394926154560\,\zeta(3)+1492742225694720\,\zeta(5)-6265138721587200\,\zeta(7)$
$\displaystyle\quad\quad+4835860807680000\,\zeta(9)+401531130\,\gamma+1718774371-401531130\,\log(2)\Bigg{)}$
$\displaystyle-\frac{15301}{3538944000\,e_{t}^{2}}\Bigg{(}2814246420480\,\pi\,\zeta(3)-505641831137280\,\pi\,\zeta(5)+15738675691455360\,\pi\,\zeta(7)$
$\displaystyle\quad\quad-113566524198701760\,\pi\,\zeta(9)+98396833403743875\,\pi\,\zeta(11)+52736\,\pi\Bigg{)}$
$\displaystyle+\frac{428}{4469590125\,e_{t}^{3}}\Bigg{(}-80247248257176\,\zeta(3)+5268405431697408\,\zeta(5)-66851431345520640\,\zeta(7)$
$\displaystyle\quad\quad+230673046606184448\,\zeta(9)-169009790268211200\,\zeta(11)+2171169\,\gamma+1653152-2171169\,\log(2)\Bigg{)}$
$\displaystyle+\mathcal{O}(1/e_{t}^{4})\Bigg{]}\,,$ (57)

where $\zeta(z)$ is Riemann-Zeta function.

##  Appendix C Total Energy radiation

In this Appendix, we collect all contributions of total energy radiation
$\Delta\mathcal{E}=\Delta\mathcal{E}_{\text{inst}}+\Delta\mathcal{E}_{\text{hered}}$
upto 3PN and $1/j^{15}$ order including both instantaneous and hereditary
contributions. Note that, to be consistent, we expand the contributions of
which exact form are known (instantaneous part and $\Delta E_{(4)}$) and
discard $1/j^{16}$ contributions of $\Delta E^{(l)}_{(2)}$ and $\Delta
E^{(l)}_{(3)}$ already computed in Eq. (B). Because of the size of the
expression, we introduce shorthand notations,

$\displaystyle p:=\sqrt{\left(1+E+\frac{1}{2}\,\nu\,E^{2}\right)^{2}-1}\,,$
(58)

($p$ is usually called the linear momentum at infinity) and $J$ (do not be
confused with the physical angular momentum $\mathcal{J}$) defined as

$\displaystyle\frac{1}{J^{n}}:=\frac{M\,\nu^{2}}{c^{n-2}}\,\frac{p^{7-n}}{j^{n}}\,,$
(59)

so that whatever value of the positive integer $n$ is, $\frac{1}{J^{n}}$ has a
dimension of energy
$\mathcal{E}\sim\text{mass}\,\frac{\text{length}^{2}}{\text{time}^{2}}$, and
is formally as small as $\mathcal{O}(\frac{1}{c^{5}})$ (i.e. 2.5PN order, the
leading order of energy radiation) and $\mathcal{O}(G^{n})$, while $p$ is
dimensionless and as small as $\mathcal{O}(\frac{1}{c})$ without entailing
$G$, hence which is going to serve as a PN parameter. Every coefficients of
$1/J^{n}$ are dimensionless and polynomials of $p$, of which the highest power
is 6 (that is, 3PN).

$\displaystyle\Delta\mathcal{E}=\frac{\,\pi}{J^{3}}\,\Bigg{[}\frac{37}{15}+p^{2}\left(\frac{1357}{840}-\frac{74\nu}{15}\right)+p^{4}\left(\frac{27953}{10080}-\frac{839\nu}{420}+\frac{37\nu^{2}}{5}\right)+p^{6}\left(-\frac{676273}{354816}-\frac{2699\nu}{504}+\frac{321\nu^{2}}{280}-\frac{148\nu^{3}}{15}\right)\Bigg{]}$
$\displaystyle+\frac{1}{J^{4}}\,\Bigg{[}\frac{1568}{45}+p^{2}\left(\frac{18608}{525}-\frac{1424\nu}{15}\right)+p^{3}\frac{3136}{45}+p^{4}\left(\frac{220348}{11025}-\frac{31036\nu}{525}+172\nu^{2}\right)$
$\displaystyle\quad+p^{5}\left(\frac{1216}{105}-\frac{2848\nu}{15}\right)+p^{6}\left(-\frac{151854}{13475}-\frac{1223594\nu}{33075}+\frac{164\nu^{2}}{3}-\frac{2366\nu^{3}}{9}\right)\Bigg{]}$
$\displaystyle+\frac{\,\pi}{J^{5}}\,\Bigg{[}\frac{122}{5}+p^{2}\left(\frac{13831}{280}-\frac{933\nu}{10}\right)+p^{3}\frac{297\,\pi^{2}}{20}+p^{4}\left(-\frac{64579}{5040}-\frac{187559\nu}{1680}+\frac{2067\nu^{2}}{10}\right)+p^{5}\left(\frac{9216}{35}-\frac{24993\pi^{2}}{1120}-\frac{15291\pi^{2}\nu}{280}\right)$
$\displaystyle\quad+p^{6}\left(\frac{29573617463}{310464000}-\frac{10593\log(\cfrac{p}{2})}{350}+\frac{99\pi^{2}}{10}+\frac{76897\nu}{480}-\frac{4059}{640}\pi^{2}\nu+\frac{12269\nu^{2}}{80}-\frac{1823\nu^{3}}{5}\right)\Bigg{]}$
$\displaystyle+\frac{1}{J^{6}}\,\Bigg{[}\frac{4672}{45}+p^{2}\left(\frac{142112}{315}-\frac{26464\nu}{45}\right)+p^{3}\left(\frac{9344}{45}+\frac{88576\pi^{2}}{675}\right)+p^{4}\left(-\frac{293992}{1701}-\frac{6732728\nu}{4725}+\frac{24424\nu^{2}}{15}\right)$
$\displaystyle\quad+p^{5}\left(\frac{56708}{105}+\frac{1024\pi^{2}}{135}+\frac{2898\zeta(3)}{5}-\frac{52928\nu}{45}-\frac{3014912\pi^{2}\nu}{4725}\right)$
$\displaystyle\quad+p^{6}\left(\frac{36589282372}{11694375}-\frac{18955264\log(2p)}{23625}+\frac{177152\pi^{2}}{675}+\frac{875976284\nu}{297675}-\frac{212216\pi^{2}\nu}{1575}+\frac{4201976\nu^{2}}{1575}-\frac{150892\nu^{3}}{45}\right)\Bigg{]}$
$\displaystyle+\frac{\,\pi}{J^{7}}\,\Bigg{[}\frac{85}{3}+p^{2}\left(\frac{2259}{8}-265\,\nu\right)+p^{3}\left(\frac{1579\,\pi^{2}}{3}-\frac{2755\,\pi^{4}}{64}\right)+p^{4}\left(\frac{19319}{378}-\frac{432805\nu}{336}+\frac{7605\nu^{2}}{8}\right)$
$\displaystyle\quad+p^{5}\left(\frac{210176}{225}-\frac{13138915\pi^{2}}{7392}+\frac{689985\pi^{4}}{3584}-\frac{23514}{7}\pi^{2}\nu+\frac{30285}{112}\pi^{4}\nu\right)$
$\displaystyle\quad+p^{6}\left(\frac{37546579757}{8467200}-\frac{337906\log(\cfrac{p}{2})}{315}+\frac{3158\pi^{2}}{9}-\frac{58957\zeta(3)}{32}+\frac{68898691\nu}{36288}-\frac{51947}{384}\pi^{2}\nu+\frac{1419153\nu^{2}}{448}-\frac{13955\nu^{3}}{6}\right)\Bigg{]}$
$\displaystyle+\frac{1}{J^{8}}\,\Bigg{[}\frac{3104}{75}+p^{2}\left(\frac{625808}{525}-\frac{21136\nu}{25}\right)+p^{3}\left(\frac{6208}{75}+\frac{68096\pi^{2}}{45}-\frac{280576\pi^{4}}{2625}\right)+p^{4}\left(\frac{29175292}{14175}-\frac{1861268\nu}{225}+\frac{104404\nu^{2}}{25}\right)$
$\displaystyle\quad+p^{5}\Bigg{(}\frac{495031}{210}+\frac{3097739\pi^{2}}{1260}+\frac{1062912\pi^{4}}{13475}+12726\zeta(3)-\frac{8883\pi^{2}\zeta(3)}{8}-\frac{118017\zeta(5)}{8}$
$\displaystyle\quad-\frac{42272\nu}{25}-\frac{6647552}{525}\pi^{2}\nu+\frac{15586304\pi^{4}\nu}{18375}\Bigg{)}$
$\displaystyle\quad+p^{6}\Bigg{(}\frac{46042764886}{1091475}-\frac{2081792\log(2p)}{225}+\frac{136192\pi^{2}}{45}-\frac{60043264\zeta(3)}{6125}$
$\displaystyle\quad-\frac{8449234\nu}{4725}-\frac{19516}{25}\pi^{2}\nu+\frac{14002412\nu^{2}}{525}-\frac{934166\nu^{3}}{75}\Bigg{)}\Bigg{]}$
$\displaystyle+\frac{\,\pi}{J^{9}}\,\Bigg{[}p^{2}\left(\frac{13447}{40}-\frac{1127\,\nu}{6}\right)+p^{3}\left(\frac{17213\,\pi^{2}}{6}-\frac{873523\,\pi^{4}}{288}+\frac{17885\,\pi^{6}}{64}\right)+p^{4}\left(\frac{11947909}{6480}-\frac{2838577\,\nu}{720}+\frac{5733\,\nu^{2}}{4}\right)$
$\displaystyle\quad+p^{5}\left(\frac{80128}{135}-\frac{1554265673\pi^{2}}{126000}+\frac{346721827097\pi^{4}}{16473600}-\frac{4078795\pi^{6}}{2048}-\frac{126707}{4}\pi^{2}\nu+\frac{34662173\pi^{4}\nu}{1152}-\frac{2811865\pi^{6}\nu}{1024}\right)$
$\displaystyle\quad+p^{6}\Bigg{(}\frac{111152851823}{3628800}-\frac{263113\log(\cfrac{p}{2})}{45}+\frac{17213\pi^{2}}{9}-\frac{93466961\zeta(3)}{720}+\frac{8474935\zeta(5)}{64}$
$\displaystyle\quad-\frac{343378331\nu}{25920}-\frac{96145}{384}\pi^{2}\nu+\frac{3277505\nu^{2}}{192}-\frac{86681\nu^{3}}{16}\Bigg{)}\Bigg{]}$
$\displaystyle+\frac{1}{J^{10}}\,\Bigg{[}-\frac{15488}{1575}+p^{2}\left(\frac{1897024}{3675}-\frac{123328\nu}{525}\right)+p^{3}\left(-\frac{30976}{1575}+\frac{3042304\pi^{2}}{675}-\frac{76079104\pi^{4}}{30375}+\frac{104726528\pi^{6}}{496125}\right)$
$\displaystyle\quad+p^{4}\left(\frac{916671536}{99225}-\frac{148687088\nu}{11025}+\frac{644624\nu^{2}}{175}\right)$
$\displaystyle\quad+p^{5}\Bigg{(}\frac{1297879}{1225}+\frac{852398347\pi^{2}}{32400}-\frac{193061153711\pi^{4}}{174636000}-\frac{985608224768\pi^{6}}{2814186375}+39102\zeta(3)-\frac{460943\pi^{2}\zeta(3)}{8}+\frac{174195\pi^{4}\zeta(3)}{64}$
$\displaystyle\quad-\frac{6123957\zeta(5)}{8}+\frac{3857175\pi^{2}\zeta(5)}{64}+\frac{47405925\zeta(7)}{64}-\frac{246656\nu}{525}-\frac{35044864}{525}\pi^{2}\nu+\frac{2186350592\pi^{4}\nu}{70875}-\frac{417267712\pi^{6}\nu}{165375}\Bigg{)}$
$\displaystyle\quad+p^{6}\Bigg{(}\frac{14145635478904}{81860625}-\frac{651053056\log(2p)}{23625}+\frac{6084608\pi^{2}}{675}-\frac{16280928256\zeta(3)}{70875}+\frac{11205738496\zeta(5)}{55125}$
$\displaystyle\quad-\frac{36786359432\nu}{297675}-\frac{9184}{225}\pi^{2}\nu+\frac{304241296\nu^{2}}{3675}-\frac{29597608\nu^{3}}{1575}\Bigg{)}\Bigg{]}$
$\displaystyle+\frac{\,\pi}{J^{11}}\,\Bigg{[}p^{3}\left(\frac{24717\,\pi^{2}}{5}-\frac{2235121\,\pi^{4}}{64}+\frac{25779537\,\pi^{6}}{800}-\frac{47703411\,\pi^{8}}{16384}\right)+p^{4}\left(\frac{5839651}{2016}-\frac{258051\,\nu}{80}+\frac{5481\,\nu^{2}}{8}\right)$
$\displaystyle\quad+p^{5}\Bigg{(}-\frac{41509919\pi^{2}}{2800}+\frac{14329698290513\pi^{4}}{45760000}-\frac{13115655312543\pi^{6}}{42997760}+\frac{3631907727\pi^{8}}{131072}-\frac{1012779}{10}\pi^{2}\nu$
$\displaystyle\quad+\frac{34063737}{64}\pi^{4}\nu-\frac{11886663789\pi^{6}\nu}{25600}+\frac{341480853\pi^{8}\nu}{8192}\Bigg{)}$
$\displaystyle\quad+p^{6}\Bigg{(}\frac{8986587257}{115200}-\frac{251878\log(\cfrac{p}{2})}{25}+\frac{16478\pi^{2}}{5}-\frac{239157947\zeta(3)}{160}+\frac{12215817747\zeta(5)}{800}-\frac{114395585661\zeta(7)}{8192}$
$\displaystyle\quad-\frac{465343901\nu}{6720}+\frac{255717}{640}\pi^{2}\nu+\frac{10035909\nu^{2}}{320}-\frac{42399\nu^{3}}{8}\Bigg{)}\Bigg{]}$
$\displaystyle+\frac{1}{J^{12}}\,\Bigg{[}\frac{928}{189}+p^{2}\left(-\frac{911024}{6615}+\frac{8464\nu}{189}\right)+p^{3}\left(\frac{1856}{189}+\frac{4238336\pi^{2}}{945}-\frac{1355776\pi^{4}}{81}+\frac{7286226944\pi^{6}}{893025}-\frac{719847424\pi^{8}}{1091475}\right)$
$\displaystyle\quad+p^{4}\left(\frac{286020724}{59535}-\frac{5744404\nu}{1323}+\frac{46468\nu^{2}}{63}\right)$
$\displaystyle\quad+p^{5}\Bigg{(}-\frac{121500019}{423360}+\frac{307739224253\pi^{2}}{3810240}-\frac{8813724626501\pi^{4}}{181440000}-\frac{2647984678792493\pi^{6}}{457686028800}+\frac{54187719655424\pi^{8}}{32081724675}$
$\displaystyle\quad+29106\zeta(3)-\frac{1908207\pi^{2}\zeta(3)}{4}+\frac{79914933\pi^{4}\zeta(3)}{320}-\frac{9183825\pi^{6}\zeta(3)}{1024}-\frac{25351893\zeta(5)}{4}+\frac{353908989\pi^{2}\zeta(5)}{64}$
$\displaystyle\quad-\frac{512457435\pi^{4}\zeta(5)}{2048}+\frac{4349655639\zeta(7)}{64}-\frac{10497111975\pi^{2}\zeta(7)}{2048}-\frac{126709233525\zeta(9)}{2048}+\frac{16928\nu}{189}$
$\displaystyle\quad-\frac{98811392}{735}\pi^{2}\nu+\frac{2685110272\pi^{4}\nu}{8505}-\frac{125715644416\pi^{6}\nu}{893025}+\frac{7749763072\pi^{8}\nu}{694575}\Bigg{)}$
$\displaystyle\quad+p^{6}\Bigg{(}\frac{10229486279182}{38201625}-\frac{907003904\log(2p)}{33075}+\frac{8476672\pi^{2}}{945}-\frac{290136064\zeta(3)}{189}+\frac{779626283008\zeta(5)}{99225}$
$\displaystyle\quad-\frac{154047348736\zeta(7)}{24255}-\frac{2203483862\nu}{8505}+\frac{696344}{315}\pi^{2}\nu+\frac{27943012\nu^{2}}{315}-\frac{2177662\nu^{3}}{189}\Bigg{)}\Bigg{]}$
$\displaystyle+\frac{\,\pi}{J^{13}}\,\Bigg{[}p^{3}\left(\frac{10549\,\pi^{2}}{4}-\frac{35497847\,\pi^{4}}{240}+\frac{297848551\,\pi^{6}}{480}-\frac{101390485757\,\pi^{8}}{204800}+\frac{2879946531\,\pi^{10}}{65536}\right)$
$\displaystyle\quad+p^{5}\Bigg{(}\frac{2089356493\pi^{2}}{116640}+\frac{9128301066877073\pi^{4}}{5987520000}-\frac{30683901515370260411\pi^{6}}{4322590272000}+\frac{51653285838124707\pi^{8}}{8968601600}-\frac{268570906065\pi^{10}}{524288}$
$\displaystyle\quad-\frac{5166909}{40}\pi^{2}\nu+\frac{2218568099}{640}\pi^{4}\nu-\frac{81672407509\pi^{6}\nu}{6400}+\frac{56346284097807\pi^{8}\nu}{5734400}-\frac{113792507829\pi^{10}\nu}{131072}\Bigg{)}$
$\displaystyle\quad+p^{6}\Bigg{(}\frac{72363445187}{1075200}-\frac{161249\log(\cfrac{p}{2})}{30}+\frac{10549\pi^{2}}{6}-\frac{3798269629\zeta(3)}{600}+\frac{141137663381\zeta(5)}{480}$
$\displaystyle\quad-\frac{243140348991507\zeta(7)}{102400}+\frac{137148534994779\zeta(9)}{65536}-\frac{3972009943\nu}{60480}+\frac{208813\pi^{2}\nu}{320}+\frac{1157409\nu^{2}}{64}-\frac{29645\nu^{3}}{16}\Bigg{)}\Bigg{]}$
$\displaystyle+\frac{1}{J^{14}}\,\Bigg{[}-\frac{10816}{3465}+p^{2}\left(\frac{1338784}{17325}-\frac{4448\nu}{231}\right)$
$\displaystyle\quad+p^{3}\left(-\frac{21632}{3465}+\frac{312832\pi^{2}}{315}-\frac{222912512\pi^{4}}{4725}+\frac{27701936128\pi^{6}}{297675}-\frac{6236143616\pi^{8}}{165375}+\frac{2021654528\pi^{10}}{693693}\right)$
$\displaystyle\quad+p^{4}\left(-\frac{1571460808}{1091475}+\frac{346442552\nu}{363825}-\frac{127592\nu^{2}}{1155}\right)$
$\displaystyle\quad+p^{5}\Bigg{(}\frac{173028797}{1108800}+\frac{97371732173\pi^{2}}{1128960}-\frac{12705804583\pi^{4}}{35280}+\frac{106040792865406033\pi^{6}}{990662400000}+\frac{5290677461059608407\pi^{8}}{81115274240000}$
$\displaystyle\quad-\frac{5322592932069376\pi^{10}}{541111756185}-\frac{27597339\pi^{2}\zeta(3)}{20}+\frac{1213492203\pi^{4}\zeta(3)}{320}-\frac{6498156357\pi^{6}\zeta(3)}{5120}+\frac{3215354373\pi^{8}\zeta(3)}{81920}$
$\displaystyle\quad-\frac{366650361\zeta(5)}{20}+\frac{5374036899\pi^{2}\zeta(5)}{64}-\frac{1812985623603\pi^{4}\zeta(5)}{51200}+\frac{10171018935\pi^{6}\zeta(5)}{8192}+\frac{66048647049\zeta(7)}{64}$
$\displaystyle\quad-\frac{7427392716051\pi^{2}\zeta(7)}{10240}+\frac{525021435477\pi^{4}\zeta(7)}{16384}-\frac{89655063257529\zeta(9)}{10240}+\frac{10562439115305\pi^{2}\zeta(9)}{16384}$
$\displaystyle\quad+\frac{126935300600955\zeta(11)}{16384}-\frac{8896\nu}{231}-\frac{25802496}{245}\pi^{2}\nu+\frac{15275511808\pi^{4}\nu}{11025}-\frac{32249151488\pi^{6}\nu}{14175}$
$\displaystyle\quad+\frac{11118153433088\pi^{8}\nu}{12733875}-\frac{969316237312\pi^{10}\nu}{14567553}\Bigg{)}$
$\displaystyle\quad+p^{6}\Bigg{(}\frac{551319893540236}{5462832375}-\frac{66946048\log(2p)}{11025}+\frac{625664\pi^{2}}{315}-\frac{47703277568\zeta(3)}{11025}+\frac{2964107165696\zeta(5)}{33075}$
$\displaystyle\quad-\frac{1334534733824\zeta(7)}{3675}+\frac{1946853310464\zeta(9)}{7007}-\frac{33565371524\nu}{363825}+984\pi^{2}\nu+\frac{103067368\nu^{2}}{4851}-\frac{6077972\nu^{3}}{3465}\Bigg{)}\Bigg{]}$
$\displaystyle+\frac{\,\pi}{J^{15}}\,\Bigg{[}p^{3}\left(-\frac{272920219\pi^{4}}{960}+\frac{11072684623\pi^{6}}{2400}-\frac{52303041073283\pi^{8}}{3686400}+\frac{266066506940663\pi^{10}}{25804800}-\frac{4738568225391\pi^{12}}{5242880}\right)$
$\displaystyle\quad+p^{5}\Bigg{(}\frac{5244099463\pi^{2}}{209952}+\frac{107378064314399\pi^{4}}{36288000}-\frac{1137117325791451946827\pi^{6}}{19070251200000}+\frac{43996398056106875941927697\pi^{8}}{231275869913088000}$
$\displaystyle\quad-\frac{5992795752733970871587\pi^{10}}{42941664460800}+\frac{513752944703427\pi^{12}}{41943040}-\frac{113399}{2}\pi^{2}\nu+\frac{35658158393\pi^{4}\nu}{3360}-\frac{2572090864343\pi^{6}\nu}{19200}$
$\displaystyle\quad+\frac{9846906270402487\pi^{8}\nu}{25804800}-\frac{195459851677530583\pi^{10}\nu}{722534400}+\frac{7743141717597\pi^{12}\nu}{327680}\Bigg{)}$
$\displaystyle\quad+p^{6}\Bigg{(}\frac{13990148029}{6912000}-\frac{29202463433\zeta(3)}{2400}+\frac{5246870699213\zeta(5)}{2400}-\frac{41808589714402511\zeta(7)}{614400}$
$\displaystyle\quad+\frac{201120530875501809\zeta(9)}{409600}-\frac{446094799380943713\zeta(11)}{1048576}\Bigg{)}\Bigg{]}+\mathcal{O}(1/c^{12},G^{16})\,.$
(60)

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