I want to evaluate $\displaystyle \int _0^{\frac{\pi }{2}}x\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx$ but it's quite difficult. I have tried to rewrite the integral as $$\int _0^{\frac{\pi }{2}}x\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx=\frac{\pi }{2}\int _0^{\frac{\pi }{2}}\tan \left(x\right)\ln ^2\left(\sin \left(x\right)\right)\:dx-\int _0^{\frac{\pi }{2}}x\tan \left(x\right)\ln ^2\left(\sin \left(x\right)\right)\:dx$$ I've also tried to integrate by parts in multiple ways yet I cant go forth with this integral, I also tried using the substitution $t=\tan{\frac{x}{2}}$ but cant get anything to work, I'll appreciate any sort of help.

I also tried using the classical expansion $$\ln \left(\cos \left(x\right)\right)=-\ln \left(2\right)-\sum _{n=1}^{\infty }\frac{\left(-1\right)^n\cos \left(2nx\right)}{n},\:-\frac{\pi }{2}<x<\frac{\pi }{2}$$ But it only gets worse.

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    $\begingroup$ The result is $$\frac{3 \pi \zeta (3)}{16}+\frac{1}{6} \pi \log ^3(2)+\frac{1}{24} \pi ^3 \log (2)$$ $\endgroup$
    – pisco
    Apr 23, 2021 at 9:26
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    $\begingroup$ @TymaGaidash Yes, you're right, there is a typo, the sign of $3\pi\zeta(3)/16$ should be negative. I apologize for the confusion. $\endgroup$
    – pisco
    Apr 23, 2021 at 22:58
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    $\begingroup$ @user Both you and Tyma Gaidash noticed the sign error, but Tyma Gaidash deleted his comment later. $\endgroup$
    – pisco
    Apr 24, 2021 at 10:40
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    $\begingroup$ @pisco Why don't you post an answer? $\endgroup$
    – user
    Apr 24, 2021 at 13:27
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    $\begingroup$ A very simple solution that exploits ideas from (Almost) Impossible Integrals, Sums, and Series is given in this post facebook.com/…. I'm far too busy now to type and arrange the whole solution in a post. $\endgroup$ May 6, 2021 at 9:00

6 Answers 6


$$I=\frac{1}{2}\int_0^{\pi/2} x^2\frac{\ln^2\cos x}{\sin^2x} \, dx+\int_0^{\pi/2} x^2{\ln\cos x}\,dx$$ integrating by parts $$\int_0^{\pi/2} x^2 \ln\cos x \, dx=\frac{\pi^3}{24}\ln2-\frac{\pi}{4}\zeta(3)$$ see Integral $\int_0^\pi \theta^2 \ln^2\big(2\cos\frac{\theta}{2}\big)d \theta$. $$I=\int_0^{\pi/2} x^2\frac{\ln^2\cos x}{\sin^2x} \, dx = \frac{1}{4} \int_0^\infty\frac{(\arctan u)^2 \log^2(1+u^2)}{u^2} \, du$$ Put $$x=\arctan u$$

Closer form for $\int_0^\infty\frac{(\arctan{x})^2\log^2({1+x^2})}{x^2}dx$

  • 5
    $\begingroup$ Funny how these hard integrals always seem to lead back to a Cleo answer $\endgroup$
    – Robert Lee
    May 4, 2021 at 21:20
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    $\begingroup$ Can you please show a way to prove that $\int _0^{\infty }\frac{\arctan ^2\left(x\right)\ln ^2\left(1+x^2\right)}{x^2}\:dx=\frac{1}{24}\int _0^{\infty }\frac{\ln ^4\left(1+x^2\right)}{x^2}\:dx+\frac{2}{3}\int _0^{\infty }\frac{\arctan ^4\left(x\right)}{x^2}\:dx$ without complex analysis? $\endgroup$
    – user815913
    May 4, 2021 at 23:01
  • $\begingroup$ Could you write $\displaystyle \int f(x)\,dx$ instead of $\displaystyle \int f(x)dx \text{ ?}$ Is it not clear why the latter usage is standard? $\endgroup$ May 15, 2021 at 15:55

My approach. $$\int _0^{\frac{\pi }{2}}x\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx$$

$$=\frac{1}{4}\int _0^{\infty }\frac{\arctan \left(x\right)\ln ^2\left(1+x^2\right)}{x\left(1+x^2\right)}\:dx=\frac{\pi }{8}\int _0^{\infty }\frac{\ln ^2\left(1+x^2\right)}{x\left(1+x^2\right)}\:dx$$ $$-\frac{1}{4}\int _0^{\infty }\frac{\arctan \left(\frac{1}{x}\right)\ln ^2\left(1+x^2\right)}{x\left(1+x^2\right)}\:dx$$

$$\int _0^{\infty }\frac{\arctan \left(\frac{1}{x}\right)\ln ^2\left(1+x^2\right)}{x\left(1+x^2\right)}\:dx=\frac{\pi }{2}\int _0^{\infty }\frac{x\ln ^2\left(\frac{x^2}{1+x^2}\right)}{1+x^2}\:dx-\int _0^{\infty }\frac{x\arctan \left(\frac{1}{x}\right)\ln ^2\left(\frac{x^2}{1+x^2}\right)}{1+x^2}\:dx$$ $$=\frac{\pi }{2}\int _0^{\infty }\frac{x\ln ^2\left(\frac{x^2}{1+x^2}\right)}{1+x^2}\:dx-\frac{4}{3}\int _0^{\infty }\frac{x\arctan ^3\left(\frac{1}{x}\right)}{1+x^2}\:dx-\frac{4}{3}\operatorname{\mathfrak{I}} \left\{\int _0^{\infty }\frac{x\ln ^3\left(\frac{x}{x-i}\right)}{1+x^2}\:dx\right\}$$ $$=\frac{\pi }{4}\int _0^1\frac{\ln ^2\left(x\right)}{1-x}\:dx+4\int _0^{\frac{\pi }{2}}x^2\ln \left(\sin \left(x\right)\right)\:dx-\frac{4}{3}\operatorname{\mathfrak{I}} \left\{3\operatorname{Li}_4\left(2\right)+i\pi \ln ^3\left(2\right)-6\zeta \left(4\right)\right\}$$ $$=\frac{5\pi }{4}\zeta \left(3\right)-\frac{\pi ^3}{6}\ln \left(2\right)-\frac{2\pi }{3}\ln ^3\left(2\right)$$

Thus. $$\frac{1}{4}\int _0^{\infty }\frac{\arctan \left(x\right)\ln ^2\left(1+x^2\right)}{x\left(1+x^2\right)}\:dx=\frac{\pi }{8}\zeta \left(3\right)-\frac{1}{4}\left(\frac{5\pi }{4}\zeta \left(3\right)-\frac{\pi ^3}{6}\ln \left(2\right)-\frac{2\pi }{3}\ln ^3\left(2\right)\right)$$ Therefore. $$\int _0^{\frac{\pi }{2}}x\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx=-\frac{3\pi }{16}\zeta \left(3\right)+\frac{\pi ^3}{24}\ln \left(2\right)+\frac{\pi }{6}\ln ^3\left(2\right)$$


This is not an answer, Cornel's solution mentioned in comment is already quite elegant. Instead I provide some remarks.


$$\int_0^{\pi/2} x^3 \cot x \log^2(\cos x) \,dx = -3 \pi \operatorname{Li}_5\left(\frac{1}{2}\right)-3 \pi \operatorname{Li}_4\left(\frac{1}{2}\right) \log (2)-\frac{3 \pi^3 \zeta (3)}{32}+\frac{141 \pi \zeta (5)}{64}-\frac{9}{16} \pi \zeta (3) \log^2(2)-\frac{1}{10} \pi \log^5(2)+\frac{1}{8} \pi^3 \log^3(2)+\frac{11}{480} \pi^5 \log (2) $$

$$\int_0^{\pi/2} x \cot x \log^4(\cos x) \, dx = -6 \pi \operatorname{Li}_5\left(\frac{1}{2}\right)-6 \pi \operatorname{Li}_4\left(\frac{1}{2}\right) \log (2)-\frac{3 \pi ^3 \zeta (3)}{32}+\frac{237 \pi \zeta (5)}{64}-\frac{9}{8} \pi \zeta (3) \log^2(2)-\frac{1}{10} \pi \log^5(2)+\frac{1}{4} \pi ^3 \log^3(2)+\frac{19}{480} \pi^5 \log (2)$$ similar evaluations also exist for $\int_0^{\pi/2} x^a \cot x \log^b(\cos x) \log^c(\sin x) \, dx$ with $a$ odd, $b,c$ positive integers with $a+b+c = 5$.

In OP's question, as well as two examples above, we observe that the results are all "divisible by $\pi$" (each term is multiplied by $\pi$). More generally, when $a$ is odd, $$\int_0^{\pi/2} x^a \cot x \log^b(\cos x) \log^c(\sin x) \, dx = \pi \times (\text{Some alternating Euler sums of weight }a+b+c)$$ When $a$ is even, then $\pi$-factor no longer appears, for example $$\int_0^{\pi /2} x^2 \cot x\ln (\cos x) \, \mathrm{d}x = - \frac{\pi^4}{720} + \frac{\ln^42}{24} - \frac{\pi^2\ln^22}{6} + \operatorname{Li}_4\left(\frac{1}{2}\right)$$


$\color{green}{\textbf{Version of 15.05.21.}}$

$\color{brown}{\textbf{Primary transformations.}}$

Substitution in the form of $$x=\arctan y,\quad \cos^2 x=\dfrac1{1+y^2},\quad \cot x = \dfrac1y,\quad \text dx = \dfrac{\text dy}{1+y^2}\tag1$$ presents the given integral in the form of $$I=\int\limits_0^{\large\frac\pi2} x \cot x\ln^2(\cos x)\,\text dx =\dfrac14\int\limits_0^\infty \arctan y \ln^2\left(1+y^2\right)\,\dfrac{\text dy}{y(1+y^2)}\tag2.$$ This allows partitionally use my answer to the similar question.

Is known the integral $$\int\limits_0^\infty\dfrac{\ln x\,\text dx}{(x+a)(x+b)} =\dfrac{\ln^2a-\ln^2b}{2a-2b},\quad(\Re a>0,\;\Re b>0).\tag3$$

Then $$\dfrac1y\arctan y\ln(1+y^2) = \dfrac i{2y}(\ln(1-iy)-\ln(1+iy))(\ln(1-iy)+\ln(1+iy))$$ $$ = 2\,\dfrac{\ln^2(1-iy)-\ln^2(1+iy)}{2(1-iy)-2(1+iy)} = 2\int\limits_0^\infty\dfrac{\ln z\,\text dz}{(z+1-iy)(z+1+iy)},$$ $$\dfrac1y\arctan y\ln(1+y^2)= 2\int\limits_0^\infty\dfrac{\ln z\,\text dz}{(z+1)^2+y^2}.\tag4$$ Besides, for $\;p>0\;$ $$\int\limits_0^\infty \dfrac{\ln(1+y^2)\,\text dy}{p^2+y^2} =\pi\,\dfrac{\ln(p+1)}{p}.\tag5$$

Taking in account $(2)-(5),$ one can get $$I = \dfrac12\int\limits_0^\infty\int\limits_0^\infty \dfrac{\ln z}{(z+1)^2+y^2} \dfrac{\ln(1+y^2)}{1+y^2}\,\text dz\,\text dy$$ $$= \dfrac12\int\limits_0^\infty\int\limits_0^\infty \left(\dfrac1{1+y^2}-\dfrac1{(z+1)^2+y^2} \right)\ln(1+y^2)\,\dfrac{\ln z}{z^2+2z}\,\text dy\,\text dz,$$ $$I= \dfrac\pi2\int\limits_0^\infty \left(\ln 2-\dfrac{\ln(z+2)}{z+1}\right)\,\dfrac{\ln z}{z^2+2z}\,\text dz.\tag6$$

Considered transformations allowed to simplify the given integral. However, the obtained integral looks non-trivial too.


Integral $(6)$ can be splitted to the six integrals. Really, taking in account $(3),$

\begin{align} &\dfrac4\pi\,I= 2\int\limits_0^\infty \dfrac{(z\ln 2-\ln(1+\frac12z))\ln z}{z(1+z)(2+z)}\,\text dz\\[4pt] &= 2\ln2\int\limits_0^\infty \dfrac{\ln z}{(1+z)(2+z)}\,\text dz -\int\limits_0^\infty \dfrac{\ln(2z) \ln(1+z)}{z(1+z)(1+2z)}\,\text dz\\[4pt] &= \ln^3 2 -\int\limits_0^\infty \dfrac{\ln(2) \ln(1+z)}{z(1+z)(1+2z)}\,\text dz -\int\limits_0^\infty \dfrac{\ln(z) \ln(1+z)}{z(1+z)}\,\text dz +2\int\limits_0^\infty \dfrac{\ln(z) \ln(1+z)}{(1+z)(1+2z)}\,\text dz\\[4pt] &= \ln^3 2 -\int\limits_0^\infty \dfrac{\ln(2) \ln(1+z)}{z(1+z)(1+2z)}\,\text dz -\int\limits_0^\infty \dfrac{\ln(z) \ln(1+z)}{z(1+z)}\,\text dz\\[4pt] &+\int\limits_0^\infty \dfrac{\ln^2(1+z)}{(1+z)(1+2z)}\,\text dz +\int\limits_0^\infty \dfrac{\ln^2(z)}{(1+z)(1+2z)}\,\text dz -\int\limits_0^\infty \dfrac{(\ln(z+1)-\ln z)^2}{(1+z)(1+2z)}\,\text dz. \end{align}

$\color{brown}{\textbf{Closed forms of the integrals.}}$

Four first interals of the five can be calculated by Wolfram Alpha immediately, \begin{align} &I_1 = -\int\limits_0^\infty \dfrac{\ln(2) \ln(1+z)}{z(1+z)(1+2z)}\,\text dz = -\ln^2 2,\\[4pt] &I_2 = -\int\limits_0^\infty \dfrac{\ln(z) \ln(1+z)}{z(1+z)}\,\text dz = -\zeta(3),\\[4pt] &I_3 = \int\limits_0^\infty \dfrac{\ln^2(1+z)}{(1+z)(1+2z)}\,\text dz = \dfrac1{12}(21\zeta(3)+4\ln^3 2 -\pi^2 \ln4),\\[4pt] &I_4 = \int\limits_0^\infty \dfrac{\ln^2(z)}{(1+z)(1+2z)}\,\text dz = \dfrac13\,\ln2(\pi^2+\ln^2 2). \end{align}

Integral I1 Integral I2 Integral I3 Integral I4

Besides, $$I_5 = -\int\limits_0^\infty \dfrac{(\ln(z+1)-\ln z)^2}{(1+z)(1+2z)}\,\text dz = -\int\limits_0^\infty \dfrac{\ln^2\left(\dfrac1z+1\right)}{z^2\left(1+\dfrac1z\right)\left(2+\dfrac1z\right)}\,\text dz,$$ $$I_5 = -\int\limits_0^\infty \dfrac{\ln^2(z+1)}{(1+z)(2+z)}\,\text dz = -\frac32\,\zeta(3).$$

Integral I5

Therefore, $$\dfrac4\pi I = \ln^3 2 - \ln^3 2 - \zeta(3) + \dfrac1{12}(21\zeta(3)+4\ln^3 2 -\pi^2 \ln4) + \dfrac13\,\ln2(\pi^2+\ln^2 2) -\dfrac32\,\zeta(3),$$ $$\color{green}{\mathbf{I = \dfrac\pi{6}\,\ln^3 2 - \dfrac{3\pi}{16}\,\zeta(3) + \dfrac{\pi^3}{24}\ln2.}}\tag7$$


A quite elegant solution.

The fact $$ \Im\left[\log^3\left(\frac{1+e^{2ix}}2\right)\right]=3x\log^2\cos x-x^3 $$

yields $$ \int_0^{\pi/2}x\cot x\log^2\cos x~d x=\frac16\Im\left[\int_{-\pi/2}^{\pi/2}\cot x\log^3\left(\frac{1+\text e^{2ix}}2\right)d x\right]+\frac13\int_0^{\pi/2}x^3\cot x~d x $$ The rest two integrals are easy. $$ \begin{align} &\int_{-\pi/2}^{\pi/2}\cot x\log^3\left(\frac{1+\text e^{2ix}}2\right)~d x \\=&\oint\frac{z+1}{z-1}\log^3\left(\frac{1+z}2\right)~\frac{d z}{2iz}\quad z=e^{2ix} \\=&2\pi~\text{Res}\left[\frac{z+1}{z-1}\log^3\left(\frac{1+z}2\right)~\frac{1}{2iz},0\right] \\=&i\pi\log^32 \end{align} $$

The contour is the unit circle with small perturbations near $z=\pm1$ . $$ \begin{align} &\int_0^{\pi/2}x^3\cot x~d x \\=&x^3\log(2\sin x)\Big|_0^{\pi/2}-3\int_0^{\pi/2}x^2\log(2\sin x)d x \\=&\frac{\pi^3}8\log2+3\int_0^{\pi/2}\sum_{n\ge1}\frac{\cos(2nx)}nx^2~d x \\=&\frac{\pi^3}8\log2+3\sum_{n\ge1}\frac{\pi(-1)^n}{4n^3} \\=&\frac{\pi^3}8\log2-\frac{9\pi}{16}\zeta (3) \end{align} $$

Where Fourier series of $\log(2\sin x)$ is applied.

Combine them and the result follows. $$ \int_0^{\pi/2}x\cot x\log^2\cos x~d x=\frac{\pi^3}{24}\log2-\frac{3\pi}{16}\zeta (3)+\frac\pi6\log^32 $$


I will be using the “make everything into a sum and then integrate” method. This means that we will call it I and manipulate the integral until it can be integrable with a gamma function: $$I=\int_0^{π/2} x\cot(x)\ln^2\cos(x)\,dx=\int_0^{π/2}x\bigg(\sum_{n=0}^∞ \bigg[c_n=\frac{(-1)^n2^{2n}B_{2n}}{(2n)!}\bigg] x^{2n-1}\bigg)\bigg(-\sum_{k=1}^∞ \frac{(-1)^k\cos^k(x)}{k}\bigg)\bigg(-\sum_{m=1}^∞\frac{(-1)^m\cos^m(x)}{m}\bigg)\,dx$$ where $B_y$ are the Bernoulli Numbers.

Because all of the indices are constant from each other, they can be treated as such and moved around without needing any cauchy products are needed, using the complex definition of the cosine function and using the binomial theorem once more gets us our simplified integral to be integrated in the next step: $$\int_0^{π/2}x\bigg(\sum_{n=0}^∞c_n=x^{2n-1}\bigg)\bigg(-\sum_{k=1}^∞ \frac{(-1)^k \cos^k(x)}{k}\bigg)\bigg(-\sum_{m=1}^∞\frac{(-1)^m\cos^m(x)}{m}\bigg)\,dx = \sum_{n=0}^∞ c_n\sum_{k=1}^∞ \sum_{m=1}^∞ \frac{(-1)^{m+k}}{mk} \int_0^{π/2}x^{2n} \frac{(e^{\mathit ix}+ e^{-\mathit ix})^{m+k}}{2^{m+k}} \, dx = \sum_{n=0}^∞ c_n\sum_{k=1}^∞ \sum_{m=1}^∞ \frac{(-1)^{m+k}}{mk} \int_0^{π/2}x^{2n}2^{-(m+k)}\sum_{j=0}^{m+k} \binom{m+k}{j}e^{-ijx} \, dx= \sum_{n=0}^∞ c_n\sum_{k=1}^∞ \sum_{m=1}^ ∞\frac{(-1)^{m+k}}{2^{m+k}mk}\sum_{j=0}^{m+k} \binom{m+k}{j} \int_0^{π/2} x^{2n} e^{-ijx}dx$$

By the way, here is proof this equals @pisco’s answer. From here on out, desmos cannot handle complex numbers. This means that if the following steps are correct, them we have a summation form of I because there are no restrictions on the binomial theorem that would give us conflicts and because this method is similar to the generalized Sophomore’s Dream integral with a sum and then a general powers of $x\ln(x)$ integration where here there is a general power of $x\cos(x)$ except with two more complicated powers.

Finally, we can integrate by using the generalized incomplete gamma function as the integral looks like the main definition of the gamma function and put it into a few forms including the summation form of the gamma function. As I am not completely as versed in substitution for finding the integral in terms of this function, a bit of machine help was used:

$$\sum_{n=0}^∞ c_n\sum_{k=1}^∞ \sum_{m=1}^∞ \frac{(-1)^{m+k}}{2^{m+k}mk} \sum_{j=0}^{m+k} \binom{m+k}{j} \int_0^{π/2} x^{2n} e^{-ijx}\,dx= \sum_{n=0}^∞ c_n\sum_{k=1}^∞ \sum_{m=1}^∞ \frac{(-1)^{m+k}}{2^{m+k}mk}\sum_{j=0}^{m+k} \binom{m+k}{j}(j^{-1}\mathit i・\mathit i^{-2n}j^{-2n}Γ(2n+1,ijx)|_0^{π/2})= \sum_{n=0}^∞ \frac{(-1)^n2^{2n}B_{2n}}{(2n)!}\sum_{k=1}^ ∞\sum_{m=1}^∞ \frac{(-1)^{m+k}}{2^{m+k}mk}\sum_{j=0}^{m+k} \binom{m+k}{j}\frac{(-1)^n\mathit iΓ(2n+1,\frac{π\mathit ij}{2},0)}{j^{2n+1}}= \sum_{n=0}^∞ \frac{B_{2n}}{(2n)!} \sum_{k=1}^ ∞\sum_{m=1}^∞ \frac{(-1)^{m+k}}{2^{k+m-2n}mk}\sum_{j=0}^{m+k} \binom{m+k}{j}\frac{\mathit iΓ(2n+1,\frac{π\mathit ij}{2},0)}{j^{2n+1}}= \sum_{n=0}^∞ \frac{2^{2n}B_{2n}}{(2n)!}\sum_{k=1}^∞ \sum_{m=1}^∞ \frac{1}{(-2)^{m+k}mk} \sum_{j=0}^{m+k} \binom{m+k}{j}\frac{\mathit i(Γ(2n+1,\frac{π\mathit ij}{2})-(2n))!}{j^{2n+1}}= \sum_{n=1}^ ∞\frac{B_{2(n-1)}}{(2(n-1))!}\sum_{k=1}^∞ \sum_{m=1}^∞ \frac{(-1)^{m+k}}{mk}\sum_{j=0}^{m+k} \binom{m+k}{j}\frac{(-\mathit i)^{j-1}}{j^{2n-1}}\sum_{p=0}^{2(n-1)} \frac{(π\mathit i)^p}{2^{k+m+2(1-n)+p}p!}$$

I will try to evaluate this expression if I can. I apologize for the unwanted solution, but this took some effort to put together.:

$$I=\sum_{n=0}^∞ \sum_{k=1}^∞ \sum_{m=1}^∞ \sum_{j=0}^{m+k} \binom{m+k}{j} \frac{(-1)^{m+k} B_{2n} \mathit iΓ(2n+1,\frac{π\mathit ij}{2},0)}{2^{k+m-2n} j^{2n+1} mk (2n)!}=\sum_{n=0}^∞ \sum_{k=1}^∞ \sum_{m=1}^∞ \sum_{j=0}^{m+k} \binom{m+k}{j}\frac{(-1)^nc_n \mathit iΓ(2n+1,\frac{π\mathit ij}{2})-(2n)!}{(-2)^{k+m} j^{2n+1} mk}= \sum_{n=0}^∞ \sum_{k=1}^∞ \sum_{m=1}^∞ \sum_{j=0}^{m+k} \binom{m+k}{j}\frac{(-1)^nc_n \mathit iΓ(2n+1,\frac{π\mathit ij}{2})}{(-2)^{k+m} j^{2n+1} mk}-\sum_{n=0}^∞ \sum_{k=1}^∞ \sum_{m=1}^∞ \sum_{j=0}^{m+k} \binom{m+k}{j}\frac{2^{2n} B_{2n}\mathit i}{(-2)^{k+m} j^{2n+1} mk} $$

Let there be a focus on the second term: $$S_1-I=S_2= \sum_{n=0}^∞ \sum_{k=1}^∞ \sum_{m=1}^∞ \sum_{j=0}^{m+k} \binom{m+k}{j} \frac{2^{2n} B_{2n}\mathit i}{(-2)^{k+m} j^{2n+1} mk} = \mathit i\sum_{n=0}^∞ 2^{2n} B_{2n} \sum_{k=1}^∞ \frac{(-1)^{k}}{2^kk} \sum_{m=1}^∞ \frac{(-1)^{m}}{2^mm} \sum_{j=0}^{m+k} \binom{m+k}{j} j^{-(2n+1)}$$

I will fix this soon enough ...

As always, there is not much of a way to check the post integration answer, so please point out any errors here and give me feedback. As a side note, the index switch was needed to expand out the generalized incomplete gamma function with the 0 in the argument and exponential for this expansion for n>0 was already simplified.

  • 2
    $\begingroup$ this is farther from a closed form than the original integral was. Your answer is not helpful or useful unless you can evaluate that sum that you end up with. $\endgroup$
    – clathratus
    May 3, 2021 at 23:24
  • $\begingroup$ @clathratus If you have any ideas of how to find the closed form of this multi sum form, please tell me. This will speed up the evaluation process unless another contributor finds the form in which case I may delete this answer. Maybe rearrange constants and rewrite as $\sum\sum...f(n)*f(k)*...f(p)$? $\endgroup$ May 3, 2021 at 23:49
  • $\begingroup$ @TymaGaidash You must have got 1 upvote and 1 downvote; each upvote gives you 10 rep and each downvote takes away 2 rep. $\endgroup$ May 4, 2021 at 20:46
  • $\begingroup$ For the first, integrate by parts,then put $$x=\tan{u}$$.For the second integrate twice in parts $\endgroup$
    – user178256
    May 5, 2021 at 13:51
  • 1
    $\begingroup$ @TymaGaidash : Some people attempt to learn how to write MathJax code by looking at posted questions and answers. $\endgroup$ May 16, 2021 at 18:20

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