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How can the following improper integral be evaluated?

$$\int_{0}^{\infty} \left( \text{coth} (x) - x \text{csch}^2 (x) \right) \left( \ln \left( \frac{4 \pi^2}{x^2} + 1 \right) \right) \, dx$$

or alternatively:

$$\int_{0}^{\infty} \frac{x \text{coth} (x) - 1}{x^2 (2 \pi + i x)} \, dx$$ Note: I am only really interested in the imaginary component of the second integral.

I've attempted multiple methods, all of which seeming unsuccessful, however, I believe contour integration may be the solution to the second integral above, which would also easily allow me to get the integral I'm interested in.

KStarGamer
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  • I could also suggest breaking apart the integral, if they converge, or simplifying and turning into summations to then simplify and integrate, even if the result is no elementary. Edit:Where is this from? Is it a PDE solution? – Тyma Gaidash May 03 '21 at 20:19
  • @TymaGaidash Sadly, the parts separately do not converge, and even if that wasn't a problem, the sums would still be difficult to do with the integrals because I don't think Fubini's Theorem would apply for me to switch the orders. It's from a problem about the Riemann Zeta Function that I'm working on also involving the sum that I also have been asking about recently. – KStarGamer May 03 '21 at 20:21
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    @TymaGaidash Furthermore, how do we evaluate the integral of $\frac{\text{coth} (x)}{x (2 \pi + i x)}$ if we choose to split up the integral? – KStarGamer May 03 '21 at 20:32

2 Answers2

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Partial solution

We can present the second integral as a series (imaginary part of the integral forms a series) $$I= \int_{0}^{\infty} \frac{x \text{coth} (x) - 1}{x^2 (2 \pi + i x)} \, dx=\int_{0}^{\infty} \frac{(x \text{coth} (x) - 1)(2\pi-ix)}{x^2 ((2 \pi)^2 + x^2)} \, dx$$ $$x\coth x=1+2\sum_{k=1}^{\infty}\frac{x^2}{\pi^2k^2+x^2}$$

$$I=2\int_{0}^{\infty}dx\frac{2\pi-ix}{(2\pi)^2+x^2}\sum_{k=1}^{\infty}\frac{1}{\pi^2k^2+x^2}=\Re I+i\Im I=I_1+I_2$$ $$I_1=2\pi\int_{0}^{\infty} \frac{x \text{coth} (x) - 1}{x^2 (4 \pi^2 + x^2)}dx=4\pi\sum_{k=1}^{\infty}\int_{0}^{\infty}\frac{1}{(2\pi)^2+x^2}\frac{1}{\pi^2k^2+x^2}dx$$$$=4\pi\sum_{k=1}^{\infty}\int_{0}^{\infty}\Bigl(\frac{1}{(2\pi)^2+x^2}-\frac{1}{\pi^2k^2+x^2}\Bigr)\frac{dx}{\pi^2k^2-4\pi^2}$$ $$=4\pi\sum_{k=1}^{\infty}\frac{1}{\pi^2k^2-4\pi^2}\Bigl(\frac{1}{2\pi}-\frac{1}{\pi k}\Bigr)\frac{\pi}{2}=2\pi^2\sum_{k=1}^{\infty}\frac{1}{2\pi^2k(2\pi+\pi k)}=\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{1}{k(k+2)}$$ $$I_1=\frac{1}{2\pi}\sum_{k=1}^{\infty}\biggl(\frac{1}{k}-\frac{1}{k+2}\biggr)=\frac{3}{4\pi}$$ $$I_2=-2i\sum_{k=1}^{\infty}\int_{0}^{\infty}\frac{1}{(2\pi)^2+x^2}\,\frac{x}{\pi^2k^2+x^2}dx$$ For $k\neq2$ $$I_2=-2i\sum_{k\neq2}^{\infty}\lim_{N\to\infty}\int_{0}^{N}\Bigl(\frac{1}{(2\pi)^2+x^2}-\frac{1}{\pi^2k^2+x^2}\Bigr)\frac{xdx}{\pi^2k^2-4\pi^2}$$ $$=-2i\sum_{k\neq2}^{\infty}\lim_{N\to\infty}\int_{0}^{N}\Bigl(\frac{d(x^2)}{(2\pi)^2(1+\frac{x^2}{(2\pi)^2})}-\frac{d(x^2)}{(\pi k)^2(1+\frac{x^2}{(\pi k)^2})}\Bigr)\frac{1}{\pi^2k^2-4\pi^2}$$ $$=-\frac{4i}{\pi^2}\sum_{k\neq2}^{\infty}\frac{\log\bigl(\frac{k}{2}\bigr)}{k^2-4}$$ For $k=2$ $$I_2=-2i\int_{0}^{\infty}\frac{xdx}{((2\pi)^2+x^2)^2}=-i\int_{0}^{\infty}\frac{dt}{((2\pi)^2+t)^2}=-\frac{i}{4\pi^2}$$

$$I=I_1+I_2=\frac{3}{4\pi}-\frac{4i}{3\pi^2}\log2-\frac{i}{4\pi^2}-\frac{4i}{\pi^2}\sum_{k>2}^{\infty}\frac{\log\bigl(\frac{k}{2}\bigr)}{k^2-4}$$

Svyatoslav
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  • I should have mentioned that I am only interested in the imaginary component of the second integral (which is where the first integral comes from after apply integration by parts and getting rid of the real part). The series you have derived is closely related to the series that I am currently stuck on: https://math.stackexchange.com/questions/4123446/how-to-evaluate-sum-k-3-infty-frac-ln-kk2-4 but this is great help! – KStarGamer May 04 '21 at 00:44
  • @KStarGamer Thank you. I agree - it also seems to me that it should be related to the derivatives of $\zeta$-function. But I cannot find out how... – Svyatoslav May 04 '21 at 00:49
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    @KStarGamer The sum $\sum_{k>2}^{\infty}\frac{\log\bigl(\frac{k}{2}\bigr)}{k^2-4}$ accurate to some constants can be expressed in terms of so called the poly-Stieltjes constants:$\sum_{k>2}^{\infty}\frac{\log\bigl(\frac{k}{2}\bigr)}{k^2-4}\sim\frac{1}{4}\sum_{k>0}^{\infty}\Bigl(\frac{\log(k+2)}{k}-\frac{\log(k)}{k+2}\Bigr)=$$\frac{1}{4}\Bigl(\gamma(2;0)-\gamma(0;2)\Bigr)$, where $\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 !N}2\right)$. For more details - in this post (section The poly-Stieltjes constants) – Svyatoslav May 04 '21 at 17:18
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    @KStarGamer Link https://math.stackexchange.com/questions/364452/evaluate-int-0-frac-pi2-frac11x21-tan-x-mathrm-dx/1381569#1381569 – Svyatoslav May 04 '21 at 17:21
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For large $x$: $$\frac{x\coth(x)-1}{x^2(2\pi+ix)}\approx\frac{x}{x^3}=\frac{1}{x^2}$$ and for small $x$: $$\frac{x\coth(x)-1}{x^2(2\pi+ix)}\approx\frac{1}{2\pi(e+e^{-1})}$$ So by comparison the integral converges. However we cannot split the top of this integral up as the two diverge for $x\to0$ unless u take a limit for the lower bound

Henry Lee
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