I am reading a paper (sorry, no e-copy) with a number of infinite series, in which each term of the series is an integral of a complicated transcendental function like the one in the title.

There are no fewer than a dozen of these infinite sums of integrals in the paper.

For instance, the authors define $$ I_1(\eta) = \sum_{n=2,3,4}^{\infty} \dfrac{n}{n^2-1} \int_{0}^{\eta} \cos nt \log \left( \dfrac{\cos(t/2) + \sqrt{\cos^2(t/2) - \cos^2(\eta/2)}}{\cos(\eta/2)}\right) dt $$ where $0 \leq \eta < \pi$. The authors then assert that $$ I_1(\eta) = \frac{\pi}{8} \left( (2\cos\eta -2) \log \sin \frac{\eta}{2} + \frac{1}{2}\cos\eta -\frac{1}{2}\right) $$

without any proof or derivation (the results are simply listed in an Appendix).

I confess I am stumped as to how the authors arrived at their results. Any advice or help would be greatly appreciated.

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    $\begingroup$ Imho, any paper stating results like this horror without proof and without any reference is not worth reading. Read the following for a good time...and some rather deep learning: jmilne.org/math/tips.html $\endgroup$ – Timbuc Dec 20 '14 at 17:12
  • $\begingroup$ @Timbuc Thanks, that link is amusing. Seeing these results asserted without any proof made me think I was missing something obvious despite racking my brains for a while. $\endgroup$ – user_of_math Dec 20 '14 at 17:53
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    $\begingroup$ For this integral I know the most marvelous proof, which this comment section is too narrow to contain$\ldots$ $\endgroup$ – Lucian Dec 21 '14 at 0:29
  • $\begingroup$ It might have been proven in another paper? Or perhaps some kind of computational engine evaluated it? $\endgroup$ – Alfred Yerger Dec 21 '14 at 3:30

$\newcommand{\s}{\sigma}$$\newcommand{\e}{\eta}$$\newcommand{\al}[1]{\begin{align}#1\end{align} }$As suggested by Norbert Schuch in the comments, it suffices to check that the derivatives with respect to $\eta$ agree. This simplifies our task somewhat by getting rid of one of the logarithms in the integrand.

Starting from the expression derived by JJacquelin, we have for the case $0 \leq \e < \pi/2$: $$ \al{ I_1'(\e) &= -\frac{\tan \frac \e 2}{8} \int_0^\e d\e \frac{ \cos \frac{t}{2} \left[2+ \cos t+2 \cos t \ln (2-2\cos t)\right]}{ \sqrt{\cos^2 \frac t 2 - \cos^2 \frac \e 2}} \\&= -\frac{\tan \frac \e 2}{8} \int_0^\e d\e\frac{ \cos \frac{t}{2} \left[2 + (1 - 2 \sin^2 \frac t 2)+2 (1 - 2 \sin^2 \frac t 2) \ln (4 \sin^2 \frac t 2)\right]}{ \sqrt{\sin^2 \frac \e 2 - \sin^2 \frac t 2}} \\&= -\frac{\tan \frac \e 2}{8} \int_0^1 dx\frac{ 2\left[3 - 2 x^2 \sin^2 \frac \e 2+2 (1 - 2 x^2 \sin^2 \frac \e 2) \ln (4 x^2 \sin^2 \frac \e 2)\right]}{ \sqrt{1 - x^2}} \\&\equiv -\frac{\tan \frac \e 2}{4} I } $$ Here I put $\sin \frac t 2 = x \sin \frac \e 2 $, and there is no boundary term because the integrand vanishes at $t = \e$.

Now we abbreviate $\s = \sin \frac \e 2$ for clarity. We have $$ I = A+B+C+D+E+F, $$

with $$ \al{ &A=\int_0^1 dx \frac{3}{\sqrt{1-x^2}}=\frac{3\pi}{2} \\&B=\int_0^1 dx \frac{-2 \s^2 x^2}{\sqrt{1-x^2}}= -\frac \pi 2 \s^2 \\&C=\int_0^1 dx \frac{2 \ln (4\s^2)}{\sqrt{1-x^2}}= \pi \ln (4\s^2) \\&D=\int_0^1 dx \frac{4 \ln x}{\sqrt{1-x^2}}= 4 \int_0^{\pi/2} dt \ln \sin t = -\pi \ln 4 \\&E=\int_0^1 dx \frac{-4x^2 \s^2 \ln(4\s^2)}{\sqrt{1-x^2}}= -\pi \s^2 \ln (4\s^2) \\&F=\int_0^1 dx \frac{-4x^2 \s^2 (2 \ln x)}{\sqrt{1-x^2}}= -8 \s^2 \int_0^{\pi/2} dt \sin^2 t\ln \sin t = \pi \s^2 (\ln 4-1). } $$ Therefore,

$$ I = \pi(1-\s^2) \left[ \frac{3}{2} + 2 \ln \s \right] = \pi \cos^2 \frac \e 2 \left[ \frac{3}{2} + 2 \ln \sin \frac \e 2\right]. $$ This gives $$ I_1'(\e) = -\frac \pi 4 \sin \frac \e 2 \cos \frac \e 2 \left[ \frac{3}{2} + 2 \ln \sin \frac \e 2\right] = -\frac \pi 8 \sin \e \left[ \frac{3}{2} + 2 \ln \sin \frac \e 2\right]. $$ This agrees with the derivative of the desired result.

The case $\pi/2 \leq \eta \leq \pi$ can be handled similarly; the only thing that changes is that now we cannot assume that $\cos^2 \frac t 2 > \cos^2 \frac \e 2$ everywhere.

  • $\begingroup$ I don't know why the bounty was given to me, since my answer was not complete. It should be given to user111187. How is it possible to transfer the bounty to him ? $\endgroup$ – JJacquelin Dec 30 '14 at 16:58
  • $\begingroup$ @JJacquelin Most likely the bounty was not given to me because I posted my answer after the deadline. If you want to transfer the bounty to me, you could put a 50 point bounty on a question where I gave the accepted answer and then award me the bounty. $\endgroup$ – user111187 Dec 30 '14 at 17:05
  • $\begingroup$ @user111187 : Well, I understand that. But I don't want to post a throwawaway question only to transfer a bounty. It is the job of the managers of the forum. If it is not possible, too bad. $\endgroup$ – JJacquelin Dec 30 '14 at 17:16

Numerical tests show a very good agreement between the two forms of $I_1(\eta)$ So, I am conviced that the assertion of the authors is correct.

By the way, it is possible to get ride of the discrete sum, thanks to : $$ \sum_{n=2}^{\infty} \dfrac{n}{n^2-1}\cos(nt)=-\frac{1}{4}\left(2+\cos(t)+2\cos(t)\ln \left( 2-2\cos(t)\right) \right) $$ leading to : $$ I_1(\eta) = -\int_{0}^{\eta}\frac{2+\cos(t)+2\cos(t)\ln \left( 2-2\cos(t)\right)}{4} \ln \left( \dfrac{\cos(\frac{t}{2}) + \sqrt{\cos^2(\frac{t}{2}) - \cos^2(\frac{\eta}{2})}}{\cos(\frac{\eta}{2})}\right) dt $$ Not sure that it will be useful.

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    $\begingroup$ Once the sum is removed, one should be able to check the validty by taking the derivative of the claimed result. $\endgroup$ – Norbert Schuch Dec 29 '14 at 15:50
  • $\begingroup$ That is progress, surely - trading in the infinite sum for a more complicated integrand might do the trick. The resulting integrand is still messy, though. $\endgroup$ – user_of_math Dec 29 '14 at 16:11

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