I am looking for a closed-form of this integral

\begin{equation} \frac{1}{2}\int_0^\infty\left[\frac{x^2\cos x}{\cosh 2x-\cos x}-\frac{2x^2}{e^{4x}-2e^{2x}\cos x+1}\right]\,dx \end{equation}

I can rewrite the integral into \begin{equation} \int_0^\infty\frac{x^2(e^{2x}\cos x-1)}{e^{4x}-2e^{2x}\cos x+1}\,dx \end{equation} But I am stuck for the next step. I have a strong feeling the integral involving gamma or beta function but I am unable to prove it. Could anyone here please help me? Any help would be greatly appreciated. Thank you.

  • $\begingroup$ Ignoring the 1/2, Mathematica gives the indefinite integral as: 1/375 (-250 x^3 + (150 + 75 I) x^2 Log[ 1 - E^((2 - I) x)] + (150 - 75 I) x^2 Log[ 1 - E^((2 + I) x)] + (90 + 120 I) x PolyLog[2, E^((2 - I) x)] + (90 - 120 I) x PolyLog[2, E^((2 + I) x)] - (12 + 66 I) PolyLog[3, E^((2 - I) x)] - (12 - 66 I) PolyLog[3, E^((2 + I) x)]) $\endgroup$ – almagest Aug 30 '14 at 19:43
  • $\begingroup$ @almagest Sorry, I've made a typo. There is no factor of 1/2. What is E? Exponential? Have you included the integral limit? $\endgroup$ – Anastasiya-Romanova 秀 Aug 30 '14 at 19:47
  • $\begingroup$ $I=\bigg(\dfrac25\bigg)^3\cdot\zeta(3)$. $\endgroup$ – Lucian Aug 30 '14 at 20:03
  • $\begingroup$ Yes. E^x is the exponential. Hmmm. I have tried putting in limits, but Mathematica comes back with "Catastrophic loss of precision". I tried upping the Precision, but something seems to go wrong when you push the upper limit above about 120. Maybe a little more thought is needed! :) $\endgroup$ – almagest Aug 30 '14 at 20:07

Edit: The previous answer was wrong because it began with the incorrect series expansion.

The correct trigonometric series is the following:




Subtracting the two gives,


With $p=e^{2x}$,

$$\begin{align} \int_{0}^{\infty}\mathrm{d}x\,\frac{x^2(e^{2x}\cos{x}-1)}{e^{4x}-2e^{2x}\cos{x}+1} &=\int_{0}^{\infty}\mathrm{d}x\,\sum_{k=1}^{\infty}x^2e^{-2kx}\cos{kx}\\ &=\sum_{k=1}^{\infty}\int_{0}^{\infty}\mathrm{d}x\,x^2e^{-2kx}\cos{kx}\\ &=\sum_{k=1}^{\infty}\left[\frac{\partial^2}{\partial k^2}\int_{0}^{\infty}\mathrm{d}x\,\frac14e^{-2kx}\cos{nx}\right]_{n=k}\\ &=\sum_{k=1}^{\infty}\left[\frac{\partial^2}{\partial k^2}\frac14\cdot\frac{2k}{4k^2+n^2}\right]_{n=k}\\ &=\sum_{k=1}^{\infty}\left[\frac{4k(4k^2-3n^2)}{(4k^2+n^2)^3}\right]_{n=k}\\ &=\sum_{k=1}^{\infty}\frac{4}{125k^3}\\ &=\frac{4}{125}\zeta{(3)}. \end{align}$$

Thanks to Tunk-Fey for pointing out the erroneous value I had beforehand.

  • $\begingroup$ I know this series. Prof. @OmranKouba gives it here math.stackexchange.com/a/816253/133248. I thought $a$ and $b$ are numbers, not function. $\endgroup$ – Anastasiya-Romanova 秀 Aug 30 '14 at 20:18
  • $\begingroup$ Thank you David! You teach me useful techniques. (>‿◠)✌ $\endgroup$ – Anastasiya-Romanova 秀 Aug 30 '14 at 20:31
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    $\begingroup$ David, perhaps I am missing something here, but I think the correct answer should be $\frac4{125}\zeta(3)$. $\endgroup$ – Tunk-Fey Aug 30 '14 at 21:03
  • $\begingroup$ @Tunk-Fey It's my error. I misquoted the trigonometric series. I accidentally used the $p<1$ sum, which is slightly different. $\endgroup$ – David H Aug 30 '14 at 21:22
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    $\begingroup$ Another way to evaluate the integral is $$\Re\left[\int_0^\infty x^2 e^{-(2k+i)x}\right]$$It looks like @V-Moy's feeling correct. +1 for your answer David. $\endgroup$ – Tunk-Fey Aug 30 '14 at 21:55

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