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How do I evaluate the following integral?

$$I=\int_{0}^{\infty}\frac{\sinh(a)k\ dk}{\cosh(k) + \cosh(a)}, \qquad a \geq0$$

Ali Caglayan
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chetan joshi
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  • Please share us your thoughts, what did you try, and where you are stuck, so that we will be able to help you . – Nizar Oct 27 '15 at 05:59
  • Being an experimental physicist it is very difficult to solve such complicated integral, so honestly i did not tried. It appears as a part of my research work, so unless i found its answer i can not proceed further. Hence i seek help of mathematician. – chetan joshi Oct 27 '15 at 06:52
  • @op how about a numerical approach? Just as good with the exception its more expensive in terms of computation. I have not checked your integral but it should converge nicely so you can truncate the upper limit. – Chinny84 Oct 27 '15 at 07:31
  • $I=\dfrac{\pi^2}6+\dfrac{a^2}2+2~\text{Li}_2\Big(-e^{-a}\Big).\quad$ See dilogarithm and polylogarithm for more information. – Lucian Oct 27 '15 at 10:31
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    This question is being voted on to be closed. Please add some thoughts or context to make this question acceptable. It's a shame as it is an interesting question. – Ali Caglayan Oct 27 '15 at 11:37
  • OP do you not have any CAS methods of solving this yourself? – Ali Caglayan Oct 27 '15 at 11:39
  • To all who intend to vote to close: we have many people posting difficult integrals/sums on M.SE. These are clearly not homework and asking for "effort" may be fruitless. The statement of the problem itself in this case is enough to make for a good question. Thus, please do not close. – Ron Gordon Oct 28 '15 at 15:07

1 Answers1

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Unfortunately, I do not know how to solve such an integral. I do however have access to Mathematica which does. I am including this as an answer so you at least have something to continue your research work off, unless you require the method.

$$ \begin{aligned} I &= \int_0^\infty\frac{\sinh(a)k}{\cosh(k)+cosh(a)}\,\mathrm{d}k \\ &= \operatorname{Li}_2(-e^{-a})-\operatorname{Li}_2\left(-e^a\right),\quad e^a\geq 0\land e^{-a}\geq-1 \end{aligned} $$

Applying the identity $\operatorname{Li}_2(z)+\operatorname{Li}_2\left(\frac1z\right)=-\frac{\pi^2}6-\frac12\ln^2(-z)$ simplifies the answer to:

$$ \frac{\pi^2}6+\frac{a^2}2+2\operatorname{Li}_2(-e^{-a}) $$

(Which I realised after looking at Lucian's comment)

T-Fowl
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  • this is not too difficult to show, a substitution $e^k=x$ will bring this very close to a point where it is obvious that result should look like this :) – tired Oct 27 '15 at 13:32