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I need a hint to evaluate the integrals:

$$ \int _{0}^{p}\tanh(\pi k)k\,dk\quad \text{and}\quad \int _{0}^{s-1/2}\tan(\pi k)k\,dk $$ Here $p$ and $s$ are real numbers.

I know I can evaluate them by power series but how else can I evaluate these two integrals ?

I know I can expand $ \tanh(x) $ and $ \tan(x) $ into a power series but I would like to get some 'closed' result in terms of $ \Gamma (x) $ function or similar.

Michael Hardy
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Jose Garcia
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    Result for the first integral returned by Mathematica 8 $$\text{ConditionalExpression}\left[\frac{\pi p \left(\pi p+2 \log \left(e^{-2 \pi p}+1\right)\right)-\text{Li}_2\left(-e^{-2 p \pi }\right)}{2 \pi ^2}-\frac{1}{24},e^{-2 \pi p}\geq -1\right]$$ – Norbert Jun 28 '12 at 20:46
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    Result for the second integral returned by Mathematica 8 $$\frac{i \left(3 \text{Li}_2\left(e^{2 i \pi s}\right)+\pi \left(\pi \left(3 s^2-3 s+1\right)+3 i (2 s-1) \log \left(1-e^{2 i \pi s}\right)\right)\right)}{6 \pi ^2}$$ – Norbert Jun 28 '12 at 20:47
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    aha thanks.. :) – Jose Garcia Jun 28 '12 at 20:50
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    That is enough for you, or you need a solution ? – Norbert Jun 28 '12 at 20:51
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    I get this for integral one: Not sure if equivalent to Norbet answers. $\int_0^p ! ktanh(\pi k)=-(1/24)(-\pi^2+12p^2\pi^2-24p\ln(1+exp(2p\pi))\pi-12polylog(2,-exp(2p\pi)))/(\pi^2)$ – night owl Jun 28 '12 at 22:28
  • @nightowl by hand? – draks ... Jun 28 '12 at 22:30
  • @nightowl 7 edits, not bad ;-) – draks ... Jun 28 '12 at 22:36
  • @draks: Of course not :D, using a CAS. Is it possible to get polylog function by hand or recognize it from doing intergrals? I just thought this was a CAS algorithmic way of deducing the problem into a more complex answer using special functions, because an elementary table look up failed. So it resorted to special solution techniques. Although $\tan k$ is considered a simple function but not sure about hyperbolic tangent :) – night owl Jun 28 '12 at 22:40
  • Damn... $ $ $ $ $ $ – draks ... Jun 28 '12 at 22:49

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