# Evaluation of these integrals

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.

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Result for the first integral returned by Mathematica 8 $$\text{ConditionalExpression}\left[\frac{\pi p \left(\pi p+2 \log \left(e^{-2 \pi p}+1\right)\right)-\text{Li}_2\left(-e^{-2 p \pi }\right)}{2 \pi ^2}-\frac{1}{24},e^{-2 \pi p}\geq -1\right]$$ –  userNaN Jun 28 '12 at 20:46
Result for the second integral returned by Mathematica 8 $$\frac{i \left(3 \text{Li}_2\left(e^{2 i \pi s}\right)+\pi \left(\pi \left(3 s^2-3 s+1\right)+3 i (2 s-1) \log \left(1-e^{2 i \pi s}\right)\right)\right)}{6 \pi ^2}$$ –  userNaN Jun 28 '12 at 20:47
aha thanks.. :) –  Jose Garcia Jun 28 '12 at 20:50
That is enough for you, or you need a solution ? –  userNaN Jun 28 '12 at 20:51
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