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I have been recently running into integrals of the form $$\int\frac{f(x)}{x^p \pm 1}dx$$ and $$\int \frac{f(x)}{e^x \pm 1} dx$$ where $f(x)$ is ususally $x^q$ or $\sin x$. For example, $f(x) = x^3$ occurs in context of Stefan's Law and Wien's Law.

I have no idea how to proceed in such cases. Wolfram is also no help. Help with non-trivial special cases or in case of definite integration from $0$ to $\infty$ will also be much appreciated.

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    $\begingroup$ Is $p$ usually an integer? And $q$? $\endgroup$ – Harald Hanche-Olsen Feb 1 '13 at 12:17
  • $\begingroup$ As I do not have a precise question, you may assume so if it helps in answering. $\endgroup$ – dexter04 Feb 1 '13 at 12:19
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    $\begingroup$ If f(x) has a anything to do with cosx, sinx, e^x etc, you can rewrite the whole integral as a function involving complex numbers. Then you can use Cauchy's Integral Theorem, Residue theorem etc. $\endgroup$ – Applied mathematician Feb 1 '13 at 12:26
  • $\begingroup$ @JoyeuseSaintValentin: An example will be much appreciated. $\endgroup$ – dexter04 Feb 1 '13 at 12:26
  • $\begingroup$ what do you know about complex analysis? $\endgroup$ – Applied mathematician Feb 1 '13 at 12:46
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The definite integral of the second type may be treated by Mellin transforms. Let $$ I(q) = \int_0^\infty \frac{x^q}{e^x-1} dx = \int_0^\infty x^q e^{-x} \sum_{k\ge 0} e^{-kx} dx = \sum_{k\ge 0} \int_0^\infty x^q e^{-(k+1)x} dx$$ with $q>1.$ The inner intgral is the Mellin transform of $e^{-(k+1)x}$ evaluated at $q+1$, so that $$ I(q) = \sum_{k\ge 0} \Gamma(q+1) \sum_{k\ge 0} \frac{1}{(k+1)^{q+1}} = \Gamma(q+1) \zeta(q+1).$$ Now suppose $q$ were an odd integer $q = 2m+1$ and recall that $$\zeta(2n) = (-1)^{n+1} \frac{B_{2n} (2\pi)^{2n}}{2 (2n)!}$$ with $B_{2n}$ the Bernoulli numbers. It follows that $$I(2m+1) = (2m+1)! (-1)^m \frac{B_{2m+2} (2\pi)^{2m+2}}{2 (2m+2)!} = \frac{(-1)^m B_{2m+2} }{2(2m+2)} (2\pi)^{2m+2}.$$

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  • $\begingroup$ +1,+1, but why two separate answers? $\endgroup$ – Pedro Tamaroff May 2 '13 at 1:25
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Applying the technique from the previous post to the definite integral of the second type we obtain Let $$ J(q) = \int_0^\infty \frac{x^q}{e^x+1} dx = \int_0^\infty x^q e^{-x} \sum_{k\ge 0} (-1)^{k+1} e^{-kx} dx = \sum_{k\ge 0} (-1)^{k+1} \int_0^\infty x^q e^{-(k+1)x} dx$$ with $q>1.$ Once again the inner integral is the Mellin transform of $e^{-(k+1)x}$ evaluated at $q+1$, so that $$ J(q) = \sum_{k\ge 0} \Gamma(q+1) \sum_{k\ge 0} \frac{(-1)^{k+1}}{(k+1)^{q+1}} = \Gamma(q+1) \zeta(q+1) \left(1 - \frac{1}{2^q}\right).$$ It follows that $$J(2m+1) = \left(1 - \frac{1}{2^{2m+1}}\right) \frac{(-1)^m B_{2m+2} }{2(2m+2)} (2\pi)^{2m+2}.$$

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