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Does anyone know a proof for the following formula ? $$\int_{0}^{\infty} \frac {1}{x^y+1} dx=\frac{\frac{\pi}{y}}{\sin(\frac{\pi}{y})}$$ for $y>1$?

If $y$ is an even positive integer than the integral can be calculated using the residue theorem. But even the case, that $y$ is an odd positive integer makes it more difficult because the integral cannot be taken from $-\infty$ to $\infty$. For arbitrary $y$, it should be even more difficult.

Jonathan Y.
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Peter
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  • Have you tried a keyhole contour? – Jonathan Y. Nov 03 '13 at 10:41
  • @JonathanY.: That keyhole contour is only useful when $y \in \mathbb{Z}$. Not useful otherwise. – Ron Gordon Nov 03 '13 at 10:47
  • I have heard from it, but I have not worked with it, seems somewhat complicated. – Peter Nov 03 '13 at 10:48
  • I calculated that the residue for some pole u is -u/y. So, the problems of choosing a suitable contour and the calculation of the roots remain. – Peter Nov 03 '13 at 10:51
  • @RonGordon yes, it seems you're right; thanks. – Jonathan Y. Nov 03 '13 at 10:51
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    You can also check this answer I once wrote : http://math.stackexchange.com/questions/432371/prove-int-0-infty-frac-mathbbdx1xn-frac-pin-sin-frac-pin/468321#468321 – Philippe Malot Nov 03 '13 at 11:34
  • You implied familiarity with the residue theorem in your problem statement, so I used it in mine. I do not wish to engage in a debate over which technique is preferable. That said, derivations of integration results involving the residue theorem tend to be short and simple compared with techniques that do not leave the real line. I think that this case is a good illustration. – Ron Gordon Nov 04 '13 at 07:56

1 Answers1

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Consider the following contour integral:

$$\oint_C \frac{dz}{z^y+1}$$

where $C$ is a wedge which has four sections: 1) along the positive real axis from $[\epsilon,R]$; 2) along an arc of radius $R$ from the positive real axis, counterclockwise to the point $z=R e^{i 2 \pi/y}$; 3) along the line segment $z=e^{i 2 \pi/y} t$, $t \in [R,\epsilon]$; 4) a small arc of radius $\epsilon$ about the origin of angle $2 \pi/y$. Thus the contour integral is equal to

$$\int_{\epsilon}^R \frac{dx}{x^y+1} + i R \int_0^{2 \pi/y} d\theta \, \frac{e^{i \theta}}{R^y e^{i y \theta}+1} + e^{i 2 \pi/y} \int_R^{\epsilon} \frac{dt}{t^y+1} + i \epsilon \int_{2 \pi/y}^0 d\phi \, \frac{ e^{i \phi}}{\epsilon^y e^{i y \phi}+1}$$

As $R \to \infty$, the second integral vanishes because $y \gt 1$; as $\epsilon \to 0$, the fourth integral vanishes. By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the residue at $z=e^{i \pi/y}$. Thus,

$$\left (1-e^{i 2 \pi/y}\right ) \int_0^{\infty} \frac{dx}{x^y+1} = \frac{i 2 \pi}{y \,e^{i \pi (y-1)/y}} = -\frac{i 2 \pi}{y} e^{i \pi/y}$$

Thus,

$$\int_0^{\infty} \frac{dx}{x^y+1} = -\frac{i 2 \pi}{y} \frac{e^{i \pi/y}}{1-e^{i 2 \pi/y}} = \frac{\pi/y}{\sin{(\pi/y)}}$$

ADDENDUM

I added the small arc over which the integral vanishes so as to avoid a branch point at $z=0$ for nonintegral $y$. Although it has no impact on the result, the derivation was not quite correct without it. It should be a habit to define integration contours so as to avoid branch points if Cauchy's theorem is to be invoked.

Ron Gordon
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