# Show that $\int_{\gamma} \frac{z^{p-1}}{z^2+1} d{z} = 2\pi i\cos\left(\frac{\pi p}{2}\right)e^{i\pi(p-1)}$

I am trying to show that

$$\int_{\gamma} \frac{z^{p-1}}{z^2+1} d{z} = 2\pi i\cos\left(\frac{\pi p}{2}\right)e^{i\pi(p-1)}\,\,,\,\gamma:=\{z\;\;;\;\;|z|=R\}$$

for $0 < p < 2$. This integral computes the path of the contour excluding the branch line along the positive real axis. So I'm really computing $\int_{\gamma}$ part in:

$$\int_{\gamma} = \int_{C_1} + \int_{C_2} + \int_{C_3} + \int_{C_4}$$

where $C_1, C_2, C_3$ and $C_4$ are all pieces of the contour.

I've computed the residues at $z=i$ and $z=-i$ but for some reason my calculations don't check out. I got $$\displaystyle\frac{e^{i\frac{\pi}{2}(p-1)}}{2i} - \frac{e^{-i\frac{\pi}{2}(p-1)}}{2i}.$$

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What's your curve of integral? – Yuchen Liu Apr 22 '12 at 5:17
$\gamma$ is a circle of radius $R$ – Low Scores Apr 22 '12 at 5:48
Well I solved my problem. I thought that this question would make sense to all of you. Funny how the thoughts in my mind don't translate well in basic English. – Low Scores Apr 22 '12 at 7:00
That's why they sell books on communication :-). What is your solution? – draks ... Apr 22 '12 at 7:51
Which radius $R$? – Did Apr 22 '12 at 9:50

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The only residues contained in $\gamma$ would be at $i$ and $-i$. \begin{align} \mathrm{Res}_{z=i}\left(\frac{z^{p-1}}{z^2+1}\right) &=\frac{e^{i\pi(p-1)/2}}{2i}\\ &=-e^{i\pi p}\frac{e^{-i\pi p/2}}{2}\tag{1} \end{align} and \begin{align} \mathrm{Res}_{z=-i}\left(\frac{z^{p-1}}{z^2+1}\right) &=\frac{e^{i3\pi(p-1)/2}}{-2i}\\ &=-e^{i\pi p}\frac{e^{i\pi p/2}}{2}\tag{2} \end{align} Thus, adding $(1)$ and $(2)$ we get $$\int_\gamma\frac{z^{p-1}}{z^2+1}\,\mathrm{d}z=-2\pi i\,e^{i\pi p}\cos\left(\frac{\pi p}{2}\right)\tag{3}$$ and $(3)$ should be valid for all $p\in\mathbb{R}$.