Residue Method for Integrating Complex Functions

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Homework Statement


\int\limits^{\infty}_{-\infty}\frac{\mbox{d}x}{x^2+1}
and i must count this using residues

The Attempt at a Solution


I=2\pi i\left(\mbox{res}_i\frac{1}{z^2+1}\right)=2\pi i\left(\lim_{z\to i}(z-i)\frac{1}{(z-i)(z+i)}\right)=2\pi i\left(-\frac12i}\right)=\pi
correct?
 
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i don't know why edit doesn't work, but now i came to that these points must be for \Im(z)\ge0 and result will be \pi now, is this solution correct now?
there should have been \Im(z)\ge0
 
Yes, the answer \pi is correct. And it seems that your method is also good!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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