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I would like to calculate $$\int_{0}^{\infty} \frac{x^3 \sin(x)}{1+x^4} dx $$ by means of the Residue Theorem. I would like to do this by calculating $$\int_{\alpha} \frac{z^3 e^{iz} }{1+z^4} , $$ (in which the image of $\alpha$ describes a half circle in $\mathbb{C}$ and joints the endpoints, such that the two singularities $z_1=\sqrt{i}$ and $z_2=\sqrt{-i}$ are within this half-circle) and considering the imaginary part of $$2 \pi i \sum_{j=1}^{2} Res(f;z_j).$$ So now I have to calculate the residues. Say I want to calculate the residue of $z_1$. We have $$Res(f;z_1) = \lim_{z \to \sqrt{i}} (z-\sqrt{i}) \frac{e^{iz} z^3}{(z+\sqrt{i})(z-\sqrt{i})(z+\sqrt{-i})(z-\sqrt{-i}) } . $$ After some algebraic manipulations this yields (I think): $$Res(f;z_1) = \frac{e^{i \sqrt{i}}}{4} .$$ My calculations of the other residue yields an even nastier expression. Do you know how I can simplify such expressions?

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If $\sqrt{i}$ is what you need note that $i=e^{i{\tfrac{\pi}{2}}}$ and therefore $\sqrt{i}$ can be taken to be $e^{i{\tfrac{\pi}{4}}}$ namely $\frac{\sqrt{2}}{2}+ i \frac{\sqrt{2}}{2}$. Multiplying this by $i$ you get the answer. You should be careful with your calculation though because $i$ has another square root, as well.

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  • $\begingroup$ Ah should've known that, thanks though! $\endgroup$ – Max Muller Jun 18 '13 at 14:11

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