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I need to use residues to evaluate $\int_0^\infty \frac{1}{1+x^5} \, dx$. Since the intergral goes from $0 \to \infty$ and not $- \infty$ to $\infty$, am I allowed to integrate over a quarter-circle instead of a semicircle? So I mean: let $C_R$ be the quarter-circle extending from $0$ to $\frac{\pi}{2}$ of radius $R$. Then can we say

$$\int_0^\infty \frac{1}{1+x^5} \, dx = \int_{C_R} \frac{1}{1+z^5} \, dz + \int_0^R \frac{1}{1+x^5} \, dx \hspace{0.2cm} ?$$

And if we cannot do this over a quarter-circle, what can I do instead? Because $\frac{1}{1+x^5}$ is not an even function and therefore I cannot write use $\int_0^\infty f(x) \,dx = 2 \int_{-\infty}^\infty f(x) \, dx$. In general, how to integrate $\int_0^\infty f(x) \, dx$ when $f(x)$ is not even?

Grigor Hakobyan
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    You can make the change $t=x^5$ first and get $I=\frac{1}{5}\int_0^\infty\frac{t^{-4/5}}{1+t}dt$. Then you can use a keyhole contour and evaluate the residue at $t=e^{\pi i}$ or, without complex integration, make the change $x=\frac{1}{1+t}$ and get Beta-function: $I=\frac{1}{5}\int_0^1\big(\frac{1-x}{x}\big)^{-4/5}\frac{dx}{x}=\frac{1}{5}\int_0^1\big(1-x\big)^{-4/5}x^{-1/5}dx$ – Svyatoslav Apr 29 '22 at 04:14
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    Instead of quarter circle, try integrate over the boundary of the circular sector $0 \le \theta \le \frac{2\pi}{5}$. – achille hui Apr 29 '22 at 06:45

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