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I am not sure if this is already asked on this site but I can't seem to find it. The integral is $$ I=\int_0^\infty \frac{x^{-3/4}}{1+x}dx. $$ My idea is to define $z^{3/4}$ using the branch cut of the positive real axis and to use the key-hole contour consisting of a large circle $C_R$ (of radius $R$ centered at origin), a small circle $C_\epsilon$ (of radius $\epsilon$), and a line right above and right below the branch cut. Then if I am correct, the contribution of the line right above the branch cut converges to $I$, the line right below the branch cut converges to $iI$ and the large circle is $2\pi R\;O(R^{-7/4})=O(R^{-3/4})\to 0$. This leaves me with the contribution from $C_{\epsilon}$ which I am not sure how to control. The last piece of information is the residue of $f$ at $z=-1$ which is $e^{3\pi i/4}$ by our branch cut. Is this the right track or should I be trying something else entirely?

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Instead of dealing with branch cuts, first try to manipulate the integral in the real domain to make it nicer. What happens if you try $u=x^{1/4}$?

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  • $\begingroup$ Oh yes that is much easier...then I can just use a more "standard" contour of a half circle and the residues at $e^{i\pi/4}$ and $e^{3i\pi/4}$. Thanks! $\endgroup$ Aug 27 '19 at 22:28

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