# Another residue integral

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?

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 $$u=x^{1/4}$$?

• 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! Aug 27 '19 at 22:28