# evaluate $\int_0^\infty \dfrac{dx}{1+x^4}$ using $\int_0^\infty \dfrac{u^{p-1}}{1+u} du$

evaluate $\int_0^\infty \dfrac{dx}{1+x^4}$using $\int_0^\infty \dfrac{u^{p-1}}{1+u} du = \dfrac{\pi}{\sin( \pi p)}$. I am having trouble finding what is $p$. I set $u = x^4$, I figure $du = 4x^3 dx$, I am unsure though how to find $p$ though. Could someone tell me what I am missing? Thanks.

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You may want to write $x = u^{1/4}$. – Sangchul Lee Feb 12 '13 at 7:51
Are you sure you have to do it this way? If you can evaluate the second integral by residues, surely you can evaluate the first by residues. – Potato Feb 12 '13 at 7:51
@Potato The point of the problem is to do it using this, for $0 < p < 1$ – user1535776 Feb 12 '13 at 7:57

If $u=x^4$ and $du=4x^3 dx$ then: $$dx = \frac{1}{4x^3} du = \frac{1}{4{(x^4)}^{3/4}} du = \frac{1}{4{u}^{3/4}} du$$ So your integral is:
$$\int_0^\infty \dfrac{dx}{1+x^4} = \frac{1}{4}\int_0^\infty \frac{u^{-3/4}du}{1+u} = \frac{\pi}{4\sin(\pi/4)}=\frac{\pi }{2 \sqrt{2}}$$