Evaluating $\int_0^\infty \frac{dx}{1+x^4}$. Can anyone give me a hint to evaluate this integral?
$$\int_0^\infty \frac{dx}{1+x^4}$$
I know it will involve the gamma function, but how?

Following is a computation that uses Gamma function:
For any real number $k > 1$, let $I_k$ be the integral:
$$I_k = \int_0^\infty \frac{dx}{1+x^k}$$
Consider two steps in changing the variable. First by $y = x^k$ and then by $z = \frac{y}{1+y}$. Notice:
$$\frac{1}{1+y} = 1 - z,\quad y = \frac{z}{1-z}\quad\text{ and }\quad dy = \frac{dz}{(1-z)^2}$$
We get:
$$\begin{align}
I_k = & \int_0^{\infty}\frac{1}{1 + y} d y^{\frac{1}{k}} = \frac{1}{k}\int_0^\infty \frac{1}{1+y}y^{\frac{1}{k}-1} dy\\
= & \frac{1}{k}\int_0^\infty (1-z) \left(\frac{z}{1-z}\right)^{\frac{1}{k}-1} \frac{dz}{(1-z)^2}
= \frac{1}{k}\int_0^\infty z^{\frac{1}{k}-1} (1-z)^{-\frac{1}{k}} dz\\
= & \frac{1}{k} \frac{\Gamma(\frac{1}{k})\Gamma(1 - \frac{1}{k})}{\Gamma(1)}
= \frac{\pi}{k \sin\frac{\pi}{k}}
\end{align}$$
For $k = 4$, we get:
$$I_4 = \int_0^\infty \frac{dx}{1+x^4} = \frac{\pi}{4\sin \frac{\pi}{4}} = \frac{\pi}{2\sqrt{2}}$$
