Evaluating $\int_0^\infty \frac{dx}{1+x^4}$. [duplicate]

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:

$$I_4 = \int_0^\infty \frac{dx}{1+x^4} = \frac{\pi}{4\sin \frac{\pi}{4}} = \frac{\pi}{2\sqrt{2}}$$

HINT:

Putting $x=\frac1y,dx=-\frac{dy}{y^2}$

$$I=\int_0^\infty\frac{dx}{1+x^4}=\int_\infty^0\frac{-dy}{y^2\left(1+\frac1{y^4}\right)}$$ $$=-\int_\infty^0\frac{y^2dy}{1+y^4}=\int_0^\infty\frac{y^2dy}{1+y^4} \text{ as } \int_a^bf(x)dx=-\int_b^af(x)dx$$

$$I=\int_0^\infty\frac{y^2dy}{1+y^4}=\int_0^\infty\frac{x^2dx}{1+x^4}$$

$$\implies 2I=\int_0^\infty\frac{dx}{1+x^4}+\int_0^\infty\frac{x^2dx}{1+x^4}=\int_0^\infty\frac{1+x^2}{1+x^4}dx=\int_0^\infty\frac{\frac1{x^2}+1}{\frac1{x^2}+x^2} dx$$

Now the idea is to express the denominator as a polynomial of $\int \left(\frac1{x^2}+1 \right)dx =x-\frac1x=u\text{(say)}$

$$\text{The denominator =}\frac1{x^2}+x^2=\left(x-\frac1x\right)^2+2=u^2+2$$

Now, complete the definite integral with $u$

$\lim_{|z|\rightarrow \infty}\frac{1}{|z^4+1|} = 0 $ and $\frac{1}{x^4+1}$ is even so you can replace the integral with $\frac{1}{2}2\pi i\sum Res$ considering only the residues in the top half of the complex plane.

Hint: $1 + x^4 = (1 + x^2)^2 - (\sqrt{2} x)^2$, so that $1 + x^4 = (1 + \sqrt{2}x + x^2)(1 - \sqrt{2}x + x^2)$. This allows you to use partial fractions in the integral.

If you know contour integration that will also work here.