How to integrate $\int_0^\infty \frac{1}{1+y^4} dy$ I tried the trigonometric substitution $y^2 = \tan \theta, sec^2\theta = 1 + y^4$
But now I'm stuck with $\frac12 \int \frac{\sqrt{\sin \theta}}{(\cos\theta)^{\frac92} } d \theta$
I ran out of imagination as what to try now
 A: $$\int_0^\infty \frac{dx}{x^4+1} = \frac{1}{2}\int_{-\infty}^\infty \frac{dx}{x^4+1}$$
The latter integral is trivial by means of contour integration.
Let $C$ be the canonical semicircle contour, along the real axis from $-R$ to $-R$ and around the semicircle $Re^{i\theta}$ for $\theta \in [0,\pi]$.  Letting $R \to \infty$, we consider the function on the arc ($|z| = R$):
$$\left|\frac{1}{x^4+1}\right|\le  \frac{1}{|x^4+1|} \le \frac{1}{|x^4|-1} \le \frac{2}{R^4} \to 0$$
so by the estimation lemma, the integral around the arc disappears.
$$\oint_C \frac{dz}{z^4+1} = \int_{-\infty}^\infty \frac{dx}{x^4+1} =2 \pi i \sum \operatorname*{Res}\frac{1}{z^4+1}$$
where the residues are of poles in the upper half plane.  These poles are $z_1=e^{i\pi/4}$ and $z_2=e^{3i\pi/4}$.  It follows that
$$b_1=\operatorname*{Res}_{z=z_1}\frac{1}{z^4+1} = -\frac{1}{4} e^{i\pi/4}$$
$$b_2=\operatorname*{Res}_{z=z_1}\frac{1}{z^4+1} = -\frac{1}{4} e^{3 i\pi/4}$$
Then
$$\int_{-\infty}^\infty \frac{dx}{x^4+1} =2 \pi i (b_1+b_2) = \frac{\pi}{\sqrt{2}}$$
and finally
$$\int_0^\infty \frac{dx}{x^4+1} =  \frac{\pi}{2\sqrt{2}}$$
A: This may be done using residue theory.  Consider
$$\oint_C \frac{dz}{1+z^4}$$
where $C$ is a closed contour that spans the perimeter of the quarter-circle in the 1st quadrant (i.e., the positive real and positive imaginary quarter plane), of radius $R$.  As $R \to \infty$, the integral over the circular arc vanishes, and we are left with this contour integral being equal to
$$(1-i) \int_0^{\infty} \frac{dx}{1+x^4}$$
This integral is equal to $ i 2 \pi$ times the residue at the pole $z=e^{i \pi/4}$.  Thus
$$(1-i) \int_0^{\infty} \frac{dx}{1+x^4} = \frac{i 2 \pi}{4 e^{i 3 \pi/4}}$$
which means that
$$\int_0^{\infty} \frac{dx}{1+x^4} =\frac{\pi}{2 \sqrt{2}}$$
A: make a change of variable $ x=y^{1/4} $ and use the identity
$$ \int_{0}^{\infty}\frac{t^{s-1}}{x+1}dt= \frac{\pi}{sin(\pi s)}$$
A: Note that
$$\int_1^{\infty} \dfrac{dy}{1+y^4} = \int_0^1 \dfrac{y^2dy}{1+y^4}$$
Hence,
$$I=\int_0^{\infty} \dfrac{dy}{1+y^4} = \int_0^1 \dfrac{1+y^2}{1+y^4}dy$$
We have $y^4+1 = (y^2+1+y\sqrt2)(y^2+1-y\sqrt2)$. Hence,
$$1+y^2= \dfrac{(y^2+1+y\sqrt2) + (y^2+1-y\sqrt2)}2$$
Hence, we get that
\begin{align}
I & = \dfrac12\int_0^1\dfrac{dy}{1+y^2-y\sqrt2} + \dfrac12\int_0^1\dfrac{dy}{1+y^2+y\sqrt2}\\
& = \dfrac12\int_0^1\dfrac{dy}{\left(y-\dfrac1{\sqrt2}\right)^2+\left(\dfrac1{\sqrt2} \right)^2} + \dfrac12\int_0^1\dfrac{dy}{\left(y+\dfrac1{\sqrt2}\right)^2+\left(\dfrac1{\sqrt2} \right)^2}\\
& = \dfrac1{\sqrt2} \left(\arctan \left(y\sqrt2-1\right) + \arctan \left(y\sqrt2+1\right) \right)_{y=0}^1\\
& = \dfrac1{\sqrt2} \left(\arctan(\sqrt2 - 1)+\arctan(\sqrt2+1)\right) = \dfrac1{\sqrt2} \left(\arctan\left(\dfrac1{1+\sqrt2} \right)+\arctan(1+\sqrt2)\right)\\
& = \dfrac1{\sqrt2} \times \dfrac{\pi}2 = \dfrac{\pi}{2 \sqrt2}
\end{align}
