How to compute $\int_0^{+\infty} \frac{dt}{1+t^4} = \frac{\pi}{2\sqrt 2}.$ 
How to compute
$$\int_0^{+\infty} \frac{dt}{1+t^4} = \frac{\pi}{2\sqrt 2}.$$
  I'm interested in more ways of computing this integral.

There is always the straight forward method to decompose into simple elements 
$\dfrac{1}{(1 + t^{4})}$ it works but it is tedious. If someone has a faster and clever method, I'm interested :)
Update :


*

*My quesion is different than this question because all the solution that state there is talk about the the straight forward method to decompose into simple elements $\dfrac{1}{(1 + t^{4})}$ as i said before i'm not intersted in that way

 A: Here is another method using purely clever substitutions. Note that 
$$
\int_0^\infty \frac{\mathrm{d}x}{1+x^4}
=
\int_{0}^\infty \frac{w^2}{1+w^4} \mathrm{d}w
$$
Where the last integral comes from the substitution $w \mapsto 1/x$. Addition now gives that
$$
\begin{align*}
\int_0^\infty \frac{\mathrm{d}x}{1+x^4}
& = \frac{1}{2}\int_0^\infty \frac{1+x^2}{1+x^4}\mathrm{d}x \\
& = \frac{1}{2}\int_0^\infty \frac{1+1/x^2}{x^2+1/x^2} \mathrm{d}x \\
& = \frac{1}{2}\int_0^\infty \frac{(x-1/x)'}{(x-1/x)^2+2} \mathrm{d}x \\
& = \frac{1}{2}\int_{-\infty}^\infty \frac{\mathrm{d}u}{u^2+2} 
  = \frac{1}{2\sqrt{2}}\int_{-\infty}^\infty \frac{\mathrm{d}y}{1+y^2}
  = \frac{\pi}{2\sqrt{2}}
\end{align*}
$$
Where the substitutions $u \mapsto x - 1/x$ and $y \mapsto \sqrt{2} u$ were used.
A: Notice first that
$$
\int_0^\infty\frac{1}{1+t^4}=\frac12\int_{-\infty}^\infty\frac{1}{1+t^4}\,dt.
$$
Applying the Residue Theorem to the function $f: \mathbb{C}\setminus\{\pm e^{\pm i\frac\pi4}\} \to \mathbb{C}, z\mapsto \frac{1}{1+z^4}$, we have, for $R>1$:
$$
\int_{-R}^Rf(t)\,dt=\frac{2\pi i}{4}\left(-e^{i\frac\pi4}+e^{-i\frac\pi4}\right)-i\int_0^\pi\frac{Re^{i\theta}}{1+R^4e^{4i\theta}}\,d\theta.
$$
Since
$$
\lim_{R\to\infty}\left|\int_0^\pi\frac{Re^{i\theta}}{1+R^4e^{4i\theta}}\,d\theta\right|\le \lim_{R\to\infty}\int_0^\pi\frac{R}{R^4-1}=\lim_{R\to\infty}\frac{\pi R}{R^4-1}=0,
$$
it follows that
$$
\lim_{R\to \infty}\int_{-R}^Rf(t)\,dt=\frac{2\pi i}{4}\left(-e^{i\frac\pi4}+e^{-i\frac\pi4}\right)=\pi\sin\left(\frac\pi4\right)=\frac{\pi}{\sqrt{2}}.
$$
Hence
$$
\int_0^\infty\frac{1}{1+t^4}\,dt=\lim_{R\to\infty}\frac12\int_{-R}^R f(t)\,dt=\frac{\pi}{2\sqrt{2}}.
$$
A: $$\int_{0}^{+\infty}\frac{dt}{1+t^4} = \frac{1}{\sqrt{2}}\int_{0}^{+\infty}\frac{dt}{1+4t^4}\tag{1} $$
and:
$$ 1+4t^4 = (1+4t^2+4t^4)-(2t)^2 = (2t^2+2t+1)(2t^2-2t+1)\tag{2} $$
from which:
$$ \frac{1}{1+4t^4} = \frac{1-t}{2-4t+4t^2}+\frac{1+t}{2+4t+4t^2}\tag{3} $$
and:
$$ \int_{0}^{+\infty}\frac{dt}{1+4t^4} = 2\int_{0}^{1}\frac{1-t}{2-4t+4t^2}\,dt = \int_{0}^{1}\frac{2t}{2-4t+4t^2}\,dt \tag{4}$$
can be explicitly computed in terms of $\arctan(1-2t)$ and $\log(1-2t+2t^2)$.
Since $\int_{0}^{1}\frac{2t}{2-4t+4t^2}\,dt=\frac{\pi}{4}$, it follows that:
$$ \int_{-\infty}^{+\infty}\frac{dt}{1+t^4} = \color{red}{\frac{\pi}{\sqrt{2}}}.\tag{5}$$
A: One may recall that the celebrated $\Gamma$ function may be defined by 

$$
\Gamma(\alpha)=\int_0^\infty u^{\alpha-1} e^{-u}\:{\rm{d}}u, \quad \alpha>0,
$$

giving
$$
\begin{align}
\int_0^\infty \frac{1}{x^4+1} \:{\rm{d}}x&=\int_0^\infty\int_0^\infty e^{-(x^4+1)t} \:{\rm{d}}t\:{\rm{d}}x
\\&=\int_0^\infty e^{-t}\int_0^\infty e^{-x^4 t} \:{\rm{d}}t\:{\rm{d}}x
\\&=\int_0^\infty e^{-t}\left(\int_0^\infty e^{-x^4 t}\:{\rm{d}}x\right)\:{\rm{d}}t
\\&=\frac14\int_0^\infty t^{-\frac14}e^{-t}\left(\int_0^\infty u^{\frac14-1} e^{-u}{\rm{d}}u\right)\:{\rm{d}}t
\\&=\frac14 \Gamma\left(1-\frac14\right)\Gamma\left(\frac14\right)
\\&=\frac{\pi}{4\sin \frac{\pi}{4}}
\\&=\color{red}{\frac{\pi}{2\sqrt{2}}}.
\end{align}
$$ 
