How can I compute the integral $\int_{0}^{\infty} \frac{dt}{1+t^4}$? I have to compute this integral $$\int_{0}^{\infty} \frac{dt}{1+t^4}$$ to solve a problem in a homework. I have tried in many ways, but I'm stuck. A search in the web reveals me that it can be do it by methods of complex analysis. But I have not taken this course yet. Thanks for any help.
 A: $1 + t^4 = (1 + 2t^2 + t^4) - 2t^2 = (1 + t^2)^2 - (\sqrt{2} t)^2$
This is a difference of squares and so can be factored.  Once you've factored it, use partial fractions.
If you don't know that trick, you can do this:
$1+t^4=0$ iff $t^4 = -1$, so $t^2 = \pm i$.  If $t^2 = i$, then $t = \pm \frac{1+i}{\sqrt{2}}$, and if $t^2 = -i$, then $t = \pm\frac{1-i}{\sqrt{2}}$.  The way you get this is that when you multiply complex numbers, you add angles and multiply lengths; hence when you take the square root of a complex number, you take the square root of the length and cut the angle in half.
So now you've got $1+t^4 = \big( (t - (1+i)/\sqrt{2})(t - (1-i)/\sqrt{2})\big)\big((t - (-1+i)/\sqrt{2})(t - (-1-i)/\sqrt{2})\big)$.  If you multiply the first two factors, you get $t^2 - t\sqrt{2} + 1$.  If you multiply the last two factors, you get $t^2 + t\sqrt{2} + 1$.
Once you've factored the denominator, use partial fractions.  You've got an irreducible quadratic factor in the denominator, so you'll get an arctangent.
A: Note that the substitution $t=1/u$ changes the integral to
$$\int_0^\infty \frac{u^2}{1+u^4}du.$$
Doesn't sound very helpful, but there was extensive discussion of that 
here.
Added: But we can do better.  Split the integral into into two parts, $0$ to $1$, and $1$ to infinity.  On the second part, let $t=1/u$.  We get $\int_0^1 \frac{u^2}{1+u^4}du$.  Now $u$ has done its duty, and is discarded for the more popular $t$. Our original integral is equal to
$$\int_0^1 \frac{1+t^2}{1+t^4} dt.$$
There is now a minor miracle. 
$$\frac{1}{1-\sqrt{2}t+t^2}+ \frac{1}{1+\sqrt{2}t+t^2}=\frac{2(1+t^2)}{1+t^4}.$$
Complete the square(s) as usual.
A: If you don't want to do a partial fraction decomposition, put $I:=\int_0^{+\infty}\frac{dt}{1+t^4}$. By the substitution $x=\frac 1t$ on $\left[0,+\infty\right)$ and $\left(-\infty,0\right]$, we get $2I = \int_{-\infty}^{+\infty}\frac{x^2}{1+x^4}\,dx$. Now we have 
\begin{align*}
4I &= \int_{-\infty}^{+\infty}\frac{t^2-\sqrt 2t+1}{(t^2-\sqrt 2t+1)(t^2+\sqrt 2t+1)}\,dt \\\
&=\int_{-\infty}^{+\infty}\frac{1}{t^2+\sqrt 2t+1}\,dt\\\
&=\int_{-\infty}^{+\infty}\frac{1}{(t+\frac{\sqrt 2}2)^2-\frac 12+1}\,dt\\\
&=\int_{-\infty}^{+\infty}\frac{du}{u^2+\frac 12}\\\
&=2\int_{-\infty}^{+\infty}\frac{du}{(\sqrt 2 u)^2+1}\\\
&=\frac 2{\sqrt 2}\arctan (\sqrt 2u)\mid_{u=-\infty}^{u=+\infty}\\\
&=\frac {2\pi}{\sqrt 2}
\end{align*}
and finally $I=\dfrac{\pi}{2\sqrt 2}$.
A: You can also do it by partial fraction decomposition. We have
$$ \frac{2\sqrt{2}}{1+t^4} = \frac{t + \sqrt{2}}{t^2 + \sqrt{2}t + 1} - \frac{t - \sqrt{2}}{t^2 - \sqrt{2}t + 1}. $$
We have
$$\int_0^\infty \frac{dt}{t^2 \pm \sqrt{2}t + 1} =
\int_0^\infty \frac{dt}{(t \pm \sqrt{2}/2)^2 + 1/2} =
\int_0^\infty \frac{2dt}{(\sqrt{2} t \pm 1)^2 + 1} =
\sqrt{2} \arctan (\sqrt{2}t \pm 1) \big|_0^\infty.$$
Continuing, we get
$$\frac{\sqrt{2}\pi}{2} - \sqrt{2} \arctan (\pm 1) =
\frac{\sqrt{2}\pi}{2} \mp \frac{\sqrt{2}}{4} =
\frac{\sqrt{2}\pi (2 \mp 1)}{4}.$$
Next, we have
$$ \int_0^\infty \frac{(2t \pm \sqrt{2}) dt}{t^2 \pm \sqrt{2}t + 1} =
\log (t^2 \pm \sqrt{2}t + 1) \big|_0^\infty. $$
The integral doesn't converge, but we can consider instead
$$ \int_0^\infty \frac{(2t + \sqrt{2}) dt}{t^2 + \sqrt{2}t + 1} - \frac{(2t - \sqrt{2}) dt}{t^2 - \sqrt{2}t + 1} =
\log \frac{t^2 + \sqrt{2}t + 1}{t^2 - \sqrt{2}t + 1} \big|_0^\infty = 0. $$
Therefore we rewrite our initial expansion
$$ \frac{4\sqrt{2}}{1+t^4} = \frac{2t + 2\sqrt{2}}{t^2 + \sqrt{2}t + 1} - \frac{2t - 2\sqrt{2}}{t^2 - \sqrt{2}t + 1} =
\frac{2t + \sqrt{2}}{t^2 + \sqrt{2}t + 1} - \frac{2t - \sqrt{2}}{t^2 - \sqrt{2}t + 1} + \frac{\sqrt{2}}{t^2 + \sqrt{2}t + 1} + \frac{\sqrt{2}}{t^2 - \sqrt{2}t + 1}. $$
Integrating, we get
$$ \int_0^\infty \frac{4\sqrt{2}}{1+t^4} =
\sqrt{2} \frac{\sqrt{2} \pi (2-1)}{4} + \sqrt{2} \frac{\sqrt{2} \pi (2+1)}{4} =
\frac{\pi}{2} + \frac{3\pi}{2} = 2\pi. $$
Therefore the integral we want is
$$ \int_0^\infty \frac{1}{1 + t^4} = \frac{2\pi}{4\sqrt{2}} = \frac{\sqrt{2}\pi}{4}. $$
A: $\newcommand{\+}{^{\dagger}}%
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With $\ds{x \equiv {1 \over 1 + t^{4}}
\quad\iff\quad t = \pars{1 - x \over x}^{1/4}}$:
\begin{align}
\color{#00f}{\large\int_{0}^{\infty}{\dd t \over 1 + t^{4}}}&=
\int_{1}^{0}x\,{1 \over 4}\,\pars{1 - x \over x}^{-3/4}
\pars{-\,{1 \over x^{2}}}\,\dd x
=
{1 \over 4}\int_{0}^{1}x^{-1/4}\pars{1 - x}^{-3/4}\,\dd x
\\[3mm]&=
{1 \over 4}\,{\rm B}\pars{{3 \over 4},{1 \over 4}}
={1 \over 4}\,{\Gamma\pars{3/4}\Gamma\pars{1/4} \over \Gamma\pars{3/4 + 1/4}}
={1 \over 4}\,{\pi \over \sin\pars{\pi\bracks{1/4}}} =
\color{#00f}{\large{\root{2} \over 4}\,\pi}
\approx 1.1107
\end{align}
A: Let the considered integral be I i.e
$$I=\int_0^{\infty} \frac{1}{1+t^4}\,dt$$
Under the transformation $t\mapsto 1/t$, the integral is:
$$I=\int_0^{\infty} \frac{t^2}{1+t^4}\,dt \Rightarrow 2I=\int_0^{\infty}\frac{1+t^2}{1+t^4}\,dt=\int_0^{\infty} \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}\,dt$$
$$2I=\int_0^{\infty} \frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2+2}\,dt$$
Next, use the substitution $t-1/t=u \Rightarrow (1+1/t^2)\,dt=du$ to get:
$$2I=\int_{-\infty}^{\infty} \frac{du}{u^2+2}\Rightarrow I=\int_0^{\infty} \frac{du}{u^2+2}=\boxed{\dfrac{\pi}{2\sqrt{2}}}$$
$\blacksquare$
A: $$
\begin{aligned}
\int_0^{\infty} \frac{d t}{1+t^2}&=\int_0^{\infty} \frac{\frac{1}{t^2}}{t^2+\frac{1}{t^2}} d t \\=& \frac{1}{2} \int_0^{\infty} \frac{\left(1+\frac{1}{t^2}\right)-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}} d t \\
=& \frac{1}{2}\left[\int_0^{\infty} \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^2+2}-\int_0^{\infty} \frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^2-2}\right]_0^{\infty} \\
=& \frac{1}{2 \sqrt{2}}\left[\tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)\right]_0^{\infty}-\frac{1}{2 \sqrt{2}}\left[\ln \left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}} \right|\right]_0^{\infty}\\
=& \frac{\pi}{2 \sqrt{2}}
\end{aligned}
$$
