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?
 A: $\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.
A: 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.
A: let $I=\int \dfrac{dx}{1+x^4}$
I=$\dfrac {1}{2}\int \dfrac {x^2+1-(x^2-1)}{1+x^4} $
splitting the fraction,
I=$\dfrac {1}{2} (\int \dfrac {x^2+1}{1+x^4} - \int \dfrac {x^2-1}{1+x^4} )$ 
let $I_1$ =$ \int \frac {x^2+1}{1+x^4}$
Dividing the numerator and denominator by $x^2$,
$I_1$= $ \int \dfrac {1+\dfrac {1}{x^2} }{x^2 +\dfrac {1}{x^2} } $
put $x-\dfrac{1}{x}=t  $ for $I_1$ (observe that the derivative of $x-\dfrac{1}{x}=t  $ is $1+\dfrac{1}{x^2}$)
Let $I_2 = \int \dfrac {x^2-1}{1+x^4}$
Dividing the numerator and denominator by $x^2$,
$I_2 = \int \dfrac {1-\dfrac{1}{x^2}}{x^2+\dfrac {1}{x^2}}$
put $x+\dfrac{1}{x}=t  $ for $I_2$ (observe that the derivative of $x+\dfrac{1}{x}=t  $ is $1-\dfrac{1}{x^2}$)
So, $ I=\dfrac {1}{2} (I_1-I_2) $
Finally, substitute the limits.
Hope this helps.. Cheers!
A: You could mess about with integration by substitution methods. 
For instance, if you change the expression to this: 
$$\frac{1}{2}\int_0^\infty \frac{(x^2+1)-(x^2-1)}{x^4+1}dx = \frac{1}{2}\int_0^\infty \frac{(x^2+1)}{x^4 + 1}-\frac{(x^2-1)}{x^4+1}dx$$
which you can then separate out as:
$$\frac{1}{2}\left[\int_0^\infty \frac{(x^2+1)}{x^4+1}dx - \int_0^\infty \frac{(x^2-1)}{x^4+1}dx\right]$$
and then divide the top and bottom in both expressions by $x^2$ to get: 
$$\frac{1}{2}\left[\int_0^\infty \frac{1+ \frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx - \int_0^\infty \frac{1- \frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx\right]$$
and then you can change some variables. If you take:
$u= x- \frac{1}{x}$ so $du= 1 + \frac{1}{x^2}$ and $v= x+\frac{1}{x}$ and $dv = 1 - \frac{1}{x^2}$ 
you can square the functions: 
$u^2 = x^2 -2 +\frac{1}{x^2}$ or $u^2 + 2 = x^2+\frac{1}{x^2}$
$v^2 = x^2 +2 +\frac{1}{x^2}$ or $v^2-2  = x^2 +\frac{1}{x^2}$
and when you substitute in you end up with 
$$\frac{1}{2}\left[\int_0^\infty \frac{du}{u^2+2} - \int_0^\infty \frac{dv}{v^2-2}\right]$$
and from there it should be easier to do, no? I think this was the way I approached it in a class last semester and I saw someone else in the interwebs had done something similar.  :-) so I could be wrong. But I think this should work. Or you could I think rearrange the above substitutions into an integration by parts for each expression. But when I did that it was a long process. 
A: This was one of the questions in my final Infinitesimal Calculus I course (one year long course first undergraduate year)...and thus one has no idea whats' going on and must then try more or less simple things, so partial fractions:
$$x^4+1=(x^2+\sqrt2\,x+1)(x^2-\sqrt2\,x+1)\implies$$
$$\frac1{x^4+1}=\frac{Ax+B}{x^2+\sqrt2\,x+1}+\frac{Cx+D}{x^2-\sqrt2\,x+1}\;,\;A,B,C,D\in\Bbb R\implies$$
$$1=(Ax+B)(x^2-\sqrt2\,x+1)+(Cx+D)(x^2+\sqrt2\,x+1)$$
Comparing respective coefficients of $\,x\,$  and checking what happens with $\,x=0\,$ and etc., we get
$$\frac1{x^4+1}=\frac{x+\sqrt2}{2\sqrt2(x^2+\sqrt2\,x+1)}-\frac{x-\sqrt2}{2\sqrt2(x^2-\sqrt2\,x+1)}$$
and
$$\int\frac{x+\sqrt2}{x^2+\sqrt2\,x+1}dx=\frac12\int\left(\frac{2x+\sqrt2}{x^2+\sqrt2\,x+1}+\frac{\sqrt2-\frac1{\sqrt2}}{x^2+\sqrt2\,x+1}\right)dx=$$
$$=\frac12\log\left(x^2+\sqrt2\,x+1\right)+\frac1{2\sqrt2}\int\frac{dx}{\left(x+\frac1{\sqrt2}\right)^2+\frac12}=$$
$$=\frac12\log\left(x^2+\sqrt2\,x+1\right)+\frac1{\sqrt2}\int\frac{dx}{1+\left(\frac x{\sqrt2}+\frac12\right)^2}=$$
$$=\frac12\log\left(x^2+\sqrt2\,x+1\right)+\arctan\left(\frac x{\sqrt2}+\frac12\right)$$
Well, take it from here also with the other integral, put limits and etc.
A: 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:
$$I_4 = \int_0^\infty \frac{dx}{1+x^4} = \frac{\pi}{4\sin \frac{\pi}{4}} = \frac{\pi}{2\sqrt{2}}$$
A: 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 $u$
