# Thread: Definite integral

1. ## Definite integral

Integrate:
$\int_{-\infty }^{\infty}\frac{x^2}{1+x^4}$
Please give me the hints only

2. ....

First show

$\int_0^{\infty} \frac{x^2}{1 + x^4}~dx$ converges

Then we can write

$\int_{-\infty}^{\infty} \frac{x^2}{1 + x^4}~dx = 2 \int_0^{\infty}\frac{x^2}{1+x^4}~dx$

For the integral $2 \int_0^{\infty}\frac{x^2}{1+x^4}~dx$

Write the numerator as $\frac{1}{2} ( (x^2+1)+(x^2-1))$

Then it equals to :

$\int_0^{\infty}\frac{(x^2+1)+(x^2-1)}{1+x^4}~dx$

$=\int_0^{\infty}\frac{x^2+1}{1+x^4}~dx + \int_0^{\infty}\frac{x^2-1}{1+x^4}~dx$

For the integral $\int_0^{\infty} \frac{x^2+1}{1+x^4}~dx$

$= \int_0^{\infty} \frac{1+1/x^2}{1/x^2+x^2}~dx$

$= \int_0^{\infty} \frac{d(x-1/x)}{ \left( x - \frac{1}{x} \right)^2 + 2 }$ then use substitution : $x-1/x = t$ ...

But the second integral should be zero : $I = \int_0^{\infty}\frac{x^2-1}{1+x^4}~dx$

After substituting $x = 1/t$ it becomes

$I = \int_0^{\infty} \frac{1- t^2}{1+t^4}~dt = - \int_0^{\infty} \frac{ t^2-1}{1+t^4}~dt = -I ~~~\implies I = 0$

3. The complex analysis offers a faster way to arrive to the result. If we apply the residue theorem we have...

$\int_{-\infty}^{+\infty} \frac{x^{2}}{1+x^{4}}\cdot dx = 2 \pi i \sum_{k} r_{k}$ (1)

... where the $r_{k}$ are the residue of the poles of $f(z) = \frac{z^{2}}{1+z^{4}}$ with positive imaginary part. If $z_{0}$ is a pole of $f(z)$ is [applying l'Hopital rule] ...

$r_{z=z_{0}} = \lim_{z \rightarrow z_{0}} (z-z_{0}) f(z) = \frac{1}{4 z_{0}}$ (2)

Now the poles of $f(z)$ with positive imaginary part are $z= \frac{i \pm 1}{\sqrt{2}}$ so that is...

$\int_{-\infty}^{+\infty} \frac{x^{2}}{1+x^{4}}\cdot dx = 2 \pi i \frac{\sqrt{2}}{4} (\frac{1}{i-1} + \frac{1}{i+1}) = \frac{\pi}{\sqrt{2}}$ (3)

Kind regards

$\chi$ $\sigma$