1. ## Complex integration question!

Hi, im really stuck on this question and keep ending up with ridiculous answers, i would be really grateful for any help. thankyou

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

For the integral J, simplify your answer until you get an expression involving real numbers only.

2. You can use contour integration. See here for a similar example.

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

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

$= \int_{0}^{\infty} \frac{x^2 + 1 }{ 1 + x^4 } ~dx - \int_{0}^{\infty} \frac{x^2-1}{ 1 + x^4 } ~dx$

By substituting $x = \frac{1}{t}$ in the second integral , we find that it is being zero !

Therefore , what you need to do is the first integral :

$\int_{0}^{\infty} \frac{x^2 + 1 }{ 1 + x^4 } ~dx$

$= \int_{0}^{\infty} \frac{1 + 1/x^2}{ x^2 + 1/x^2 } ~dx$

$\int_{0}^{\infty} \frac{d(x- 1/x) }{ (x-1/x)^2 + 2 }$

$= \int_{-\infty}^{\infty} \frac{du}{u^2 + 2 }$

$= \frac{\pi}{\sqrt{2}}$