Question

Prove that ${\int }_0^{\infty }\frac{dx}{1+x^4}={\int }_0^{\infty }\frac{x^{2}dx}{1+x^4}=\frac{\pi }{2\sqrt{2}}$.

Answer 1

${\int }_0^{\infty }\frac{dx}{1+x^4}=I$ (say)

Put $x=\frac{1}{t},dx=\frac{-1}{t^2}dt$

So, ${\int }_{\infty }^0\frac{\frac{-1}{t^2}dt}{\frac{t^4+1}{t^4}}={\int }_{\infty }^0\frac{-t^{2}dt}{1+t^4}={\int }_0^{\infty }\frac{t^2}{1+t^4}dtI={\int }_{\infty }^0\frac{x^2}{1+x^4}dx-\left(i\right)I={\int }_0^{\infty }\frac{1}{1+x^4}dx-\left(ii\right)$

add $\left(i\right)$ and $\left(ii\right)$ , we get,

$2I={\int }_0^{\infty }\frac{1+x^2}{1+x^4}dx={\int }_0^{\infty }\frac{(1+\frac{1}{x^2})}{x^2+\frac{1}{x^2}}dx2I={\int }_0^{\infty }\frac{(1+\frac{1}{x^2})}{{(x-\frac{1}{x})}^2+2}dx$

Let $x-\frac{1}{x}=z(1+\frac{1}{x^2})dx=dz$

$2I={\int }_{-\infty }^{\infty }\frac{dz}{z^2+2}=\frac{\pi }{\sqrt{2}}I=\frac{\pi }{2\sqrt{2}}$