Question

Evaluate $\underset{0}{\overset{\infty }{\int }}\frac{dx}{\left(x^2+a^2\right)\left(x^2+b^2\right)}$

Answer 1

${\int }_0^{\infty }\frac{dx}{(x^2+a^2)(x^2+b^2)}$

$={\int }_0^{\infty }\frac{[(x^2+a^2)-(x^2+b^2)]dx}{(x^2+a^2)(x^2+b^2)}$

$=\frac{1}{a^2-b^2}{\int }_0^{\infty }\frac{(x^2+a^2)dx}{(x^2+a^2)(x^2+b^2)}-\frac{1}{a^2-b^2}{\int }_0^{\infty }\frac{(x^2+b^2)dx}{(x^2+a^2)(x^2+b^2)}$

$=\frac{1}{a^2-b^2}{\int }_0^{\infty }\frac{dx}{x^2+b^2}-\frac{1}{a^2-b^2}{\int }_0^{\infty }\frac{dx}{x^2+a^2}$

$=\frac{1}{a^2-b^2}\times \frac{1}{b}{\left[{\tan }^{-1}\frac{x}{b}\right]}_0^{\infty }-\frac{1}{a^2-b^2}\times \frac{1}{a}{\left[{\tan }^{-1}\frac{x}{a}\right]}_0^{\infty }$

$=\frac{1}{a^2-b^2}\{\frac{1}{b}[\frac{\pi }{2}-0]-\frac{1}{a}[\frac{\pi }{2}-0]\}$

$=\frac{\pi }{2(a^2-b^2)}(\frac{1}{b}-\frac{1}{a})$

$=\frac{\pi }{2(a^2-b^2)}\left(\frac{a-b}{ab}\right)$

$=\frac{\pi }{2(a-b)(a+b)}\left(\frac{a-b}{ab}\right)$

$=\frac{\pi }{2ab(a+b)}$