Question
Evaluate the following definite integrals:
$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}$
$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}$
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Answer
We have,
$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}=\frac{1}{\text{b}^2}\int_{0}^\limits{\infty}\frac{1}{\big(\frac{\text{a}}{\text{b}}\big)^2+\text{x}^2}\text{ dx}$
We have that,
$\int\frac{1}{\text{a}^2+\text{x}^2}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{\text{a}}$
$\therefore\ \frac{1}{\text{b}^2}\int_{0}^\limits{\infty}\frac{1}{\big(\frac{\text{a}}{\text{b}}\big)^2+\text{x}^2}\text{ dx}=\frac{1}{\text{b}^2}\Big[\frac{\text{b}}{\text{a}}\tan^{-1}\Big(\frac{\text{bx}}{\text{a}}\Big)\Big]^{\infty}_0$
$=\frac{1}{\text{ab}}\Big[\tan^{-1}\Big(\frac{\text{bx}}{\text{a}}\Big)\Big]^{\infty}_0$
$=\frac{1}{\text{ab}}\big[\tan^{-1}\infty-\tan^{-1}0\big]$
$=\frac{1}{\text{ab}}\Big[\frac{\pi}{2}-0\Big]$
$=\frac{\pi}{2\text{ab}}$
$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}=\frac{\pi}{2\text{ab}}$
$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}=\frac{1}{\text{b}^2}\int_{0}^\limits{\infty}\frac{1}{\big(\frac{\text{a}}{\text{b}}\big)^2+\text{x}^2}\text{ dx}$
We have that,
$\int\frac{1}{\text{a}^2+\text{x}^2}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{\text{a}}$
$\therefore\ \frac{1}{\text{b}^2}\int_{0}^\limits{\infty}\frac{1}{\big(\frac{\text{a}}{\text{b}}\big)^2+\text{x}^2}\text{ dx}=\frac{1}{\text{b}^2}\Big[\frac{\text{b}}{\text{a}}\tan^{-1}\Big(\frac{\text{bx}}{\text{a}}\Big)\Big]^{\infty}_0$
$=\frac{1}{\text{ab}}\Big[\tan^{-1}\Big(\frac{\text{bx}}{\text{a}}\Big)\Big]^{\infty}_0$
$=\frac{1}{\text{ab}}\big[\tan^{-1}\infty-\tan^{-1}0\big]$
$=\frac{1}{\text{ab}}\Big[\frac{\pi}{2}-0\Big]$
$=\frac{\pi}{2\text{ab}}$
$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}=\frac{\pi}{2\text{ab}}$
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