Mathematics

Evaluate the following integral $$\int _{ 0 }^{ \infty  }{ \dfrac { dx }{ \left( { x }^{ 2 }+{ a }^{ 2 } \right) \left( { x }^{ 2 }+{ b }^{ 2 } \right)  }  } =$$


ANSWER

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


SOLUTION
$$ \displaystyle \int_{0}^{\infty }\frac{dx}{(x^{2}+a^{2})(x^{2}+b^{2})}$$
$$ \displaystyle\Rightarrow \frac{1}{(a^{2}-b^{2})} \int_{0}^{\infty} \frac{(a^{2}-b^{2})dx}{(x^{2}+a^{2})(x^{2}+b^{2})}$$
$$ \displaystyle \Rightarrow \frac{1}{(a^{2}-b^{2})} \int_{0}^{\infty}\left [ \frac{(x^{2}+a^{2})-(x^{2}+b^{2})}{(x^{2}+a^{2})(x^{2}+b^{2})} \right ]dx $$
$$ \displaystyle = \frac{1}{(a^{2}-b^{2})} \left [ \int_{0}^{\infty} \frac{dx}{(x^{2}+b^{2})} -\int_{0}^{\infty} \dfrac{dx}{(x^{2}+a^{2})}\right ]$$
$$ \displaystyle = \dfrac{1}{(a^{2}-b^{2})} \left [ \frac{1}{b} tan^{-1}\frac{x}{b}-\frac{1}{a} tan^{-1}\frac{x}{a} \right ]_{0}^{\infty }$$
$$ \displaystyle = \frac{1}{(a^{2}-b^{2})} \left [ \frac{1}{b}tan^{-1}\infty  - \frac{1}{a}tan^{-1} \infty ] -(0-0) \right ]$$
$$ \displaystyle = \dfrac{1}{(a^{2}-b^{2})} \left [ \dfrac{1}{b} \times  \dfrac{\pi}{2}-\dfrac{1}{a} \times \dfrac{\pi}{2} \right ]$$
$$ = \dfrac{\pi}{2(a-b)(a+b)} \left [ \dfrac{(a-b)}{ab} \right ]$$
$$ = \dfrac{\pi}{2ab (a+b)}$$
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Single Correct Medium Published on 17th 09, 2020
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