Mathematics

If $$ \displaystyle \int_{0}^{\infty }\frac{e^{-ax}\sin bx}{x}\:dx= \tan^{-1}\left ( b/a \right ) $$ then the value of $$ \displaystyle \int_{0}^{\infty }\frac{\sin ^2ax}{x^2}\:dx $$ equals, $$ a\neq 0 $$


ANSWER

$$ \displaystyle \frac{\pi a}{2} $$


SOLUTION
$$ \displaystyle \int_{0}^{\infty }\frac{\sin ^2ax}{x^2}\:dx $$ 

let $$u=ax$$

$$dx=\dfrac{du}{a}$$

$$x=\dfrac{u}{a}$$

$$ \displaystyle a\int_{0}^{\infty }\frac{\sin ^2ax}{u^2}\:dx $$ 

$$\displaystyle \Rightarrow a\int^{\infty}_0\left(\dfrac{1}{u^2}-\dfrac{\cos^2u}{u^2}\right)du$$.

$$\displaystyle \Rightarrow a\left[\dfrac{-1}{u}\right]^{\infty}_0-a\left[\dfrac{-1}{u}.\cos^2 u-\int \left(\dfrac{-1}{u}\right)(-\sin 2u)du\right]^{\infty}_0$$

$$\Rightarrow +\dfrac{\pi}{2} a-a\sin (2a\cdot 0)$$

$$\Rightarrow \boxed{\dfrac{\pi \, a }{2}}$$
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Single Correct Hard Published on 17th 09, 2020
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