Question

If $a,b>0$ and $\underset{0}{\overset{\infty }{\int }}\frac{\ln \left(bx\right)}{x^2+a^2}dx=\frac{\pi }{2a},$ then the value of $ab$ is

• A. $1$
• B. $\frac{1}{e}$
• C. $e$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $e$

Let $I=\underset{0}{\overset{\infty }{\int }}\frac{\ln \left(bx\right)}{x^2+a^2}dx$
Assuming $x=a\tan \theta$
$\Rightarrow dx=a{\sec }^2\theta d\theta$

Now, $I=\underset{0}{\overset{\pi /2}{\int }}\frac{\ln \left(ab\tan \theta \right)a{\sec }^2\theta d\theta }{a^2{\sec }^2\theta }$
$\Rightarrow I=\frac{1}{a}\underset{0}{\overset{\pi /2}{\int }}(\ln ab+\ln \tan \theta )d\theta \Rightarrow I=\frac{\pi }{2a}\ln ab+\underset{0}{\overset{\pi /2}{\int }}\ln \tan \theta d\theta$
Replacing $\theta \to \frac{\pi }{2}-\theta$, we get
$I=\frac{\pi }{2a}\ln ab+\underset{0}{\overset{\pi /2}{\int }}\ln \cot \theta d\theta$
Adding both, we get
$2I=\frac{\pi }{a}\ln ab+\underset{0}{\overset{\pi /2}{\int }}[\ln \tan \theta +\ln \cot \theta ]d\theta \Rightarrow 2I=\frac{\pi }{a}\ln ab+\underset{0}{\overset{\pi /2}{\int }}\left[\ln 1\right]d\theta \Rightarrow \frac{\pi }{2a}=\frac{\pi }{2a}\ln ab\Rightarrow \ln \left(ab\right)=1\therefore ab=e$