$$\int_0^\infty \frac{(\tan^{-1}(3x)-\tan^{-1}(x))}{x}\,dx.$$
I tried to evaluate this integral in the following way but got wrong answer ,the actual answer is $\frac{\pi \ln3}{2}$ but i am getting the answer as 0 using following method –
We can write this as $$\int_0^\infty \frac{\tan^{-1}(3x)3\,dx}{3x}-\int_0^\infty \frac{\tan^{-1}(x)\,dx}{x}.$$ If we substitute 3x=t in first integral , we get the answer as $$\int_0^\infty \frac{\tan^{-1}(t)\,dt}{t}-\int_0^\infty \frac{\tan^{-1}(x)\,dx}{x} =0$$ .
Why does this happen ?
Best Answer
This is simply a Frullani integral with the choice $f(x) = \tan^{-1} x$, $a = 3$, $b = 1$. You cannot separate the integrand in the way you have done, because the individual terms are not convergent on their own, like how $$\sum_{n=1}^\infty \frac{1}{n} - \frac{1}{n+1} \ne \sum_{n=1}^\infty \frac{1}{n} - \sum_{n=1}^\infty \frac{1}{n+1}.$$