I am facing difficulty in evaluation of the following improper integral:

$$\int_{0}^{\infty} \frac{\tan^{-1}(\alpha x) \tan^{-1}(\beta x)}{ x^2} dx,\ \ \alpha, \beta>0$$

I thought of applying Leibniz rule of differentiation under integral sign. But here we have two parameters $\alpha$ and $\beta$. I do not know how to apply Leibniz in such cases. I think I can first differentiate the given integral first w.r.t. first parameter $\alpha$ and denote that as $I'(\alpha)$ which is as follows:

$$I'(\alpha)= \int_{0}^{\infty} \frac{\tan^{-1}(\beta x)}{(1+\alpha^2 x^2) x }dx$$

Next I differentiate it w.r.t. $\beta$ and denote the result as $J(\beta)$ which is as follows:

$$J(\beta) =\int_{0}^{\infty} \frac{1}{(1+\alpha^2 x^2)(1+\beta^2 x^2)} dx.$$ This integral is quite easy to evaluate. But my question is that am I allowed to do all this. What are the other methods to evaluate this integral? Please suggest.


1 Answer 1


Yes, you can apply Differentiation Under the Integral Sign as I show you bellow. Just a quick note, I'm applying a slightly modified version of Leibniz Rule, but in essence it's pretty much the same.

$$f(\alpha,\beta)=\int_{0}^{\infty}\frac{\arctan(\alpha x)\arctan(\beta x)}{x^2}dx$$





$$\boxed{f(\alpha,\beta)=\int_{0}^{\infty}\frac{\arctan(\alpha x)\arctan(\beta x)}{x^2}dx=\frac{\pi}{2}log\left(\frac{\left(\alpha+\beta\right)^{\alpha+\beta}}{\alpha^\alpha\beta^\beta}\right)}$$


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