# Improper integral $\int_{0}^{\infty}\tan^{-1}(\alpha x) \tan^{-1}(\beta x)/ x^2 dx, \alpha, \beta>0.$

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.

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.

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

$$=\int_{0}^{\infty}\int_{0}^{\alpha}\int_{0}^{\beta}\frac{dydzdx}{(1+y^2x^2)(1+z^2x^2)}=\int_{0}^{\alpha}\int_{0}^{\beta}dydz\int_{0}^{\infty}\frac{dx}{(1+y^2x^2)(1+z^2x^2)}$$

$$=\int_{0}^{\alpha}\int_{0}^{\beta}\frac{dydz}{y^2-z^2}\int_{0}^{\infty}\left(\frac{y^2}{1+y^2x^2}-\frac{z^2}{1+z^2x^2}\right)dx$$

$$=\int_{0}^{\alpha}\int_{0}^{\beta}dydz\left[\frac{y\arctan(yx)-z\arctan(zx)}{(y-z)(y+z)}\right]^{\infty}_{0}=\frac{\pi}{2}\int_{0}^{\alpha}\int_{0}^{\beta}\frac{dydz}{y+z}$$

$$=\frac{\pi}{2}\int_{0}^{\alpha}dz[log(z+\beta)-log(z)]=\frac{\pi}{2}[(\beta+z)log(\beta+z)-z-zlog(z)+z]^{\alpha}_{0}$$

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