Improper integration problem

Question

$$ \int \limits^{\infty }_{0}\frac{\tan^{-1}\left( x\right) + \tan^{-1}\left( \frac{\alpha x +\beta }{\beta x -\alpha } \right) }{x^{2}+1} dx$$. For $ \alpha, \beta >0$... My question is how we can evaluate this improper integration above?. Actually I tried partial fraction and many other ways but I couldn't complete. How I can find the value of this integration?. Thanks

Answer 1

Remember that the derivative of $\tan^{-1}(x)$ is $\dfrac{1}{x^2+1}$.

This might make you curious as to what the derivative of the second inverse tangent term is: $$\dfrac{d}{dx}\left[ \tan^{-1}\left(\frac{\alpha x+\beta}{\beta x - \alpha}\right)\right] = \dfrac{1}{1+((\alpha x+\beta)/(\beta x-\alpha))^2}\cdot\frac{\alpha(\beta x-\alpha)-(\alpha x+\beta)\beta}{(\beta x-\alpha)^2}$$

$$=\dfrac{1}{\dfrac{(\beta x-\alpha)^2+(\alpha x+\beta)^2}{(\beta x-\alpha)^2}} \cdot \dfrac{-\alpha^2-\beta^2}{(\beta x-\alpha)^2} = -\dfrac{\alpha^2+\beta^2}{(\alpha^2+\beta^2)(x^2+1)} = -\dfrac{1}{x^2+1}$$

This means you can integrate: $\dfrac{\tan^{-1}(x)}{x^2+1}$ using a $u$-substitution: $u=\tan^{-1}(x)$ and then integrate $\dfrac{\tan^{-1}\left(\frac{\alpha x+\beta}{\beta x - \alpha}\right)}{x^2+1}$ using a $u$-substitution: $u=\tan^{-1}\left(\frac{\alpha x+\beta}{\beta x - \alpha}\right)$.

I hope this helps get you started! :)

Answer 2

Sketch:

$$\arctan(a) + \arctan(b) = \arctan \bigg(\frac{a+b}{1-ab}\bigg) $$

Plug in our values to get $\arctan \big(-\frac{ \beta}{\alpha}\big )$.

Hope this helps.