$I(\alpha)=\int_{0}^{1}\frac{\arctan(\alpha x)}{x\sqrt{1-x^2}}dx$

Question

I have to find $$I(\alpha)=\int_{0}^{1}\frac{\arctan(\alpha x)}{x\sqrt{1-x^2}}dx$$ Can someone help me to solve it?

Have you tried differentiating under the integral sign? What have you tried? — Aaron Maroja, May 04 '16 at 14:10
@Tuanlee, you should probably add some context and elaborate on your own attempts — Yuriy S, May 04 '16 at 14:22
The answer should be $$\large \frac{\pi}{2}\operatorname{arcsinh}(\alpha).$$ It also follows from here with $a\mapsto i a$. — Noam Shalev - nospoon, May 04 '16 at 14:22

score 2 · Answer 1 · answered May 04 '16 at 14:27

$$I'(\alpha) = \int_{0}^{1}\frac{dx}{(1+\alpha^2 x^2)\sqrt{1-x^2}} = \int_{0}^{\pi/2}\frac{d\theta}{1+\alpha^2\sin^2\theta}=\frac{\pi}{2\sqrt{1+\alpha^2}} $$ by the tangent half-angle substitution. By integrating with respect to $\alpha$:

$$ I(\alpha)=\int_{0}^{1}\frac{\arctan(\alpha x)}{x\sqrt{1-x^2}}\,dx = \color{red}{\frac{\pi}{2}\text{arcsinh}(\alpha)}.$$

$I(\alpha)=\int_{0}^{1}\frac{\arctan(\alpha x)}{x\sqrt{1-x^2}}dx$

1 Answers1