7
$\begingroup$

What is your best idea for this integral? No need for any full solution, this is optional. I'm curious
about the core idea you might like to use to make all very simple.

Calculate in closed-form

$$\int_0^{1} \frac{\displaystyle \arctan^2\left(\frac{\sqrt{x}}{\sqrt{4-x}}\right)}{x} \, dx$$

Supplementary questions:

$$a) \ \int_0^{1} \frac{\displaystyle \arctan^3\left(\frac{\sqrt{x}}{\sqrt{4-x}}\right)}{x} \, dx$$

$$b) \ \int_0^{1} \frac{\displaystyle \arctan^3\left(\frac{\sqrt{x}}{\sqrt{4-x}}\right)\log(x)}{x} \, dx.$$

EDIT: Maybe it helps to write down the closed-form I got for the first integral, which is

$$\int_0^{1} \frac{\displaystyle \arctan^2\left(\frac{\sqrt{x}}{\sqrt{4-x}}\right)}{x} \, dx$$

$$=\frac{1}{216} \left(\sqrt{3} \pi \left(\psi ^{(1)}\left(\frac{1}{3}\right)-\psi ^{(1)}\left(\frac{2}{3}\right)+\psi ^{(1)}\left(\frac{1}{6}\right)-\psi ^{(1)}\left(\frac{5}{6}\right)\right)-144 \zeta (3)\right).$$

$\endgroup$
  • 2
    $\begingroup$ You might enjoy reading this paper on inverse trig integrals by Rogers: arxiv.org/abs/math/0601082 . Your integral is equivalent to a special case of the third family of integrals evaluated there. $\endgroup$ – David H Feb 18 '16 at 13:26
  • $\begingroup$ @DavidH thank you. Indeed, nice paper (+1). $\endgroup$ – user 1357113 Feb 18 '16 at 13:47
7
$\begingroup$

This is not even an idea, it's an evidence: set $$x=4\sin^2\phi.$$

$\endgroup$
  • $\begingroup$ Lovely stuff +1 $\endgroup$ – complexmanifold Feb 18 '16 at 13:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.