How to show that the following integral: $$\int _0^1\int _0^1\frac{\log \left(x^2+y^2\right)}{\sqrt{x+y}}dydx$$ is equal to $$-\frac{128 a}{9}+\frac{16}{3} \sqrt{2 a} \tan ^{-1}\left(\sqrt{\frac{a}{2}}\right)+\frac{8}{3} \sqrt{2} \log (2)$$ where $a=\sqrt{2}-1$? While discussing about difficult problems on integration on another (local) website, one user posted this result without a proof. Because of the complicated form of the answer, I haven't figured out a reasonable approach (Note that mathematica can give a extreme complicated result due to primitive cauculation, which coincide to the original one after simplification, but that's not ingenious after all). Now I'd like you to help me with this, and I'll appreciate all kinds of method other than calculating primitives. Thank you!

  • $\begingroup$ I believe that $u=\sqrt{x+y}$ works $\endgroup$ – Sina Babaei Zadeh Oct 22 '19 at 3:20
  • $\begingroup$ Integration by parts works, even the indefinite Integral exists. I guess one also find the integral in Gradsteyn $\endgroup$ – stocha Oct 22 '19 at 8:12
  • $\begingroup$ @stocha Indeed, the result can be shown by brute force with the help of CAS, and I've just verified this. But I'd still like to see some cleverer thoughts. (By the way, this doesn't come from Gradshteyn's table. In fact I've scanned and gave proof to almost all logarithmic integrals there, but none of them is similar to this) $\endgroup$ – Display name Oct 22 '19 at 8:36
  • $\begingroup$ @stocha Thank you for your suggestion. I edited this question a bit. $\endgroup$ – Display name Oct 22 '19 at 8:39
  • $\begingroup$ @Fengshan Xiong: I understand, I was just playing with your integral in Mathematica. $\endgroup$ – stocha Oct 22 '19 at 10:16

One can exploit a symmetry and split the integral in half along the line $y=x$ then use the following modified polar coordinates:

$$x = s^{\frac{2}{3}}\cos\theta \hspace{10 pt} y = s^{\frac{2}{3}}\sin\theta$$

$$\implies 2\cdot \frac{2}{3}\int_0^{\frac{\pi}{4}} \int_0^{\sec^{\frac{3}{2}}\theta} \frac{\log\left(s^{\frac{4}{3}}\right)}{\sqrt{\cos\theta+\sin\theta}}dsd\theta = \frac{16}{9}\int_0^{\frac{\pi}{4}} \frac{\sec^{\frac{3}{2}}\theta}{\sqrt{\cos\theta+\sin\theta}}\left(\frac{3}{2}\log(\sec\theta)-1\right)d\theta$$

then let $x=\tan\theta$

$$ \implies \frac{4}{3}\int_0^1 \frac{\log(1+x^2)}{\sqrt{1+x}}dx - \frac{16}{9}\int_0^1 \frac{1}{\sqrt{1+x}}dx$$

The integral on the right evaluates to $\frac{32}{9}(\sqrt{2}-1) \equiv \frac{32}{9}a$. The integral on the left becomes

$$ = \frac{8}{3}\sqrt{2}\log 2 - \frac{16}{3}\int_0^1 \frac{x\sqrt{1+x}}{1+x^2}dx$$

We have one and a half of the three terms the user posted, but this last integral is tricky and isn't yielding to a variety of methods. I will try to finish later, but in the meantime if anyone has any clever suggestions for this last integral I will be happy try it.

  • $\begingroup$ For the last integral, you can just substitute $u=\sqrt{1+x}$, then use partial fractions. $\endgroup$ – Prasiortle Oct 22 '19 at 10:40
  • $\begingroup$ @Prasiortle that doesn't work nicely as the denominator $u^4-2u^2+2$ is not factorable over $\mathbb{R}$ as far as I can tell. And I would prefer to avoid complex logs, they are not elegant. $\endgroup$ – Ninad Munshi Oct 22 '19 at 10:42
  • $\begingroup$ If you only want to work over $\mathbb{R}$ then I don't think you can solve this integral analytically, since there appears to be no way to express the antiderivative of $\frac{x\sqrt{1+x}}{1+x^2}$ without using complex numbers. $\endgroup$ – Prasiortle Oct 22 '19 at 10:48
  • $\begingroup$ @Prasiortle we don't need to compute the antiderivative. We have a definite integral. Not to mention just because wolfram doesn't spit out an answer without complex numbers does not mean it cannot be simplified to not have them. But again, this is a definite integral, no antiderivative necessary. $\endgroup$ – Ninad Munshi Oct 22 '19 at 10:49
  • $\begingroup$ Correct, but that doesn't help. You've already said complex methods are out, and there's no trick like DUTIS that can be used here (as far as I can tell). $\endgroup$ – Prasiortle Oct 22 '19 at 10:53

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