Evaluate $\int _0^1\int _0^1\frac{\log \left(x^2+y^2\right)}{\sqrt{x+y}}dydx$

Question

How to show $$\int _0^1\int _0^1\frac{\log \left(x^2+y^2\right)}{\sqrt{x+y}}d\sigma=-\frac{128 a}{9}+\frac{16}{3} \sqrt{2 a} \tan ^{-1}\left(\sqrt{\frac{a}{2}}\right)+\frac{8}{3} \sqrt{2} \log (2)$$ Where $a=\sqrt{2}-1$? Any kind of help will be appreciated.

Answer 1

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.