Hint to solve this integral: $\int_{0}^{1}\int_{0}^{1}(x^4+x^2y^2+y^4)\sqrt{x^2+y^2}\arctan{\sqrt{x^2+y^2}}dxdy$

Question

I am trying to evaluate this integral:$$I=\int_{0}^{1}\int_{0}^{1}(x^4+x^2y^2+y^4)\sqrt{x^2+y^2}\arctan{\sqrt{x^2+y^2}}dxdy$$

But I am getting stuck to exploit the symmetry. First try, convert this integral to polar coordinate system by: $x=r\cos{\phi}$ and $y=r\sin{\phi}$ then $I$ become:

$$I=2\int_{0}^{\frac{\pi}{4}}\int_{0}^{\frac{1}{\cos{\phi}}}r^6(\cos^4{\phi}+\sin^2{\phi})\arctan{r}drd\phi=2\int_{0}^{\frac{\pi}{4}}(\cos^4{\phi}+\sin^2{\phi})d\phi\int_{0}^{\frac{1}{\cos{\phi}}}r^6\arctan{r}dr $$ and IBP for integral with variable $r$, and it became a mess.

Second try, rewrite: $$I=\int_{0}^{1}\int_{0}^{1}(x^4+x^2y^2+y^4)\sqrt{x^2+y^2}\arctan{\sqrt{x^2+y^2}}dxdy=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}(x^4+x^2y^2+y^4)\frac{x^2+y^2}{1+z^2(x^2+y^2)}dzdxdy$$

But it seems more difficult, my experience told me that has a way to exploit the symmetry, but i can't figure it out yet.

I will be thankful if everyone can give me some hints. Thank you for reading and good day.

Answer 1

The last integral $$I=\int r^6 \tan ^{-1}(r)\,dr$$ is not too difficult. Integration by parts gives $$I=\frac 17 r^7 \tan ^{-1}(r) -\frac 17\int \frac{r^7}{r^2+1}\,dr$$ Now $$u=r^2+1 \quad \implies \quad \int \frac{r^7}{r^2+1}\,dr=\frac 12\int \frac{(u-1)^3}{ u}\,du$$ Expand the cube, integrate and go back to $r$ to have $$I=\frac 1 {84}\left(12 r^7 \tan ^{-1}(r)-2 r^6+3 r^4-6 r^2+6 \log \left(r^2+1\right)\right)$$ Using the bounds $$J(\phi)=\int_0^{\sec (\phi )} r^6 \tan ^{-1}(r)\,dr$$ $$J(\phi)=\frac{1}{84} \left(-2 \sec ^6(\phi )+3 \sec ^4(\phi )-6 \sec ^2(\phi )+6 \log \left(\sec ^2(\phi )+1\right)+12 \sec ^7(\phi ) \tan ^{-1}(\sec (\phi ))\right)$$ and now we need to compute $$2\int_{0}^{\frac{\pi}{4}}(\cos^4(\phi)+\sin^2(\phi))\,J(\phi)\,d\phi$$ and we face a few terrible integrals. Computing as many terms as possible, I am left with $$\frac 1{28}\int_0^{\frac \pi 4}(\cos (4 \phi )+7) \sec ^7(\phi ) \tan ^{-1}(\sec (\phi ))\,d\phi+\frac{K}{20160}$$

$K$ is explicit (not nice) invloving a bunch of complex terms, logarithms, arctangents, polylogarithms and Catalan constant. Its numerical value is $$K=-740.32725033387336851113397130373214440722799293632\cdots$$

But remains the problem of $$\int_0^{\frac \pi 4}(\cos (4 \phi )+7) \sec ^7(\phi ) \tan ^{-1}(\sec (\phi ))\,d\phi$$ which I solved using series expansion around $\phi=0$

$$(\cos (4 \phi )+7) \sec ^7(\phi ) \tan ^{-1}(\sec (\phi ))=\sum_{n=0}^\infty a_n \, \phi^{2n}$$ the first coefficients being $$\left\{2 \pi ,2+5 \pi ,\frac{64+109 \pi }{12} ,\frac{64296+85410 \pi }{6480},\frac{296224+318649 \pi }{20160},\frac{32635952+29536685 \pi }{1814400}\right\}$$

Using twelve terms, we find $12.8151$ while numerical integration gives $12.8156$