Help me find the triple improper integral

Question

I don't understand how to calculate the integral $$\int\limits_{ \mathbb{R}^{3} }\exp\left(-\frac{x^{2}_{1} + x^{2}_{2} + x^{2}_{3} }{ a }\right)\operatorname{arctg}\left( x_{1}+x_{2}+x_{3} + 1\right)dx_{1}dx_{2}dx_{3}$$

Help please, thanks :)

Answer 1

Convert to spherical coordinates

$$\int_0^{2\pi}\int_0^\pi\int_0^\infty r^2\sin\theta\exp\left(-\frac{r^2}{a}\right)\tan^{-1}(r\sin\theta(\cos\phi+\sin\phi)+r\cos\theta+1)drd\theta d\phi$$

Now let $\sqrt{a}t = r$

$$a^{\frac{3}{2}}\int_0^{2\pi}\int_0^\pi\int_0^\infty t^2\sin\theta \:e^{-t^2}\tan^{-1}(\sqrt{a}t\sin\theta(\cos\phi+\sin\phi)+\sqrt{a}t\cos\theta+1)dtd\theta d\phi$$

By dominated convergence the limit $a\to0^+$ is boring and goes to $0$, but the limit of $a^{-\frac{3}{2}}\int df$ is given by

$$\frac{\pi}{4}\int_0^{2\pi}\int_0^\pi\int_0^\infty t^2\sin\theta \:e^{-t^2}\:dtd\theta d\phi = \frac{\pi^{\frac{5}{2}}}{4}$$