Find $\int_0^\pi\int_0^{2\pi}\exp{\bigg[x\cos(\phi)\sin(\theta)+y\sin(\phi)\sin(\theta)+z\cos(\theta))\bigg]}\sin(\theta)\,d\phi \,d\theta$

Question

I want to solve the following integration $$I = \int_0^\pi\int_0^{2\pi}\exp{\bigg[x\cos(\phi)\sin(\theta)+y\sin(\phi)\sin(\theta)+z\cos(\theta))\bigg]}\sin(\theta)\,d\phi \,d\theta $$ My Attempt:

First solve the $\phi$ part $$I = \int_0^\pi\exp{[z\cos(\theta)}]\sin(\theta) \Bigg[\int_0^{2\pi} \exp\bigg[x\cos(\phi)\sin(\theta)+y\sin(\phi)\sin(\theta))\bigg] \, d\phi\Bigg] \, d\theta\\ I = \int_0^\pi\exp[z\cos(\theta)]\sin(\theta) \, d\theta I_2 $$ where $$I_2 = \int_0^{2\pi} \exp\bigg[x\cos(\phi)\sin(\theta)+y\sin(\phi)\sin(\theta))\bigg] \, d\phi$$ Nothing seems to work here. I have tried integration by parts and substitution method but both just keeps expanding the terms. How can I solve this. Please help.

Answer 1

You may notice that the integral is in fact $$ \iint_{S} e^{(\vec{v} \cdot \vec{r})} d\vec{r}, \quad \vec{v} = \{x;y;z\}, $$ where integration is performed over a sphere of unit radius. You can use the symmetry of the problem and rotate everything so that $\vec{v}$ points along $Z$ axis. This simplifies the expression drastically $$ \int_0^{2\pi} \int_0^{\pi} e^{v\, cos(\theta)} \sin(\theta) d\theta d\phi, \quad v = \sqrt{x^2 + y^2 + z^2}. $$ The net result should be something like $$ \frac{4 \pi \sinh(\sqrt{x^2+y^2+z^2})}{\sqrt{x^2+y^2+z^2}}. $$