Hi everybody,

I have a function:

$\displaystyle h(x,y,z)=\frac{z(x^2+2y^2)}{8x^4\sqrt{x^2+y^2}}ata n\left(\frac{z}{\sqrt{x^2+y^2}}\right)$,

and I would like to integrate it twice against x to obtain $\displaystyle h_2(x,y,z)=\iint h(x,y,z)dx dx$.

More precisely, I need to compute

$\displaystyle \int_{x_1=0}^{a_1}dx_1\int_{x=d+a1-x_1}^{d+a_1+a_2-x_1}dx\,h(x,y,z)$,

which can also be written

$\displaystyle \left[h_2(x,y,z)\right]_{x=d+a_1}^{x=d+a_1+a_2}+\left[h_2(x,y,z)\right]_{x=d+a_2}^{x=d}$.

However there are no symmetries that can be exploited. $\displaystyle d, a_1, a_2$ can have any positive real value.

Until now I have obtained a quite complicated expression for $\displaystyle h_2$ with explicit terms and two simple integrals which I can't figure out how to compute, namely :

$\displaystyle \int \frac{\sqrt{x^2+y^2}}{x}atan\left(\frac{z}{\sqrt{x ^2+y^2}}\right)dx$,

and

$\displaystyle \int \frac{\sqrt{x^2+y^2}}{x^3}atan\left(\frac{z}{\sqrt {x^2+y^2}}\right)dx$.

Any idea to integrate this function ? I have not been lucky with variable changes...

Thanks,

Kind regards,

Arixo

PS: for those interested, it is an integration involved in the computation of the van der Waals interaction energy between two solid cubes of size a1 and a2 and separated by a distance d in the x direction...