Integration of mixed function at infinity limit

[raymound]

Hi,

I am trying to calculate second virial coefficient from interaction potentias and I have to Integrate at Infinity level and it seems that Integrate doesn't converge.

can you help me integrate this function. Integrate is attached as an image.

Regards
Raymond

 

[mathman]

It looks like it would be difficult to integrate analytically. To check its behavior near r = 0, expand the numerator as a power series in r. The first term may be ~ r², in which case you are OK. However I am not sure - it is messy.

 

[JJacquelin]

If rs is not nul, the integral is not convergent.
So the integral can be computed only if rs=0

 

[mathman]

If rs is not nul, the integral is not convergent.
So the integral can be computed only if rs=0

Why?

 

[JJacquelin]

If rs is not nul, the function to be integrated is equivalent to c/r² close to r=0, where c is a constant (Expand the function around r=0).
The integral of c/r² is divergent for r tending to 0.
if rs=0 then c=0 and one can see from the expansion that the next term is integrable. So, there is no integration problem around r=0 in this particular case of rs=0.
All this concerns the question of convergence around r=0 only.

 

[mathman]

If rs is not nul, the function to be integrated is equivalent to c/r² close to r=0, where c is a constant (Expand the function around r=0).
The integral of c/r² is divergent for r tending to 0.
if rs=0 then c=0 and one can see from the expansion that the next term is integrable. So, there is no integration problem around r=0 in this particular case of rs=0.
All this concerns the question of convergence around r=0 only.

My estimate is that the numerator is ~ r² near r = 0, unless s = 1.

 

[JJacquelin]

My estimate is that the numerator is ~ r² near r = 0, unless s = 1.

Does "rs" means r multiplied by s, or does "rs" is the symbol of a parameter (i.e. is constant) ?
In the first case you are right. In the second case I am right.