An integral that Mathematica won't solve

Hi all,

I have found that Mathematica 6.0 refuses to solve the following integration:

Integrate[ Exp[-i t u] (1-2 i t a)^(-0.5) (1 - 2 i t b)^(-0.5) , {t,-inf,+inf}, Assumptions->u>0,a>0,b>0 ]

where i = SQRT(-1)

I cannot understand why the integral cannot be done, does anyone know of a reason why Mathematicca won't solve this problem?

Natski

 

This integration is already an inverse FT, if I FT it I will just end with where I started...

 

Using InverseFT doesn't help either, exact same result, where Mathematica merely reprints the question. It's an extremely troubling problem...

 

Note that setting a = b or a=0 or b=0 allows Mathematica to produce an analytical equations. The fact there is a solution for a=b leads me tot suspect that it should be possible!

 

Well actually I entered it into Mathematica as neither i nor I but the special imaginary symbol which is provided in the palette. This really isn't a problem of me entering this wrong!

I have become convinced the problem must be solved using contour integration, but have yet been able to actually solve it...

 

Another approach which I can confirm doesn't work is setting the limits to some finite value, say z, and then attempting to tend z to infinity once a solution is acquired. Unfortunately, this integral is also not feasible in either integral form or InverseFourierTransform functional form.

I also tried doing a convolution of the Fourier Transform of each denominator, but this doesn't work either. Nor does employing KroneckerDeltas in a double integral. I also tried all of these approaches with finite limit of z and attempting to tend to infinity but once again, none of these approaches worked.

Any other ideas?? I refuse to accept that this integral is impossible.

 

I think one of the main problems here is that if I try to do this by hand, then I cannot compute the residue at a certain point in my derivation...

Take the function Exp[-i u z]/(SQRT(1 - 2 i z a).

One might assume a pole lies at z = 1/(2 i a) but when you try to compute the residue at this point you get a complex infinity as the denominator, basically because of the SQRT bit.

How can you compute the residue of this function?

 

After some reading, I think I understand now that function has multiple branch points, rather than multiple poles, and hence residue theorem cannot work for such integrals...

Can anyone recommend any books on how to integrate functions with multiple branches? I tried Arfken but it didn't offer much.

Another question, all functions with poles can be integrated by residue theorem, but can all functions with multiple branches always be necessarily integrated? If there is no closed form to this equation, can it be proved that there is no closed form?

Natski