An integral that Mathematica won't solve

natski · Apr 29, 2008

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

Pere Callahan · Apr 29, 2008

Maybe it doesn't converge by reasonable standards. Try FourierTransform instead

natski · Apr 29, 2008

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

_Mayday_ · Apr 29, 2008

Try [itex]\int \sin(x^2)dx[/itex]

Pere Callahan · Apr 29, 2008

natski said: ↑

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

I know that this is an Fourier integral, that's why I suggested using mathematica's built-in function FourierTransform[] (or InverseFourierTransform[]) instead of having it evaluate the defining integral using Integrate[].

The argument for FourierTransform would of course NOT include the complex exponential ...

natski · Apr 29, 2008

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

natski · May 1, 2008

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!

djeitnstine · May 5, 2008

natski said: ↑

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!

try wx maxima, if the integral is possible i.e. converges or diverges it will do it.

Its free, google it.

chroot · May 5, 2008

In Mathematica, capital I refers to sqrt(-1), not lowercase i.

- Warren

natski · May 7, 2008

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...

natski · May 9, 2008

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.

djeitnstine · May 9, 2008

How about plotting the graph and taking its area the old fashioned way?

Eidos · May 9, 2008

Another approach all together is to use pencil and paper.

You'll probably have to use the Laurent series expansion to find the residues though...
I'll try it out and see if I can come up with anything.

Pere Callahan · May 9, 2008

The integrand is not holomorphic in a neighborhood of the poles so what do you want to use the residues for?;-)

Eidos · May 9, 2008

To evaluate the contour integral :P
If the contour encloses any poles then the integral is equal to 2pi i times the sum of residues.

Pere Callahan · May 10, 2008

Eidos said: ↑

If the contour encloses any poles then the integral is equal to 2pi i times the sum of residues.

No the residue theorem requires that the function to be integrated be holomorphic in a neighbirhood of the poles.

Try for example to integrahte the function z-> Log (z) along the unit circle..

Eidos · May 10, 2008

Pere Callahan said: ↑

No the residue theorem requires that the function to be integrated be holomorphic in a neighbourhood of the poles.

Try for example to integrate the function z-> Log (z) along the unit circle..

I was worried about branch lines, they are essential singularities no? That means that the principal part of the Laurent series expansion is infinite. My understanding though is that the co-efficient [tex]a_{-1}[/tex] of the series:
[tex]\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n[/tex] is still the residue, I've probably lead myself astray though :P

I seem to recall having to redefine the contour so as to exclude any branch lines. In that case I am at a loss as to how to solve the OP's problem. My approach was to make use of the binomial series of the denominators and try to get the [tex]a_{-1}[/tex] term that way.

natski · Jul 16, 2008

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?

natski · Jul 24, 2008

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

Eidos · Jul 24, 2008

My understanding about how to do a contour integral with branch lines is you define a "cut" in the contour which completely eliminates the branch altogether. You then split up your contour so that it looks like:

[tex]\oint=\int_{cut} + \int_{loop\, sans\, cut} [/tex]

The closed path is easy since we can use the residue theorem/Cauchy's integral formula's. You just have to choose the loop without the cut in such a way that it is amenable to solving (like making it all real or some such thing).

I hope that that is of some use. Im sure someone else can be more concise.
Good luck!

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