Convergence of Integral with Real and Imaginary Parameters

[fermi]

The integral given below is to be computed as a function of real variables x and s. Even a partial answer only for s>0 is very useful. Here is the integral:

$$\int_{0}^{\infty}{dk \frac{k^2 e^{-k^2 x^2}}{(k^2 + s)^{3/2}}}$$

Thank you for your help.

 

[ShayanJ]

You should show some attempt at solving it.

 

[fermi]

I tried integration by parts to isolate the exponential in a definite integral; that did not work. I also tried to change variables from k to z, with k=sqrt(s) * tan(z), which greatly simplifies the expression and gets rid of the nasty square root, but this time I have a trigonometric exponential to integrate with. I also noted that the integrand is an even function of k, and the integral can be expanded to be on the entire real axis. I tried doing the new integral as a Contour integral, but again it did not work on account of the exponential term and also on account of the nasty branch cut from the square root.

 

[DarthMatter]

Hi. Mathematica does not know a general integral, so maybe you should look somewhere else. How about doing the old integral as a contour integral?

 

[FactChecker]

Do you have a reason to think that it has a closed form solution?

 

[DarthMatter]

What I meant that Mathematica could not give an antiderivative for this integrand, but the definite integral still may be calculatable using complex analysis.

 

[ShayanJ]

I expanded the exponential and then interchanged the order of summation and integration, and I got:
$$
\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!} \int_{0}^\infty \frac{k^{2n+2}}{(k^2+s)^{\frac 3 2}} dk
$$
But I'm not sure how to deal with the integrals!
As you can see, its an even function of k so maybe it can be contour integrated but I have problem with it.

 

[DarthMatter]

I'm not sure. I think the problem is that the exponential explodes when you insert imaginary numbers. Maybe you could try not countour-integrating ## \exp(-z^2 x^2) ## but ## \exp(-|z|^2 x^2) ##. On the real axis, which is the part you are interested in, it should not matter.

 

[FactChecker]

I think the e^-k2x2 needs to stay inside the integration to make it converge.

 

[FactChecker]

I'm not sure. I think the problem is that the exponential explodes when you insert imaginary numbers. Maybe you could try not countour-integrating ## \exp(-z^2 x^2) ## but ## \exp(-|z|^2 x^2) ##. On the real axis, which is the part you are interested in, it should not matter.

But I think ## \exp(-|z|^2 x^2) ## will not be an analytic function, so the contour integration would not be valid.

 

[DarthMatter]

Probably. But maybe a better analytic function can be found.

 

[fermi]

I expanded the exponential and then interchanged the order of summation and integration, and I got:
$$
\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!} \int_{0}^\infty \frac{k^{2n+2}}{(k^2+s)^{\frac 3 2}} dk
$$
But I'm not sure how to deal with the integrals!
As you can see, its an even function of k so maybe it can be contour integrated but I have problem with it.

I am happy to get an answer in terms of a power series if a closed form answer cannot be found. However, the power series you suggested has a problem for n=0 term in the series. For n=0, the integral:
$$
\int_{0}^\infty \frac{k^{2n+2}}{(k^2+s)^{\frac 3 2}} dk
$$
does not converge. This sort of divergence can happen when you expand the integrand in power series, even when the whole integral is strongly convergent. The integral I am trying to evaluate is strongly convergent for s > 0 (That's easy to prove). In fact, it is likely to be convergent for all Real s, by allowing the variable 's' to acquire a small positive imaginary part, and taking the limit that the imaginary part goes to zero after integration. (But that's much harder to prove. It feels intuitively right, but I have no proof yet for s<=0.)