# I need help evaluating this integral

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1. Dec 23, 2014

### 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:

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

2. Dec 23, 2014

### ShayanJ

You should show some attempt at solving it.

3. Dec 23, 2014

### 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.

4. Dec 24, 2014

### 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?

5. Dec 24, 2014

### FactChecker

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

6. Dec 24, 2014

### 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.

7. Dec 24, 2014

### 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.

8. Dec 24, 2014

### 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.

9. Dec 24, 2014

### FactChecker

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

10. Dec 24, 2014

### FactChecker

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

11. Dec 24, 2014

### DarthMatter

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

12. Dec 24, 2014

### fermi

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.)