What is the Integral of Modified Bessel Function using Integral Representation?

[qasdc]

Homework Statement

I need to evaluate the following integral:
\int_{0}^{\infty} dk K_{0}(kr)
, where K_{0}(x) is the modified Bessel, using the integral representation:
K_{0}(x)=\int_{0}^{\infty} dt \frac{cos (xt)}{ \sqrt{t^2 +1}}

Homework Equations

The Attempt at a Solution

 

[fzero]

This is a standard integral that can be found in tables. If you want to explicitly do the integral, you should write

\cos xt = \frac{1}{2} \lim_{\epsilon \rightarrow 0} \left( e^{-\epsilon x^2 +i xt} + e^{-\epsilon x^2 -i xt} \right)

to improve the convergence of the x integral. You will be able to set \epsilon = 0 after making a change of variables for the t integral without encountering any divergences.

 

[qasdc]

fzero thank you for your reply.
Yes, I need to do the intergal step by step, otherwise I could get it from mathematica.
The answer I get form mathematica is \frac{\pi}{2r}
In your reply, I suppose that the first term in each exponential is - \epsilon t^2 rather than - \epsilon x^2.
How should I proceed after the substitution for \cos (xt)?

 

[fzero]

fzero thank you for your reply.
Yes, I need to do the intergal step by step, otherwise I could get it from mathematica.
The answer I get form mathematica is \frac{\pi}{2r}
In your reply, I suppose that the first term in each exponential is - \epsilon t^2 rather than - \epsilon x^2.
How should I proceed after the substitution for \cos (xt)?

I did the x integration first, so I wanted those to be Gaussian integrals, hence you need - \epsilon x^2, not - \epsilon t^2. The x integral is then a Gaussian (you need to complete a square), you write the results down in terms of \epsilon. The t-integral also becomes a Gaussian, but you can simplify it by changing variables and taking \epsilon to zero.

 

[qasdc]

So,
\int_{0}^{\infty} dk K_{0}(kr)=\frac{1}{r}\int_{0}^{\infty} dx K_{0}(x)=\frac{1}{r}\int_{0}^{\infty} dx \int_{0}^{\infty} dt \frac{\cos (xt)}{\sqrt{t^2 +1}}=
=\frac{1}{4r}\int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dt \frac{\cos (xt)}{\sqrt{t^2 +1}}= \frac{1}{4r}\int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dt \frac{1}{\sqrt{t^2 +1}}\frac{1}{2} \lim_{\epsilon \rightarrow 0} \left( e^{-\epsilon x^2 +i xt} + e^{-\epsilon x^2 -i xt} \right)=
=\frac{1}{4r} \lim_{\epsilon \rightarrow 0} \frac{\sqrt{\pi}}{\sqrt{\epsilon}}\int_{-\infty}^{\infty}dt \frac{e^{\frac{-t^2}{4\epsilon}}}{\sqrt{t^2+1}}

Now, at this point I do not know how to proceed. Any help?

 

[fzero]

Change variables to s=t/(2\sqrt{\epsilon}).

 

[qasdc]

I found it.
Using the following definition of the delta function:
\lim_{\epsilon \rightarrow 0} \frac{1}{\sqrt{\epsilon}}e^{-\frac{t^2}{4\epsilon}} =2\sqrt{\pi}\delta(t)
we find that,
<br /> =\frac{1}{4r} \lim_{\epsilon \rightarrow 0} \frac{\sqrt{\pi}}{\sqrt{\epsilon}}\int_{-\infty}^{\infty}dt \frac{e^{\frac{-t^2}{4\epsilon}}}{\sqrt{t^2+1}} =\frac{\sqrt{\pi}}{4r} \int_{-\infty}^{\infty}dt \frac{1}{\sqrt{t^2+1}}2\sqrt{\pi}\delta(t) = \frac{\pi}{2r} <br />

Thanks a lot for your help fzero!