# Tricky integration (maybe delta function)

1. Mar 8, 2016

### Kara386

1. The problem statement, all variables and given/known data
I need to integrate

$\frac{A}{2a\sqrt{2\pi}} \int_{-\infty}^{\infty} \frac{e^{ik(x+x')}}{(b^2+k^2)}dk$

I have tried substitution and integration by parts and that hasn't worked. I can see that part of it is the delta function, but I don't really know how to use that fact! I think the delta function is

$\delta(x+x') = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{ik(x+x')}dk$

Which is almost exactly what I have. If it helps, this is from trying to reverse Fourier transform a function. How can I integrate this?

2. Relevant equations

3. The attempt at a solution

2. Mar 8, 2016

### vela

Staff Emeritus
Try a contour integration in the complex plane.

3. Mar 8, 2016

### Kara386

That sounds well beyond anything we've been taught! I've never heard of contour integration, in the complex plane or otherwise... Is that the only method?

4. Mar 8, 2016

### Kara386

Maybe I have the integral wrong, let me double check.

5. Mar 8, 2016

### Ray Vickson

Let $x+x'= y$, so you want to evaluate
$$I(y) = \int_{-\infty}^{\infty} \frac{\exp(iky)}{k^2+b^2} \, dk.$$
Write $\exp(iky) = \cos(ky) + i \sin(ky)$, and note that in the integral the denominator is even in $k$, while $\sin(ky)$ is odd in $k$. Therefore, without any need for computation we conclude that the imaginary part = 0 because the "$\sin$" term integrates to zero for any fixed $y$. Thus, we have
$$I(y) = 2 \int_0^{\infty} \frac{\cos(ky)}{k^2+b^2} \, dk.$$

Assuming we can legitimately differentiate wrt $y$ under the integral sign, we have
$$\begin{array}{rcl}\frac{d^2}{dy^2} I(y) &= &2 \int_0^{\infty} \frac{-k^2 \cos(ky)}{k^2+b^2} \, dk \\ &=& -2 \int_0^{\infty} \frac{(k^2 + b^2 - b^2)\cos(ky)}{k^2 + b^2} \\ &=& - 2 \int_0^{\infty} \cos(ky) \, dy + 2 b^2 \int_0^{\infty} \frac{\cos(ky)}{k^2+b^2} \, dk \\ &=& - 2 \int_0^{\infty} \cos(ky) \, dy + b^2 I(y) \end{array}$$
Thus, you have a DE $I''(y) - b^2 I(y) = f(y)$, where $f(y) = -2 \int_0^{\infty} \cos(ky) \, dk$. Do you recognize $f(y)$ from somewhere?

Of course, to get the full $I(y)$ you need two boundary conditions on $I(y)$, such as values of $I(0)$ and $I'(0)$, and both of these can be obtained from results you have seen already (or can derive on your own).

There remains a serious issue, however: we just carelessly differentiated under the integral sign and further split the resulting integral into two separate integrations. Those manipulations need justification, but for now I will leave it.

6. Mar 8, 2016

### vela

Staff Emeritus
Are you allowed to just look the transform in a table?

7. Mar 9, 2016

### Kara386

Brilliant, thank you!

8. Mar 9, 2016

### Kara386

Maybe! I don't think we've been taught the right techniques to actually evaluate it. I'll ask my lecturer about the tables. Thanks for helping! :)