Tricky integration (maybe delta function)

[Kara386]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

 

[vela]

Try a contour integration in the complex plane.

 

[Kara386]

Try a contour integration in the complex plane.

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?

 

[Kara386]

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

 

[Ray Vickson]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

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.

 

[vela]

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?

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

 

[Kara386]

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.

Brilliant, thank you!

 

[Kara386]

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

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! :)