# Fourier transform integration using well-known result

1. Nov 11, 2016

### binbagsss

Problem

F denotes a forward fourier transform, the variables I'm transforming between are x and k
- See attachment

Relevant equations

So first of all I note I am given a result for a forward fourier transform and need to use it for the inverse one.

The result I am given to use, written out is :

$\int e^{-ikx} e^{-\alpha x^{2}} dx = \frac{1}{2\alpha} e^{-k^{2}/4\alpha}$

I note that $F(k) C(t)$ gives me a function of k, so I apply $F^{-1}$ on this I get a function of x.

Attempt:

My thoughts are I'm looking to change signs in my exponential terms so that it is , effectively a forward transform and then use the result, so just thinking of it as a integration result rather than a particular fourier transform.

However if I do this my exponential terms are:

$e^{-(-ikx+\alpha k^{2})}$

, I don't know how I can then apply the result, completing the square looks like the only candidate to me , but this seems like it will be too scrappy, in particular with 'i' terms, how do I deal with this $e^{ikx}$ term, if I'm on the right lines, is completion of the square necessary or is there some other approach to being able to use this result ?

Many thanks in advance.

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2. Nov 11, 2016

### blue_leaf77

Not too scrappy, it will only give you an extra constant phase factor.
Anyway, you can just use the forward transform as a "template" with which you do the back transform. Note that in the RHS of the equation you are given, the expression is real so the sign of i should not matter.

3. Nov 12, 2016

### binbagsss

No sorry I'm still totally stuck, I think I see the logic in your arguement that the sign of i shouldn't matter but I don't see how I can apply the result given if the sign is wrong.

Any more hints anyone?

Cheers

4. Nov 12, 2016

### vela

Staff Emeritus
How about making the change of variables $k' = -k$?

5. Nov 12, 2016

### lurflurf

From your given transform you should see at one something reminiscent of the quadratic formula for Re(a)>0
Which as you say relates to completing the square
$$\int_{-\infty}^\infty\! \exp\left(-a x^2+2b x+c\right) \,\mathrm{d}x=\\ \exp\left(\frac{b^2}{a}+c\right)\int_{-\infty}^\infty\! \exp\left(-a (x-b/a)^2\right) \,\mathrm{d}x=\\ \exp\left(\frac{b^2}{a}+c\right)\sqrt{\frac{\pi}{a}}$$
Notice the first integral cares what b is, but the second does not since the effect of b (and c) has been moved outside.
See that since b is squared in the result -(-i)^2= -i^2=1 and the sign does not matter

Last edited: Nov 12, 2016
6. Nov 12, 2016

### blue_leaf77

Even if you were to change the sign of $i$ or equivalently the sign of $k$ as suggested by vela above, you will get
$$\int e^{ikx} e^{-\alpha x^{2}} dx = \frac{1}{2\alpha} e^{-k^{2}/4\alpha}$$
If you are still wondering why this is true, try expanding $e^{ikx}$ into Cartesian form. The imaginary part of the integral that contains $\sin kx$ vanishes because the product with $e^{-\alpha x^{2}}$ results in an odd function and you integrate it over entire axis.

7. Nov 12, 2016

### binbagsss

yeh that's all fine.

I don't know what result you've used in the last equality though, it wasn't using the result quoted in the OP which I'm trying to get at.
Anyway so completing the square I get:

$e^{(-1/D)\frac{x^{2}D^{2}}{4}} \int e^{-1/D(k-ixD/2)^{2}} dk$

In order to then use the result given I need to write it in a similar complete square form in order to make a comparison...pretty sure this isn't what my lecturer was hinting at...

8. Nov 12, 2016

### blue_leaf77

What about changing the integration variable?

9. Nov 13, 2016

### lurflurf

• Member warned about providing too much help
I don't know what Fourier convention is in use looks like a typo
I have
$$F(k)\left[e^{-\alpha x^2}\right]= \int_{-\infty}^\infty\! \exp\left(-i k x-\alpha x^2\right) \,\mathrm{d}x= \sqrt{\frac{\pi}{\alpha}}\exp\left(\frac{-k^2}{4\alpha}\right)$$
let 1/alpha=4D t
to see
$$\frac{1}{\sqrt{4D \pi t}}F(k)\left[e^{-\frac{ x^2}{4D t}}\right]= \frac{1}{\sqrt{4D \pi t}}\int_{-\infty}^\infty\! \exp\left(-i k x-\frac{ x^2}{4D t}\right) \,\mathrm{d}x= \exp\left(-D k^2t\right)$$
substitute it in
$$\frac{Q}{2\pi}\int_{-\infty}^\infty\! \exp\left(i k x-D k^2t\right) \,\mathrm{d}x=QF^{-1}(x)\left[e^{-D k^2t}\right]=QF^{-1}(x)\left[\frac{1}{\sqrt{4D \pi t}}F(k)\left[e^{-\frac{ x^2}{4D t}}\right]\right]$$
cancel the forward and inverse transforms to obtain the result

10. Nov 13, 2016

### binbagsss

exactly what i was looking for,- thank you.