Fourier transform integration using well-known result

[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.

 

[blue_leaf77]

but this seems like it will be too scrappy,

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.

 

[binbagsss]

No sorry I'm still totally stuck, I think I see the logic in your argument 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

 

[vela]

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

 

[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

 

[blue_leaf77]

No sorry I'm still totally stuck, I think I see the logic in your argument 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

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.

 

[binbagsss]

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

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...

 

[blue_leaf77]

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...

What about changing the integration variable?

 

[lurflurf]

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

 

[binbagsss]

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

exactly what i was looking for,- thank you.