Can You Solve This Fourier Transform Problem?

[Hart]

Homework Statement

The Fourier transform of a function f(x) is given by the product of the Fourier transforms of $$cos(\alpha x)$$ and $$e^{-|x|}$$;

$$f^{~} = F^{~}\left[cos[\alpha x]\right]F^{~}\left[e^{-|x|}\right]$$

Find f(x) and show that it can be written as a real function.

Note: Do not use the Convolution Theorem, instead calculate f(x) by inverse Fourier transforming f(k).

Homework Equations

Through use of Dirac Delta Function:

• $$\delta(\alpha - k) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ix(\alpha-k)}dx$$
  
  
• $$\delta(-\alpha - k) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-ix(\alpha+k)}dx$$

The Attempt at a Solution

Since the solution is the product of two Fourier transforms, can calculate them seperately (?!)

Firstly..

$$F\left[cos(\alpha x)\right] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}cos(\alpha x)e^{ikx}dx$$

From Euler's Formula:

$$cos(\alpha x) = \frac{1}{2}\left[e^{i\alpha x}+e^{-i\alpha x}\right]$$

Therefore:

$$F\left[cos(\alpha x)\right] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{1}{2}\left[e^{i\alpha x}+e^{-i\alpha x}\right]e^{ikx}dx$$

Hence:

$$F\left[cos(\alpha x)\right] = \frac{1}{2\sqrt{2\pi}}\int_{-\infty}^{\infty}\left[e^{i(\alpha -k) x}+e^{-i(\alpha +k) x}\right]dx$$

Then by using Dirac Delta Functions (already stated):

$$F\left[cos(\alpha x)\right] = \frac{1}{2 \sqrt{2 \pi}}\left[ 2 \pi \delta(\alpha -k) + 2 \pi \delta (-\alpha -k)\right]dx$$

.. and after some rearranging:

$$F\left[cos(\alpha x)\right] = \sqrt{\frac{\pi}{2}}\left[\delta(\alpha -k) + \delta (\alpha +k)\right]dx$$

Secondly..

$$F\left[e^{-|x|}\right] = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$$

$$F\left[e^{-|x|}\right]= \frac{1}{\sqrt{2 \pi}}\left[\int_{-\infty}^{0}f(x)e^{(1+ik)x}dx+\int_{0}^{\infty}f(x)e^{-(1+ik)x}dx \right]$$

Which then after input of integration limits:

$$F\left[e^{-|x|}\right]=\frac{1}{\sqrt(2\pi)}\left[-e^{(1+ik)x} + e^{-(1+ik)x}\right]$$

.. and simplifies to:

$$F\left[e^{-|x|}\right]= -\sqrt{\frac{2}{\pi}}sinh\left((1+ik)x\right)$$

..

So that's the calculations of the two parts (hopefully correctly), not sure how to now use these tho to calculate the actual value of the overall Fourier transform f(x)?

Any help and advice with any or all of this would be great! =D

 

[gabbagabbahey]

$$F\left[cos(\alpha x)\right] = \sqrt{\frac{\pi}{2}}\left[\delta(\alpha -k) + \delta (\alpha +k)\right]dx$$

Good...

Secondly..

$$F\left[e^{-|x|}\right] = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$$

$$F\left[e^{-|x|}\right]= \frac{1}{\sqrt{2 \pi}}\left[\int_{-\infty}^{0}f(x)e^{(1+ik)x}dx+\int_{0}^{\infty}f(x)e^{-(1+ik)x}dx \right]$$

You've already got a mistake here.

$$e^{-|x|}=\left\{ \begin{array}{lr} e^{x} & x<0 \\ e^{-x} & x>0 \end{array}\.$$

So,

$$F\left[e^{-|x|}\right]= \frac{1}{\sqrt{2 \pi}}\left[\int_{-\infty}^{0}e^{(1-ik)x}dx+\int_{0}^{\infty}e^{-(1+ik)x}dx \right]$$

Which then after input of integration limits:

$$F\left[e^{-|x|}\right]=\frac{1}{\sqrt(2\pi)}\left[-e^{(1+ik)x} + e^{-(1+ik)x}\right]$$

.. and simplifies to:

$$F\left[e^{-|x|}\right]= -\sqrt{\frac{2}{\pi}}sinh\left((1+ik)x\right)$$

If you've substituted your integration limits, why are there still $x$'s in this expression? Also, shouldn't there be some pre-factors in front of your exponentials?

$$\frac{d}{dx} e^{(1-ik)x}=(1-ik)e^{(1-ik)x} \neq e^{(1-ik)x}$$

So that's the calculations of the two parts (hopefully correctly), not sure how to now use these tho to calculate the actual value of the overall Fourier transform f(x)?

Any help and advice with any or all of this would be great! =D

Well, you are told that $f(k)= F\left[cos(\alpha x)\right]F\left[e^{-|x|}\right]$ and you know that

$$f(x)=F^{-1}[f(k)]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(k)e^{ikx}dk$$

Soooo...

 

[Hart]

Attempting the second Fourier transform again..


$$F\left[e^{-|x|}\right] = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$$

$$F\left[e^{-|x|}\right] = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}e^{-|x|}e^{-ikx}dx$$


$$F\left[e^{-|x|}\right]= \frac{1}{\sqrt{2 \pi}}\left[\int_{-\infty}^{0}e^{(1-ik)x}dx+\int_{0}^{\infty}f(x)e^{-(1+ik)x}dx \right]$$

Then integrating:

$$F\left[e^{-|x|}\right]= \frac{1}{\sqrt{2 \pi}}\left[\left(\frac{e^{(1-ik)x}}{(1-ik)}\right)\right|^{0}_{-\infty} + \left(\frac{e^{-(1+ik)x}}{-(1+ik)}\right)\right|^{\infty}_{0} \right]$$

Which then after input of integration limits:

$$F\left[e^{-|x|}\right]= \frac{1}{\sqrt{2 \pi}}\left[\left(\frac{1}{(1-ik)}\right) + \left(\frac{-1}{-(1+ik)}\right)$$

Then after some rearranging..

$$F\left[e^{-|x|}\right]= \frac{1}{\sqrt{2 \pi}}\left(\frac{2}{(1+k^{2})}\right)$$

Therefore:

$$F\left[e^{-|x|}\right]= \sqrt{\frac{2}{\pi}}\left(\frac{1}{(1+k^{2})}\right)$$

.. which I think is more correct now?


So:

$$f(k)= F\left[cos(\alpha x)\right]F\left[e^{-|x|} \right]$$

$$f(k)= \left[\sqrt{\frac{\pi}{2}}\left[\delta(\alpha -k) + \delta (\alpha +k)\right]dx \right] \left[\sqrt{\frac{2}{\pi}}\left(\frac{1}{(1+k^{2})}\right) \right]$$

$$f(k)= \left(\frac{\delta(\alpha -k) + \delta (\alpha +k)}{(1+k^{2})}\right)dx$$

Then:
$$f(x)=F^{-1}[f(k)]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(k)e^{ikx}dk$$

$$f(x)=F^{-1} \left[ \left(\frac{\delta(\alpha -k) + \delta (\alpha +k)}{(1+k^{2})}\right)dx \right]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(\left(\frac{\delta(\alpha -k) + \delta (\alpha +k)}{(1+k^{2})}\right)dx\right)e^{ikx}dk$$

$$\int_{-\infty}^{\infty}\left(\left(\frac{\delta(\alpha -k) + \delta (\alpha +k)}{(1+k^{2})}\right)dx\right)e^{ikx}dk$$

.. not sure what I do with this expression now :S

 

[gabbagabbahey]

$$F\left[e^{-|x|}\right]= \sqrt{\frac{2}{\pi}}\left(\frac{1}{(1+k^{2})}\right)$$

.. which I think is more correct now?

Good

So:

$$f(k)= F\left[cos(\alpha x)\right]F\left[e^{-|x|} \right]$$

$$f(k)= \left[\sqrt{\frac{\pi}{2}}\left[\delta(\alpha -k) + \delta (\alpha +k)\right]dx \right] \left[\sqrt{\frac{2}{\pi}}\left(\frac{1}{(1+k^{2})}\right) \right]$$

$$f(k)= \left(\frac{\delta(\alpha -k) + \delta (\alpha +k)}{(1+k^{2})}\right)dx$$

Why is there a $dx$ in there? $F[\cos(\alpha x)]$ is a function, not a differential. (I missed that you had a dx in your original post, it shouldn't be there)

Then:
$$f(x)=F^{-1}[f(k)]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(k)e^{ikx}dk$$

$$f(x)=F^{-1} \left[ \left(\frac{\delta(\alpha -k) + \delta (\alpha +k)}{(1+k^{2})}\right)dx \right]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(\left(\frac{\delta(\alpha -k) + \delta (\alpha +k)}{(1+k^{2})}\right)dx\right)e^{ikx}dk$$

.. not sure what I do with this expression now :S

Without the errant $dx$, you have

$$f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(\frac{\delta(\alpha -k) + \delta (\alpha +k)}{(1+k^{2})}\right)e^{ikx}dk$$

Now, split the integral into two and use the defining property of the delta function to integrate each piece.

 

[Hart]

I get about splitting that integral up into two parts, hence the integrals, i.e.

$$f(x)=\frac{1}{\sqrt{2\pi}}\left[\int_{-\infty}^{0}\left(\frac{\delta(\alpha -k) + \delta (\alpha +k)}{(1+k^{2})}\right)e^{ikx}dk + \int_{0}^{\infty}\left(\frac{\delta(\alpha -k) + \delta (\alpha +k)}{(1+k^{2})}\right)e^{ikx}dk \right]$$


Defining properties of delta function I have found:

[1] $$\delta(x) = 0$$ for $$x \neq 0$$

[2] $$\int_{-\infty}^{\infty} f(x) \delta \left( x-a \right) dx = f(a)$$

[3] $$\int_{-\infty}^{\infty} \delta \left( x-a \right) dx = 1$$

[4] $$\delta \left(-x \right) = \delta \left(x \right)$$

.. can't see how these are to be used though, sorry bit more help?

 

[gabbagabbahey]

That's not what I meant by splitting up the integral;

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(\frac{\delta(\alpha -k)+\delta(\alpha +k)}{(1+k^{2})}\right)e^{ikx}dk=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(\frac{ \delta (\alpha -k)}{(1+k^{2})}\right)e^{ikx}dk+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(\frac{\delta (\alpha +k)}{(1+k^{2})}\right)e^{ikx}dk$$

Use the second property you've listed above to evaluate each integral.

$$\int_{-\infty}^{\infty} f(x) \delta \left( x-a \right) dx = f(a)\implies\int_{-\infty}^{\infty} g(k) \delta \left( k-a \right) dk = g(a)$$

 

[Hart]

Still not quite sure, but will give it a go! ..

$$\int_{-\infty}^{\infty} f(x) \delta \left( x-a \right) dx = f(a)\implies\int_{-\infty}^{\infty} g(k) \delta \left( k-a \right) dk = g(a)$$

So for the first integral:

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(\frac{ \delta (\alpha -k)}{(1+k^{2})}\right)e^{ikx}dk$$

$$\int_{-\infty}^{\infty} f(x) \delta \left( x-a \right) dx = f(a)\implies\int_{-\infty}^{\infty} \delta(k- \alpha) \left( \frac{e^{ixk}}{(1+k^{2})} \right)dk = f(\alpha)$$

.. is this what you mean to do? :S

 

[gabbagabbahey]

You don't seem to be getting it...For the first integral, you are integrating an ordinary function of $k$, $$\inline{\frac{e^{ixk}}{(1+k^{2})}}$$ , times a delta function of $k$, $\delta(k-\alpha)$ , over all values of $k$...Property (2) of the delta function (as listed a few posts ago) tell you that the result is simply that function $$\inline{\frac{e^{ixk}}{(1+k^{2})}}$$ evaluated at $k=\alpha$.

In other words,

$$\int_{-\infty}^{\infty} g(k) \delta \left( k-a \right) dk = g(a)\implies\int_{-\infty}^{\infty} \delta(k- \alpha) \left( \frac{e^{ixk}}{(1+k^{2})} \right)dk = \frac{e^{i\alpha x}}{(1+\alpha^{2})}$$

 

[Hart]

I didn't get how it changes from $$\delta(\alpha-k)$$ to $$\delta(k-\alpha)$$.

Right, so:

$$\frac{1}{\sqrt{(2 \pi)}}\int_{-\infty}^{\infty}\left( \frac{e^{ikx}}{(1+k^{2})} \right)\delta(k- \alpha)dk = \frac{e^{i\alpha x}}{\sqrt{2 \pi}(1+\alpha^{2})}$$

Then for the second integral, a similar method, as the only difference is the $$(\alpha+k})$$ term instead of $$(\alpha-k})$$ term.

But then know that:

$$\delta \left(-x \right) = \delta \left(x \right)$$

So presumably can then use definition [2] as used in the first integral, thus giving the same result as for the first integral, but just negative:

$$- \frac{e^{i\alpha x}}{\sqrt{2 \pi}(1+\alpha^{2})}$$

.. but then that will give an overall result of 0.. ah :|

 

[gabbagabbahey]

I didn't get how it changes from $$\delta(\alpha-k)$$ to $$\delta(k-\alpha)$$.

You said it yourself:

But then know that:

$$\delta \left(-x \right) = \delta \left(x \right)$$

$$\delta \left(-x \right) = \delta \left(x \right) \implies \delta(\alpha-k)=\delta(-(\alpha-k))=\delta(k-\alpha)$$

So presumably can then use definition [2] as used in the first integral, thus giving the same result as for the first integral, but just negative:

$$- \frac{e^{i\alpha x}}{\sqrt{2 \pi}(1+\alpha^{2})}$$

No, that's not what you should get; $\delta(k+\alpha)=\delta(k-(-\alpha))$.

 

[Hart]

Thanks for just clarifying that minor confusion, I just overlooked it before but I do understand that now.

So for the second integral:

$$\frac{1}{\sqrt{(2 \pi)}}\int_{-\infty}^{\infty}\left( \frac{e^{ikx}}{(1+k^{2})} \right)\delta(k+ \alpha)dk = \frac{1}{\sqrt{(2 \pi)}}\int_{-\infty}^{\infty}\left( \frac{e^{ikx}}{(1+k^{2})} \right)\delta(k - (-\alpha))dk = \frac{e^{-i\alpha x}}{\sqrt{2 \pi}(1+\alpha^{2})}$$

.. getting somewhere?!

 

[gabbagabbahey]

Thanks for just clarifying that minor confusion, I just overlooked it before but I do understand that now.

So for the second integral:

$$\frac{1}{\sqrt{(2 \pi)}}\int_{-\infty}^{\infty}\left( \frac{e^{ikx}}{(1+k^{2})} \right)\delta(k+ \alpha)dk = \frac{1}{\sqrt{(2 \pi)}}\int_{-\infty}^{\infty}\left( \frac{e^{ikx}}{(1+k^{2})} \right)\delta(k - (-\alpha))dk = \frac{e^{-i\alpha x}}{\sqrt{2 \pi}(1+\alpha^{2})}$$

.. getting somewhere?!

Looks good to me!

So, your final result is...?

 

[Hart]

$$\frac{e^{i\alpha x}}{\sqrt{2 \pi}(1+\alpha^{2})} + \frac{e^{-i\alpha x}}{\sqrt{2 \pi}(1+\alpha^{2})} = \frac{1}{\sqrt{2\pi}} \left(\frac{e^{0}}{1+\alpha^{2}}\right) = \frac{1}{\sqrt{2\pi}(1+\alpha^{2})} = f(x)$$

??

 

[gabbagabbahey]

$$\frac{e^{i\alpha x}}{\sqrt{2 \pi}(1+\alpha^{2})} + \frac{e^{-i\alpha x}}{\sqrt{2 \pi}(1+\alpha^{2})} = \frac{1}{\sqrt{2\pi}} \left(\frac{e^{0}}{1+\alpha^{2}}\right) = \frac{1}{\sqrt{2\pi}(1+\alpha^{2})} = f(x)$$

??

Huh? Where did $e^0$ come from?

[tex]e^{i\alpha x}+e^{-i\alpha x}=2\cos(\alpha x)\neq e^0[/itex]