# Fourier Transform Calculations

1. Feb 25, 2010

### Hart

1. The problem statement, all variables and given/known data

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).

2. Relevant equations

Through use of Dirac Delta Function:

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

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

3. 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

2. Feb 25, 2010

### gabbagabbahey

Good....

You've already got a mistake here.

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

So,

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

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?

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

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.....

3. Feb 26, 2010

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

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

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

4. Feb 26, 2010

### gabbagabbahey

Good

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)

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.

5. Feb 26, 2010

### 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?

6. Feb 26, 2010

### 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.

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

7. Feb 26, 2010

### 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

8. Feb 26, 2010

### 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,

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

9. Feb 26, 2010

### 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 :|

10. Feb 26, 2010

### gabbagabbahey

You said it yourself:

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

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

11. Feb 27, 2010

### 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?!

12. Feb 27, 2010

### gabbagabbahey

Looks good to me!

13. Feb 27, 2010

### Hart

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

??

Last edited: Feb 28, 2010
14. Mar 1, 2010

### gabbagabbahey

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

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