# Integrating derivate of Dirac: Where is my fault?

1. Jul 6, 2010

### divB

Hi,

I want to do the following calculation:

$$c_{m,n} = \int_{-\infty}^{\infty} t^m \phi(t-n)\,dt$$

I know three things:
• I know the values of $$c_{m,n}$$ for m={0,1} and the first 4 for m={2,3}
• I know that my $$\phi$$ is symmetric, i.e. $$\phi(t) = \phi(-t)$$
• The fourier transform of $$\phi(t)$$, i.e. $$\Phi(\omega)$$

This is how I tried solving the problem:

$$c_{m,n} = \int t^m \phi(t-n)\,dt = \int t^m \phi(-(n-t))\,dt$$

because of the symmetry

$$= \int t^m \phi(n-t)\,dt$$

...is exactly the definition of a convolution...

$$(t^m * \phi)(n) = \mathcal{F}^{-1}\left\{ \mathcal{F}\left\{t^m\right\} \Phi(\omega)\right\} = \mathcal{F}^{-1}\left\{ j^m \sqrt{2\pi} \delta^{(n)}(\omega) \Phi(\omega) \right\} = j^m \sqrt{2\pi} \mathcal{F}^{-1}\left\{\delta^{(n)}(\omega) \Phi(\omega) \right\}$$

Now, writing the inverse fourier transform gives:

$$j^m \sqrt{2\pi} \frac{1}{\sqrt{2\pi}} \int_{-\pi}^{\pi} \delta^{(n)}(\omega) \Phi(\omega) e^{j\omega n}\,d\omega = j^m \int_{-\infty}^{\infty} \delta^{(n)}(\omega) \underbrace{\Phi(\omega) e^{j\omega n}}_{f(\omega)}\,d\omega$$

Now it is well known that

$$\int \delta^{(n)}(x) f(x)\,dx = (-1)^n f^{(n)}(0)$$

so that

$$j^m \int_{-\infty}^{\infty} \delta^{(n)}(\omega) f(\omega)\,d\omega = j^m (-1)^m \left. \frac{\partial^m}{\partial \omega^m} f(\omega) \right|_{\omega = 0}$$

So the whole procedure should reduce to calculate the derivative of $$f(\omega)$$ and set the result to zero. I fill in my known $$\Phi(\omega)$$:

$$f(\omega) = \mathrm{sinc}^2 (\omega/2) \sqrt{ 1 - \frac{2}{3} \sin^2(w/2)} \cdot e^{j\omega n}$$

Problem 1: Even the 0th derivative is only computable via a limit. Calculating the limit with $$\omega \rightarrow 0$$ gives the result:

$$\sqrt{\frac{2}{2+\cos(2)}}$$

But this is not true. The result should be 1 for $$m=0$$.

The result for $$m=1$$ should be linear: {0,1,2,3,...}. But as you can verifiy the result with my approach is constant.

Also the result for m={2,3} should give a a quadratic and cubic function. But with my approach it stays constant; Even worse, the solution is not always real!

.... So there is anywhere a mistake. Can anybody help me where?

Thank you,
divB

2. Jul 6, 2010

### Mute

You calculated the limit of $f(\omega)$ incorrectly. f(0) = 1, as is easily seen by just plugging in $\omega=0$. (sinc(x) has a removable singularity at x = 0, so sinc(0) = 1, and you don't have to bother with any limit calculations). I would suspect that you're making similar mistakes when calculating the derivatives and setting $\omega=0$. Your reasoning up to that point otherwise looked okay.

The first derivative admittedly doesn't look pretty, and you may have to take some limits to evaluate it and the other; if so, since there will probably just be an isolated term that is indeterminate, you can calculate its limit separately from the other stuff. (e.g., if you were to calculate f(0) by a limit, you can safely set $\omega=0$ in the exponential and sine, and then use L'Hopital on the sinc function).

I used wolframalpha to do the first derivative, and it looks like the result will indeed be n.

Last edited: Jul 6, 2010
3. Jul 7, 2010

### divB

Indeed, thank you for the hint!

For m=0:

But actually this is what I tried first: Just plugging in the 0 gave 1. But then I recalled my Mathematic courses: You must not do it this way. If you encounter any $\infty$ or 0 in the denominator, you better should calculate limits...

So I did this. Can you tell me why Mathematica gives me a different result? I thought when taking the limit the result should be the same; and if it is not, there is a problem with the direct way...

Additionally I plotted the function in Mathematica (without the exponential because this will be 1 anway). As you can see the value at 0 is not 1 but 1.37 ... the same stuff as Mathematica gave symbolically...

I have attached a screenshot.

Now to m=1: Indeed, when I use Wolfram Alpha there is a huge amount of expressions but when I just plug in zero, there seems to stay just n.

Now I tried also something which I did not yesterday: I did the second derivative in Mathematica. Again, just setting $$\omega=0$$ gives an error. But using the limit, I get

$$\lim_{\omega \rightarrow 0} \frac{\partial}{\partial\omega} f(\omega) = j n \sqrt{\frac{3}{2+\cos(2)}}$$

The j cancels out, so there stays only the n. But there is again the root which should be 1 ....

Who is right? And why? Does really Mathematica do a mistake when calculating a limit?

Regards,
divB

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4. Jul 7, 2010

### losiu99

Haven't looked at the whole problem, but it seems the argument of sine in the first screen is "1" instead of "w/2". If so, Mathematica is right.
Moreover, in the plot command, there's "Sin[w/w]" instead of "Sin[w/2]".

5. Jul 7, 2010

### divB

Yessss! That was it! Thank you very much!! Plot is not @1.

And:

Code (Text):

In[11]:=
Limit[D[ToDiff[w],{w,0}],w\[Rule]0]

Out[11]=
1

In[12]:=
Limit[D[ToDiff[w],{w,1}],w\[Rule]0]

Out[12]=
j n

In[13]:=
Limit[D[ToDiff[w],{w,2}],w\[Rule]0]

Out[13]=
\!$$j\^2\ n\^2$$

In[14]:=
Limit[D[ToDiff[w],{w,3}],w\[Rule]0]

Out[14]=
\!$$j\^3\ n\^3$$

[...]

Thank you very much!!! :-)

Regards
divB