# Fourier transform of integral

1. Jun 17, 2011

### lost87

Hey, this is my first post, great forum!! You've really helped me a lot of times.
I have a problem solving an integro-differential equation. It involves a term of the form: integration over [t, +infinity) of f(s)*exp(t-s)ds.
I have to solve the equation using Fourier transform, and most of the other terms are ok, but I have a problem with this one. I know that if the limits of integration where (-infinity, +infinity) it would be a convolution and it's fourier transform would be simple, but I don't know what to do in this case. I tried to find the Fourier transform using the logic of the proof of the convolution theorem, but I end up with a result with t and s. Do you have any ideas on what to do? Any ideas and help would be really welcome!
Thanks!!!

2. Jun 17, 2011

### Hootenanny

Staff Emeritus
Hi lost87 and welcome to Physics Forums.

Could you perhaps post the full integro-differential equation so that we can take a look at it?

3. Jun 17, 2011

### MisterX

So, what you are describing is like convolution, only integrating over s>=t. I think there is way you could restate that integral as an equivalent integral from negative infinity to infinitiy by using a different function in place of f(s).

4. Jun 17, 2011

### lost87

Hey, the equation has the form: f'=af(t)+b(integral of f(s)*exp(t-s) ds from t to +infinity) +cf''+e(integral of f(s)*exp(-(t-s)) ds from -infinty to t). I can find the Fourier transform for most of the terms, but I don't know what to do with the terms with the integral. I hope this helps.

5. Jun 17, 2011

### lost87

Hey, thanks for your reply, but I am not sure that I understand what you are suggesting. Will that involve a change of the variables s, t?

6. Jun 17, 2011

### MisterX

No, I was suggesting replacing f(s) with a different function of s, so that the bounds of the integral could be changed to from -infinity to infinity, and the integral would be equal to the integral you described.

Is this a homework problem?

Last edited: Jun 17, 2011
7. Jun 18, 2011

### lost87

Hey, thanks again for your reply. No it's not a homework problem, it's just something I am working on in my dissertation

8. Jun 18, 2011

### MisterX

u(s-t) evaluates to zero when s < t

$\int^{\infty}_{-\infty}f(s)u(s-t)e^{t-s}ds = \int^{t}_{-\infty}f(s)u(s-t)e^{t-s}ds + \int^{\infty}_{t}f(s)u(s-t)e^{t-s} = \int^{\infty}_{t}f(s)e^{t-s}ds$
as
$\int^{t}_{-\infty}f(s)u(s-t)e^{t-s}ds = 0$

Now, to get the Fourier transform of $\int^{\infty}_{-\infty}f(s)u(s-t)e^{t-s}ds$, we have to figure out what two functions [of t] are being convolved. I hadn't thought this all the way through so I may have been misleading when I suggested replacing f(s). Actually, we'll have to group the unit step with the exponential to use the convolution theorem.

Let
$g(\tau) = u(-\tau)e^{\tau}$
so that
$g(t-s) = u(s-t)e^{t-s}$

and
$\int^{\infty}_{-\infty}f(s)u(s-t)e^{t-s}ds = \int^{\infty}_{-\infty}f(s)g(t-s)ds$

so

$\int^{\infty}_{-\infty}( \int^{\infty}_{t}f(s)e^{t-s}ds)e^{-j\omega t}dt = \mathcal{F}\left\{ \int^{\infty}_{t}f(s)e^{t-s}ds\right\} = \mathcal{F}\left\{ f(t) * g(t) \right\}$

$= \mathcal{F}\left\{ f(t) \right\}\mathcal{F}\left\{ g(t) \right\} = \mathcal{F}\left\{ f(t) \right\} \mathcal{F}\left\{ u(-t)e^{t} \right\}$

$\mathcal{F}\left\{ u(t)e^{-t} \right\} = \frac{1}{1 + j\omega}$
so
$\mathcal{F}\left\{ u(-t)e^{t} \right\} = \frac{1}{1 - j\omega}$

finally

$\mathcal{F}\left\{ \int^{\infty}_{t}f(s)e^{t-s}ds\right\} = \frac{\mathcal{F}\left\{ f(t) \right\} }{1 - j\omega}$

9. Jun 19, 2011

### lost87

This is great! I can't explain how helpful this has been! Thanks for everything!

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