How can I easily compute the Fourier Transform of a convolution integral?

[muzialis]

Hi there,
I am trying to get some practice with Fourier Transforms, there is a long way to go.
For example, let me consider the function $$ \gamma (t) = \int_{-\infty}^{t} C(t-\tau) \sigma(\tau) \mathrm{d}{\tau}$$
Defining the Fourier Transform as
$$ \gamma(\omega) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \gamma(t) e^{i\omega t} \mathrm{d}t$$
I am supposed to compute with ease that
$$ \gamma(\omega) = \sigma(\omega) \int_0 ^{\infty} C(t) e^{i\omega t} \mathrm{d}t$$,
but I am struggling, because I can not apply the convolution theorem (as the first equation is a convolution only to actual time).
I tried to use the definition writing
$$\gamma(\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{t} C(t-\tau) \sigma(\tau) \mathrm{d}{\tau} e^{i\omega t} \mathrm{d}t$$
hoping to invertt integration order and express the inner integral as a Fourier Transform, but again I am not getting anywhere. I tried a variable change $$t-\tau = u$$ and that helps in changing the integration limits to 0 and infinity, but still does not bring me to the desired result, any advice or hint?
Many thanks as usual

 

[mathman]

You are almost there. e^iωt = e^iω(t-u) e^iωu.
After the change of variables and the order of integration switch you have the result, noting that the integral (-∞,∞) is the Fourier transform of σ.

Further suggestion don't use the same symbol for a function and its Fourier transform.

 

[muzialis]

Mathman,
thanks for your help. I understand your suggestion, I will continue only for this post with the previous notation just not to confuse things further.
I am getting what I would like to, but in a little bit of a shaky way, so I post the calc for a check...
Here is the calculation:

$$\gamma(\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{t} C(t-\tau) \sigma(\tau) \mathrm{d}{\tau} e^{i\omega t} \mathrm{d}t$$
Using the change of variable $$ t - \tau = u$$ one obtains
$$\gamma(\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{0}^{\infty} C(u) \sigma(t-u) \mathrm{d}{u} \, e^{i\omega (t-u)} e^{i \omega u} \mathrm{d}t$$
inverting the order of integration
$$\gamma(\omega) = \frac{1}{2\pi} \int_{0}^{\infty} \int_{-\infty}^{\infty} C(u) \sigma(t-u) \, e^{i\omega (t-u)} e^{i \omega u} \mathrm{d}t \mathrm{d}{u}$$
and the integrals can be separated now, the one between infinite limits of integration leading the Fourier transform of $\sigma$ (upon a trivial change of variables $$ t - u = z$$ at constant u, is this corrrect?), the final result being as desired
$$\gamma (\omega) = \sigma(\omega) \int_{0}^{\infty} C(u)e^{i\omega u} \mathrm{d}u$$ Many thanks

 

[mathman]

It is correct.

(Note - your Latex didn't work in one line).

 

[muzialis]

Mathman, many thanks much appreciated.