My professor said that:
$$x(t)\otimes \delta (t-t_0)=x(t_0)$$
How can I prove it? I tried to apply the definition of convolution $q(x)w(x)=\int_{-\infty}^{+\infty}{q(\tau)w(x-\tau)d\tau}$:
$$x(t)\otimes \delta (t-t_0)=\int_{-\infty}^{+\infty}{x(\tau)\delta (t-t_0-\tau)d\tau}$$
Then I applied the sifting property $\int_{-\infty}^{+\infty}{q(x)\delta(x-a)dx=q(a)}$:
$$x(t)\otimes \delta (t-t_0)=x(t_0+\tau)$$
What is wrong?
Thank you very much.
$\tau$ is your integration variable, so it should disappear in your result. And your convolution is not defined properly. I think what is meant is you look at the "functions" $$ f(t) = x(t), \quad g(t) = \delta(t - t_{0}) $$ and then at the convolution $(f \otimes g)(t)$.
I would be surprised if the result was independent of $t$ as your professor said, since if we do not shift by $t_{0}$ we get $$ (x \otimes \delta )(t) = \int_{\mathbb{R}} x(\tau) \delta(t - \tau) d \tau = x(t), $$ by the definition of the $\delta$ distribution and its "symmetry" $\delta(-x) = \delta(x)$.
For the more general case we have \begin{align*} (f \otimes g)(t) &= \int_{\mathbb{R}} x(\tau) \delta ((t - \tau) - t_{0}) d \tau = \int_{\mathbb{R}} x(\tau) \delta ((t - t_{0}) - \tau) d \tau \\ &= \int_{\mathbb{R}} x(\tau) \delta (\tau- (t - t_{0})) d \tau= x(t - t_{0}). \end{align*}