# Dirac delta and convolution

Is there a way to simplify the following equation?

$$\int_{-\infty}^{\infty} f(q) \delta(q-k) * g(q) dq =\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(q) \delta(q-k) g(q-\tau) d\tau dq$$

The $$*$$ is the convolution symbol and $$\delta$$ is the Dirac delta function.

Is it allowed to first do the integration and then the convolution to obtain:

$$\int_{-\infty}^{\infty} f(q) \delta(q-k) * g(q) dq \stackrel{?}{=} f(k)*g(k)$$

• the convolution is not well-defined, what is the variable of $\delta *g$ there? Convolution is defined by $f*g(x):=\int f(x-y)g(y)\,\mathrm d y$ – Masacroso Nov 14 at 19:56
• I changed the question to make it more clear. I also see the answer now. By changing the variables over which to integrate, you end up with $f(k)*g(k)$. – Fre Nov 14 at 20:34