# Convolution with Dirac delta

How to solve this expression: $$\int_{-\infty}^{\infty} \left[ \delta(k-k_0)f(k)\right]*f(k)dk=?$$

Here $$\delta$$ represents the Dirac delta function and $$*$$ represents the convolution over the $$k$$ variable. What I think: $$\int_{-\infty}^{\infty} \left[ \delta(k-k_0)f(k)\right]*f(k)dk=\int_{-\infty}^{\infty} \delta(k-k_0)*\left[f(k)f(k)\right]dk = \int_{-\infty}^{\infty} f(k-k_0)^2dk =\int_{-\infty}^{\infty}f(k)^2dk$$

However, I have doubts about the solution as the influence of the Dirac function seems to disappear?

• Why do you believe that the step $(ab)*c = a*(bc)$ was justified? Commented Jun 24, 2020 at 14:53
• Oh, you are right! Is it then possible to simplify the expression?
– Fre
Commented Jun 24, 2020 at 14:56
• Okay, I worked it out and think that it simplifies to $f(k_0)*f(k_0)$?
– Fre
Commented Jun 24, 2020 at 15:00

Two approaches. Option 1: $$\int_{-\infty}^{\infty} \left[ \delta(k-k_0)f(k)\right]*f(k)dk=\\ \int_{-\infty}^{\infty} \int_{-\infty}^\infty \left[ \delta(\tau-k_0)f(\tau)\right]f(k-\tau)\, d\tau \,dk = \\ \int_{-\infty}^{\infty} \left[\int_{-\infty}^\infty [f(\tau)f(k-\tau)]\,\delta(\tau-k_0)\, d\tau\right] \,dk = \\ \int_{-\infty}^{\infty} f(k_0)\,f(k-k_0) \,dk = \\ f(k_0)\int_{-\infty}^\infty f(u)\,du.$$ Option 2: note that $$\delta(k - k_0)f(k) = f(k_0) \delta(k - k_0)$$. It follows that $$\left[ \delta(k-k_0)f(k)\right]*f(k) = f(k_0) [\delta(k - k_0) * f(k)] = f(k_0)f(k-k_0).$$ That brings us to $$\int_{-\infty}^{\infty} f(k_0)\,f(k-k_0) \,dk$$, like before.
$$\left[\delta\left(k-k_0\right)\ f(k)\ *\ f(k)\right](x)=f\left(k_0\right)\ f\left(x-k_0\right)$$
$$\int\limits_{-\infty}^{\infty}\left[\delta\left(k-k_0\right)\ f(k)\ *\ f(k)\right](x)\ dx=\int\limits_{-\infty}^{\infty}f\left(k_0\right)\ f\left(x-k_0\right)\ dx$$