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?
Best Answer
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.