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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?

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

2 Answers 2

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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.

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$$\left[\delta\left(k-k_0\right)\ f(k)\ *\ f(k)\right](x)=f\left(k_0\right)\ f\left(x-k_0\right)$$

$$\int\limits_{-\infty}^{\infty}\left[\delta\left(k-k_0\right)\ f(k)\ *\ f(k)\right](x)\ dx=\int\limits_{-\infty}^{\infty}f\left(k_0\right)\ f\left(x-k_0\right)\ dx$$

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