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I'm looking to evaluate the following integral:

$\int_{- \infty}^{\infty} f(x) \delta(x-a) \delta(x-a)dx$

If instead of two identical Dirac deltas we had:

$\int_{- \infty}^{\infty} f(x) \delta(x-a) \delta(x-b)dx$

I believe we would have

$\int_{- \infty}^{\infty} f(x) \delta(x-a) \delta(x-b)dx = f(a) \delta(a-b)$

which with $a=b$ leaves the result undefined. I'm tempted to say the result of the integral is just $f(a)$, but I feel that that may not exactly be rigorous.

Edit:

I should note that this problem came up in computing the power spectrum of a Langevin equation, so the integral is actually:

$\int_{- \infty}^{\infty} f(x) \delta^*(x-a) \delta(x-a)dx$

where * denotes the complex conjugate. I'm not sure if the conjugate of the Dirac delta has any meaning.

user38729
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  • If recalled correctly it can be shown that $\delta^{2}(t-a) = \delta(a) , \delta(t-a)$ and would lead to the result $$\int_{-\infty}^{\infty} f(x) , \delta^{2}(x-a) , dx = \delta(a) , \int_{-\infty}^{\infty} f(x) \delta(x-a) , dx = f(a) , \delta(a).$$ – Leucippus Jun 27 '18 at 21:37
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    There is no meaning to the distribution $\delta^2(x)$. – Mark Viola Jun 28 '18 at 03:08
  • Possible duplicates: https://math.stackexchange.com/q/12944/11127 and links therein. – Qmechanic Jun 30 '18 at 17:49

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