# How can I compute the integral of the product of two Dirac delta and a polynomial?

$$\int_{0}^{\infty}dk~k^{d-2}\delta(k-a)\delta(k-b).$$

I tried substituting $$k^{d-2}\delta(k-a)$$ with other espressions such as $$\frac{d}{dk}\biggl[k^{d-2}\Theta(k-a)\biggr]-(d-2)k^{d-3}\Theta(k-a)$$ and integrating by parts but it doesn't lead to the solution.

• What is the context? Are you sure you have the right expression? Feb 19, 2020 at 14:53
• If $a \ne b$, the integral is just zero. Feb 19, 2020 at 17:32
• As others have noted, if $a\not=b$, the literal functional/integral is $0$. But, in contrast to some claims, if $a=b$ it is simply undefined, in a strong sense, since $\delta^2$ "is not a thing". This leads me to strongly wonder whether you're really asking the question you intend to ask... Feb 19, 2020 at 23:31

OP's distribution simplifies to $$\int_{\mathbb{R}}\mathrm{d}k~\theta(k)~k^{d-2}~\delta(a-k)~\delta(k-b)~=~\theta(a)~a^{d-2}~\delta(a-b),$$ cf. e.g. this related Math.SE post.
The contribution of the integrand is zero wherever $$k\ne a$$ or $$k \ne b$$. If $$a\ne b$$, then the integral is zero.
If $$a=b \ge 0$$, then the integral is equal to $$b^{d-2}$$ else zero. Thanks to @Qmechanic for pointing that out (essentially a step function multiplying this term).