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

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.

Qmechanic
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Filippo Caleca
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    What is the context? Are you sure you have the right expression? – Spencer Feb 19 '20 at 14:53
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    If $a \ne b$, the integral is just zero. – vonbrand Feb 19 '20 at 17:32
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    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... – paul garrett Feb 19 '20 at 23:31

2 Answers2

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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.

Qmechanic
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  • Okay, you are right. Didn't see that the integral's lower limit is zero, so we need to include a step function. Otherwise our answers agree. – mjw Feb 19 '20 at 23:21
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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).

mjw
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