Ok, I do not have much experiencing using (or even integrating) Heaviside-step functions, so I am looking for a little help. The integral in question is, \begin{equation} 6m^2\int_0^\infty d\mu \frac{1}{\mu^3}\sqrt{1 - \frac{4m^2}{\mu^2}}\theta(\mu-2m) \end{equation} where $\theta$ is the step-function. Any help direct help or hints is of course appreciated. (The reason it is actually in here in the first place is that without it the integral is divergent). I was thinking about performing integration by parts to turn the step-function into a delta-function $\delta(\mu-2m)$ but then the integral in the end would just give 0 when I know this integral is equal to $1/2$ (I am just trying to show that).


1 Answer 1


Notice that $$ \theta(\mu-2m)= \begin{cases} 0\,\,\text{if}\,\,\mu<2m, \\ 1\,\,\text{if}\,\,\mu>2m. \end{cases} $$ Therefore, $$ \int_0^\infty d\mu \frac{1}{\mu^3}\sqrt{1 - \frac{4m^2}{\mu^2}}\theta(\mu-2m)= \int_{2m}^\infty d\mu \frac{1}{\mu^3}\sqrt{1 - \frac{4m^2}{\mu^2}}. $$

  • $\begingroup$ Oh.... Ok, that helps. Then the integration is just $\frac{1}{12m^2}\left(1-\frac{4m^2}{\mu^2}\right)^{3/2}$ which at the bounds becomes just $\frac{1}{12m^2}$ with the factor of $6m^2$ becomes simply $\frac{1}{2}$. Thank you! $\endgroup$
    – MathZilla
    Commented Jun 27, 2023 at 3:20

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