Integral $ \int_\sigma^\infty r^2 {e^{-A/r^6}} dr $

Question

Below integral can be calculated by using taylor expansion for the $ e^{-A/r^6} $ term. I want to know how to solve this integral analytically? $$ \int_\sigma^\infty r^2 {e^{-A/r^6}} dr $$

Hint: I think part of the answer contains error function.

Answer 1

The integral $$ \int_\sigma^\infty r^2 {e^{-A/r^6}} \;dr, \qquad A>0 $$ diverges. Indeed, $r^2 {e^{-A/r^6}} \to +\infty$ as $r \to +\infty$.

Using Maple, I have an antiderivative:

$$ \int r^2 {e^{-A/r^6}}\;dr = \frac{\sqrt{A\pi}}{3}\;\text{erf}\left(\frac{\sqrt{A}}{r^3}\right)+\frac{r^3}{3}\;\exp\left(\frac{-A}{r^6}\right) $$

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added
A step-by-step version of the antiderivative.

Substutute $y=\sqrt{A}/r^3$ $$ \int r^2 e^{-A/r^6} dr = \frac{-\sqrt{A}}{3}\int \frac{e^{-y^2}}{y^2}\;dy $$ Then integrate by parts for this one $$ \frac{-\sqrt{A}}{3}\int \frac{e^{-y^2}}{y^2}\;dy = \frac{\sqrt{A} e^{-y^2}}{3y} +\frac{2\sqrt{A}}{3} \int e^{-y^2}\;dy $$

This integral can be recognized using the erf function.

Answer 2

I see from the comments that you are actually trying to compute the integral $$\int_\sigma ^\infty dr\, r^2\left( 1-e^{-A/r^6}\right)$$ Instead of splitting it into 2 divergent integrals, you can expand the exponential into its power series to obtain $$\int_\sigma ^\infty dr\,r^2\sum_{n\ge 1}(-1)^{n+1} \frac{(A/r^6)^n}{n!} $$Switching sum and integral, we obtain $$\sum_{n\ge 1}(-1)^{n+1}\frac{A^n}{n!}\int_\sigma^\infty dr\, r^{2-6n}$$ Integrating, your final result can be expressed as $$\frac{\sigma^3}{3}\sum_{n\ge 1} \frac{(-1)^{n+1}}{2n-1} \frac{\left(A/\sigma^6\right)^n}{n!}$$