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Someone here helped me out with $\int_0^\infty \frac{1}{2r^2}e^{-\sqrt{v}/r}dv=1$. I would like to calculate $\int_0^\infty v\cdot \frac{1}{2r^2}e^{-\sqrt{v}/r}dv$. Let $m(v)=v$ and $n(v)=\frac{1}{2r^2}e^{-\sqrt{v}/r}$ (as $r$ is treated as a constant here). The integral becomes $\int_0^\infty m(v)\cdot n(v)dv=v\cdot1-\int_0^\infty 1\cdot (1)dv$. This diverges, but I'm pretty sure the initial integral is convergent. Can anyone explain me why integration by parts doesn't work here?

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  • $\begingroup$ your calculation is wrong ! $\endgroup$
    – alkabary
    Jun 5 '15 at 12:23
  • $\begingroup$ because you can not evaluate the integral inside the second integral, it has to be kept as an primitive. $\endgroup$
    – Mann
    Jun 5 '15 at 12:24
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    $\begingroup$ You should do the integration by parts with the antiderivatives and evaluate at the end. $\endgroup$ Jun 5 '15 at 12:29
  • $\begingroup$ In general, $~\displaystyle\int_0^\infty\exp\big(-\sqrt[\large^n]x\big)~dx ~=~ n!$ $\endgroup$
    – Lucian
    Jun 5 '15 at 14:48
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$$v=r^2u^2$$

$$r^2\int _0^\infty u^3e^{-u}du$$

This is a gamma function integral: https://en.wikipedia.org/wiki/Gamma_function

Result is

$$6r^2$$

Alternatively you can find the antiderivative by hand using the feynman trick: https://math.stackexchange.com/a/943152/219783

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If $r > 0$ then the result follows. Since $$\frac{1}{2r^{2}}\int e^{-\sqrt{v}/r}\ dv = \frac{1}{r^{2}}\int e^{-u/r}u\ du\big|_{v := u^{2}} = \frac{-1}{r}\int u\ de^{-u/r} = \frac{-1}{r}\big( e^{-u/r}u - \int e^{-u/r}\ du + C_{1} \big) = \frac{-1}{r}\big( e^{-u/r}u + re^{-u/r} + C_{2} \big) = \frac{-1}{r}\big( e^{-\sqrt{v}/{r}}\sqrt{v} + re^{-\sqrt{v}/r} + C_{2}\big),$$ it follows that $$\int_{0}^{x}\frac{1}{2r^{2}}e^{-\sqrt{v}/r}\ dv = \frac{-1}{r}e^{-\sqrt{x}/r}\sqrt{x} - e^{-\sqrt{x}/r} + 1 \to 1$$ as $x \to +\infty.$

If $r < 0$, the improper integral diverges.

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