As part of a larger derivation in my notes, I am trying to find the
$$ \int_{0}^{\infty} \frac{1}{a/b}\exp\left(-\frac{x^{2}}{2(a/b)}\right)db $$

However I've been having a hard time seeing how to make any progress with the formulation, I tried substituting with $\frac{a}{b}$ but could not simplify to a known integration technique for me. How should I approach this integral most easily?

  • $\begingroup$ Substitute $x^2$ with f, and then consider the derivative of the exponential term with respect to f. Replace your integrand with this term, suitably adjusted with a factor of two, and go from there. $\endgroup$ – Pablo Oct 13 '16 at 1:04
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    $\begingroup$ $\int x e^{-y x}dx=?$ $\endgroup$ – tired Oct 13 '16 at 1:08

Rewriting your expression $$I=\int \frac{1}{a/b}\exp\left(-\frac{x^{2}}{2(a/b)}\right)\,db$$ $$I=\int\frac{b e^{-\frac{b x^2}{2 a}}}{a}\,db$$ Now, in order to simplify the exponential term, change variable $$b=\frac{2 a}{x^2}t \implies db=\frac{2 a}{x^2}\,dt$$ which makes $$I=\int \frac{4 a e^{-t} t}{x^4}\,dt=\frac{4 a }{x^4}\int te^{-t}\,d t$$ The integral now is quite simple (one integration by parts).

For the definite integral, as already same in comments and answers, they do not have any reaso to change.


@tired made a comment above that is on the right track, I believe. If you try to simplify the integral using some substitution to get rid of excess, it brings about a nice form that is solvable using integration by parts.

I suggest using something like the following (variables names are arbitrary):

$$\sigma = \frac{1}{a/b} = b/a$$


This also implies:

$$\frac{d\sigma}{db} = 1/a \rightarrow db = a d\sigma$$

You now have the integral:

$$\int_{0}^{\infty} a\sigma e^{-\alpha\sigma} d\sigma = a \int_{0}^{\infty}\sigma e^{-\alpha\sigma} d\sigma$$

This is the form @tired has mentioned above, in case you didn't follow where that was coming from. The integral bounds are the same (you can check by looking at the definition of sigma and how it relates to b). Simple integration by parts will complete this problem. Don't forget to substitute back in at the end of the integration the values for sigma and alpha if necessary.


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