Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'd like to show the following equality (at least Mathematica claims it is an equality): \begin{multline*} \int_0^\infty x^p \exp(-(ax - b)^2)\, dx = \frac{1}{2} e^{-b^2} a^{-p-1} \left(\Gamma \left(\frac{p+1}{2}\right) \, _1F_1\left(\frac{p+1}{2};\frac{1}{2};b^2\right)+\ b p \Gamma \left(\frac{p}{2}\right) \, _1F_1\left(\frac{p}{2}+1;\frac{3}{2};b^2\right)\right) \end{multline*} Here $a,b>0$ and $p > 1$ (if it matters).

This looks a lot like the expression given on Wikipedia for the uncentered moments of a Gaussian, except the integral is over $[0,\infty)$ rather than all of $\mathbb{R}.$

Any suggestions on how to proceed? I haven't been able to find this in any table of integrals.

share|improve this question
add comment

1 Answer

The point of this derivation is to expand the exponent, and convert the integral into an infinite sum of integrals, each of which is basically a $\Gamma$-function. Then the sum can be written as a sum of two hypergeometric series. I'm pretty sure this is what Mathematica does.

Call the integral $I$, and write the exponential as $$ \exp(-(ax-b)^2) = e^{-b^2-a^2x^2}e^{2abx} = e^{-b^2-a^2x^2}\sum_{k\geq0} (2ab)^k\frac{x^k}{k!}. $$ Then, using Mathematica, we have $$\int_0^\infty e^{-a^2x^2}x^{k+p}dx = \frac{a^{-1-k-p}}{2} \Gamma\left(\frac{1+k+p}{2}\right). $$ After a little manipulation the integral becomes $$ I = \frac{1}{2}e^{-b^2}a^{-1-p}\sum_{k\geq0} (2b)^k \Gamma\left(\frac{k}{2} + \frac{1+p}{2}\right) \frac{1}{k!}, $$ call the sum $S$.

Now, by definition we have $$ F(a;b;x) = \sum_{k\geq0}\frac{\Gamma(a+k)/\Gamma(a)}{\Gamma(b+k)/\Gamma(b)}\frac{x^k}{k!}, $$ where $\Gamma(a+k)/\Gamma(a)$ are the rising powers of $a$, $$ \frac{\Gamma(a+k)}{\Gamma(a)} = a(a+1)(a+2)\cdots(a+k-1). $$

Split $S$ into two sums, one over even $k$, the other over odd $k$; this gets rid of $k/2$ in the argument of $\Gamma$. Then $$ S = \sum_{k\geq0}\left((2b)^{2k} \frac{\Gamma\left(k+\frac{1+p}{2}\right)}{(2k)!} + (2b)^{2k+1}\frac{\Gamma\left(k+\frac{2+p}{2}\right)}{(2k+1)!}\right). $$ Using $$ \frac{2^{2k}}{(2k)!} = \frac{1}{k!}\frac{1}{\Gamma(\frac{1}{2}+k)/\Gamma(\frac{1}{2})}, \qquad \frac{2^{2k}}{(2k+1)!} = \frac{1}{k!}\frac{1}{\Gamma(\frac{3}{2}+k)/\Gamma(\frac{3}{2})} $$ Find that $$ S = \Gamma\left(\frac{1+p}{2}\right) F\left(\frac{1+p}{2};\frac12;b^2\right) + 2b\Gamma\left(1+\frac{p}{2}\right)F\left(1+\frac{p}{2};\frac32;b^2\right). $$ Substitute back, use $\Gamma(1+p/2)=(p/2)\Gamma(p/2)$, and you get the required expression for $I$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.