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I have the following integral involving a confluent hypergeometric function:

$$\int_{0}^{\infty}x^3e^{-ax^2}{}_1F_1(1+n,1,bx^2)dx$$

where $a>b>0$ are real constants, and $n\geq 0$ is an integer.

Wolfram Mathematica returns the following solution: $\frac{a^{n-1}(a+bn)}{(a-b)^{n+2}}$. However, I can't figure out how it arrived at it (I always try to check the solutions "on paper" that Mathematica gives me -- or at least using Gradshteyn and Ryzhik). Can anyone help?

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You can probably resolve it by recognizing that $_{1}F_{1}(1+n,1,bx^{2})$ is the Confluent HyperGeometric function $M(1+n,1,bx^{2})$ substitute $y=x^{2}$. You have to keep the b for the next step. Render the first term negative by the first Kummer transform $M(1+n,1,by)=e^{y}M(-n,1,-by)$ and set $-b=k$. This makes the series representation have a finite ($n$) number of terms if you want; it might not be necessary .
Now we can use http://dlmf.nist.gov/13.10.E3 for a direct integration/series or we can recognize that $M(-n,1,-by)$ is the 0 th Laguerre Polynomial like so:
$M(-n,1,-by)= \frac{n!}{(1+n)_{n}} L^{0}_{n}(-by)$ Either way you have a finite series to sum. I leave summing the series as an exercise for the reader (no I haven't done it!).

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  • $\begingroup$ I was just thinking about this question the other day, and, as you said, the trick here is the summation representation of ${}_1F_1$ (which I didn't think about when I posted the question -- it came up as a step in a long series of calculations in Mathematica). Anyway, thanks for your answer (I edited tiny bit of formatting for mathematical symbols). $\endgroup$ – M.B.M. Mar 28 '14 at 17:25
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Let's start with the hypergeometric function. We have: \begin{eqnarray} F_{2,1}[1+n,1;b x^2] &=& \sum\limits_{m=0}^\infty \frac{(1+n)^{(m)}}{m!} \cdot \frac{(b x^2)^m}{m!} \\ &=& \sum\limits_{m=0}^\infty \frac{(m+1)^{(n)}}{n!} \cdot \frac{(b x^2)^m}{m!} \\ &=&\left. \frac{1}{n!} \frac{d^n}{d t^n} \left( t^n \cdot e^{b x^2 \cdot t} \right)\right|_{t=1} \\ &=& \frac{1}{n!} \sum\limits_{p=0}^n \binom{n}{p} n_{(p)} (b x^2) ^{n-p} \cdot e^{b x^2} \end{eqnarray}

Therefore the integral in question reads: \begin{eqnarray} &&\int\limits_0^\infty x^3 e^{-a x^2} F_{2,1}[1+n,1;b x^2] dx =\\ && \frac{1}{2} \sum\limits_{p=0}^n \frac{n^{(p)}}{p! (n-p)!} b^{n-p} \frac{(n-p+1)!}{(a-b)^{n-p+2}} \\ &&\frac{1}{2} \frac{1}{(a-b)^{n+2}} \sum\limits_{p=0}^n \binom{n}{p} \frac{(n-p+1)^{(p)}}{(n-p+2)^{(p-1)}} \cdot b^{n-p} (a-b)^p \\ && \frac{1}{2} \frac{1}{(a-b)^{n+2}} \sum\limits_{p=0}^n \binom{n}{p} (n-p+1) \cdot b^{n-p} (a-b)^p \\ && \frac{1}{2} \frac{1}{(a-b)^{n+2}} a^{n-1} (a+b n) \end{eqnarray}

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