The Incomplete Gamma and Confluent Hypergeometric Functions (Part 4)

In the previous post, I show that the curious integral

$\displaystyle \int_0^z t^{a-1} e^{-t} \, dt = \displaystyle e^{-z} \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s}$ ,

where the right-hand side is a special case of the confluent hypergeometric function when $a$ is a positive integer, can be confirmed if I can show that

$\displaystyle \sum_{s=0}^n \frac{(-1)^s (a-1)!}{s! (a+n-s)!} =^? \frac{(-1)^n}{n! (a+n)}$ .

I’m using the symbol $=^?$ to emphasize that I haven’t proven that this equality is true. To be honest, I didn’t immediately believe that this worked; however, I was psychologically convinced after using Mathematica to compute this sum for about a dozen values of $n$ and $a$ .

To attempt a proof, we first note that if $n=0$ , then

$\displaystyle \sum_{s=0}^0 \frac{(-1)^s (a-1)!}{s! (a+0-s)!} = (-1)^0 \frac{(a-1)!}{0!a!} = \frac{(-1)^0}{0! (a+0)}$ ,

and so the equality works if $n=0$ . So, for the following, we will assume that $n \ge 1$ . I tried replacing $n$ with $n-a$ in the above equation to hopefully simplify the summation a little bit:

$\displaystyle \sum_{s=0}^{n-a} \frac{(-1)^s (a-1)!}{s! (a+n-a-s)!} =^? \frac{(-1)^{n-a}}{(n-a)! (a+n-a)}$

$\displaystyle \sum_{s=0}^{n-a} \frac{(-1)^s (a-1)!}{s! (n-s)!} =^? \frac{(-1)^{n-a}}{n \cdot (n-a)!}$

$\displaystyle \sum_{s=0}^{n-a} \frac{(-1)^s}{s! (n-s)!} =^? \frac{(-1)^{n-a}}{n \cdot (a-1)!(n-a)!}$

The left-hand side looks like a binomial coefficient, which suggest multiplying both sides by $n!$ :

$\displaystyle \sum_{s=0}^{n-a} (-1)^s \frac{n!}{s! (n-s)!} =^? \frac{(-1)^{n-a} n!}{n \cdot (a-1)!(n-a)!}$

$\displaystyle \sum_{s=0}^{n-a} (-1)^s \binom{n}{s} =^? \frac{(-1)^{n-a} (n-1)!}{(a-1)!(n-a)!}$

Surprise, surprise: the right-hand side is also a binomial coefficient since $(a-1)+(n-a) = n-1$ :

$\displaystyle \sum_{s=0}^{n-a} (-1)^s \binom{n}{s} =^? (-1)^{n-a} \binom{n-1}{n-a}$ .

Now we’re getting somewhere. To again make a sum a little simpler, let’s replace $a$ with $n-a$ :

$\displaystyle \sum_{s=0}^{n-(n-a)} (-1)^s \binom{n}{s} =^? (-1)^{n-(n-a)} \binom{n-1}{n-(n-a)}$

$\displaystyle \sum_{s=0}^a (-1)^s \binom{n}{s} =^? (-1)^a \binom{n-1}{a}$

Therefore, it appears that the confirming the complicated integral at the top of this post reduces to this equality involving binomial coefficients. In the next post, I’ll directly confirm this equality.

The Incomplete Gamma and Confluent Hypergeometric Functions (Part 4)

Published by John Quintanilla

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Published by John Quintanilla

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