The Incomplete Gamma and Confluent Hypergeometric Functions (Part 3)

Calculus 2 Minutes

In the previous post, I confirmed 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, by differentiating the right-hand side. However, the confirmation psychologically felt very unsatisfactory — we basically guessed the answer and then confirmed that it worked.

A seemingly better way to approach the integral is to use the Taylor series representation of e^{-t} to integrate the left-hand side term-by-term:

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

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

= \displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{n!} \frac{z^{a+n}}{a+n}.

Well, that doesn’t look like the right-hand side of the top equation. However, the right-hand side of the top equation also has a e^{-z} in it. Let’s also convert that to its Taylor series expansion and then use the formula for multiplying two infinite series:

\displaystyle e^{-z} \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s} = \left( \sum_{s=0}^\infty \frac{(-z)^s}{s!} \right) \left( \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s} \right)

= \displaystyle z^a \left( \sum_{s=0}^\infty \frac{(-1)^s z^s}{s!} \right) \left( \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^s \right)

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

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

Summarizing, apparently the following two infinite series are supposed to be equal:

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

or, matching coefficients of z^{a+n},

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

When I first came to this equality, my immediate reaction was to throw up my hands and assume I made a calculation error someplace — I had a hard time believing that this sum from s=0 to s=n was true. However, after using Mathematica to evaluate this sum for about a dozen different values of n and a, I was able to psychologically assure myself that this identity was somehow true.

But why does this awkward summation work? This is no longer a question about integration: it’s a question about a finite sum with factorials. I continue this exploration in the next post.

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Published by John Quintanilla

I'm a Professor of Mathematics and a University Distinguished Teaching Professor at the University of North Texas. For eight years, I was co-director of Teach North Texas, UNT's program for preparing secondary teachers of mathematics and science.

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