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 3)

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.

The Incomplete Gamma and Confluent Hypergeometric Functions (Part 2)

In this series of posts, I confirm this curious integral:

$\displaystyle \int_0^z t^{a-1} e^{-t} \, dt = \displaystyle \frac{z^a e^{-z}}{a} M(1, 1+a, z)$ ,

where the confluent hypergeometric function $M(a,b,z)$ is

$M(a,b,z) = \displaystyle 1+\sum_{s=1}^\infty \frac{a(a+1)\dots(a+s-1)}{b(b+1)\dots (b+s-1)} \frac{z^s}{s!}$ .

This integral can be confirmed — unsatisfactorily confirmed, but confirmed — by differentiating the right-hand side. For the sake of simplicity, I restrict my attention to the case when $a$ is a positive integer. To begin, the right-hand side is

$\displaystyle \frac{z^a e^{-z}}{a} M(1, 1+a, z) = \displaystyle \frac{z^a e^{-z}}{a} \left[1 + \sum_{s=1}^\infty \frac{1 \cdot 2 \cdot \dots \cdot s}{(a+1)(a+2)\dots (a+s)} \frac{z^s}{s!} \right]$

$= \displaystyle \frac{z^a e^{-z}}{a} \left[1 + \sum_{s=1}^\infty \frac{1}{(a+1)(a+2)\dots (a+s)} z^s \right]$

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

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

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

We now differentiate, first by using the Product Rule and then differentiating the series term-by-term (blatantly ignoring the need to confirm that term-by-term differentiation applies to this series):

$\displaystyle \frac{d}{dz} \left[\frac{z^a e^{-z}}{a} M(1, 1+a, z) \right] = \displaystyle \frac{d}{dz} \left[ e^{-z} \sum_{s=0}^\infty \frac{(a-1)!}{(a+s)!} z^{a+s} \right]$

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

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

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

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

We now shift the index of the first series:

$\displaystyle \frac{d}{dz} \left[\frac{z^a e^{-z}}{a} M(1, 1+a, z) \right] =-e^{-z} \sum_{s=1}^\infty \frac{(a-1)!}{(a+s-1)!} z^{a+s-1} + e^{-z} \sum_{s=0}^\infty \frac{(a-1)!}{(a+s-1)!} z^{a+s-1}$ .

By separating the $s=0$ term of the second series, the right-hand side becomes:

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

since the two infinite series cancel. We have thus shown that

$\displaystyle \frac{d}{dz} \left[\frac{z^a e^{-z}}{a} M(1, 1+a, z) \right] = \frac{e^{-z} z^{a-1}}{a}$ .

Therefore, we may integrate the right-hand side:

$\displaystyle \int_0^z t^{a-1} e^{-t} \, dt = \left[\frac{t^a e^{-t}}{a} M(1, 1+a, t) \right]_0^z$

$\displaystyle = \frac{z^a e^{-z}}{a} M(1, 1+a, z) - \frac{0^a e^{0}}{a} M(1, 1+a, 0)$

$\displaystyle = \frac{z^a e^{-z}}{a} M(1, 1+a, z)$ .

While this confirms the equality, this derivation still feels very unsatisfactory — we basically guessed the answer and then confirmed that it worked. In the next few posts, I’ll consider the direct verification of this series.

The Incomplete Gamma and Confluent Hypergeometric Functions (Part 1)

Yes, the title of this post is a mouthful.

While working on a research project, a trail of citations led me to this curious equality in the Digital Library of Mathematical Functions:

$\gamma(a,z) = \displaystyle \frac{z^a e^{-z}}{a} M(1, 1+a, z)$ ,

where the incomplete gamma function $\gamma(a,z)$ is

$\gamma(a,z) = \displaystyle \int_0^z t^{a-1} e^{-t} \, dt$

and the confluent hypergeometric function $M(a,b,z)$ is

$M(a,b,z) = \displaystyle 1+\sum_{s=1}^\infty \frac{a(a+1)\dots(a+s-1)}{b(b+1)\dots (b+s-1)} \frac{z^s}{s!}$ .

While I didn’t doubt that this was true — I don’t doubt this has been long established — I had an annoying problem: I didn’t really believe it. The gamma function

$\Gamma(a) = \displaystyle \int_0^\infty t^{a-1} e^{-t} \, dt$

is a well-known function with the famous property that

$\Gamma(n+1) = n!$

for non-negative integers $n$ ; this is often seen in calculus textbooks as an advanced challenge using integration by parts. The incomplete gamma function $\gamma(a,z)$ has the same look as $\Gamma(a)$ , except that the range of integration is from $0$ to $z$ (and not $\infty$ ). The gamma function appears all over the place in mathematics courses.

The confluent hypergeometric function, on the other hand, typically arises in mathematical physics as the solution of the differential equation

$z f''(z) + (b-z) f'(z) - af(z) = 0$ .

As I’m not a mathematical physicist, I won’t presume to state why this particular differential equation is important — except that it appears to be a niche equation that arises in very specialized applications.

So I had a hard time psychologically accepting that these two functions were in any way related.

While ultimately unimportant for advancing mathematics, this series will be about the journey I took to directly confirm the above equality.

Unit Circle

Source: https://xkcd.com/3041/