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 2)

Published by John Quintanilla

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Published by John Quintanilla

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