I have been studying differential equation, in particular special functions.

Euler's Gamma function, and Gauss's Pi function are essentially the same, differing only by an offset of one unit.

for $z\in\mathbb C,\qquad {\frak R} (z)>0$

$$\Gamma(z)=\int\limits_0^\infty x^{z-1}e^{-x}\,\mathrm dx$$

$$\Pi(z)=\Gamma(z+1)=\int\limits_0^\infty x^ze^{-x}\,\mathrm dx$$

Both extend the notion of the factorial (which is only defined for positive integers).

$$\Gamma(z+1)=\Pi(z)=z!,\qquad z\in\mathbb Z \geq0$$

The Pi function appears to be a more natural analog of the factorial (It dosen't introduce the unit offset). My text book exclusively uses the Gamma function, and dosen't mention the Pi function at all. I was wondering if there is any good reasons to focus on the Gamma function (presumably it makes some calculations simpler further down the line).

The best reason I can come up with on my own is that for Laplace transforms

$$\mathcal L\big\{ t^r\big\}=\frac{\Pi(r)}{s^{r+1}}=\frac{\Gamma(r+1)}{s^{r+1}},\qquad r\geq-1 \in\mathbb R$$ Using the Gamma function here preserves some symmetry. I am not sure it this is the reason or if there are some subtleties I am completely missing.


2 Answers 2


Since you mention Laplace transforms, in its current form $\Gamma(s)$ is the Mellin transform of $e^{-x}$.

Here is another reason which is perhaps the most convincing. The Haar measure of a subset $S\subset \mathbb{R}^\times$ of the multiplicative group of real numbers is $\int_{x\in S} \frac{dt}{t}$, so the measure $\frac{dt}{t}$ over the real line is natural. The Gamma function is an analogue of a Gauss sum, and is the integral of multiplicative function $x^s$ against the additive function $e^{-x}$ over the measure of the group.

This problem was posed on Math Overflow, and received a large number of upvotes there. Take a look at the answers appearing on this thread: https://mathoverflow.net/questions/20960/why-is-the-gamma-function-shifted-from-the-factorial-by-1

  • $\begingroup$ I can't understand most of the information on that page, . . . something to do with choosing the location of complex poles. Care to elaborate? $\endgroup$ Dec 31, 2012 at 6:16
  • $\begingroup$ I am still in the process of completing a Sophomore year at university and I haven't touched on such things as: Mellin transform; Haar measure; Gauss sum. So to say I can't properly comprehend your response is a massive understatement. I have checked out the wiki pages on such things but these have proved just as confusing/beyond my level. I still appreciate your response and I've accepted it because have no reason to doubt. One day I will understand these concepts and ill re-read your post and it will make more sense to me then. $\endgroup$ Jan 26, 2013 at 14:37
  • 3
    $\begingroup$ @UnkleRhaukus, "Haar measure" just means that $\int_0^\infty f(t)\,{dt\over t}$ is unchanged under change of variables $t\to c\cdot t$ with $c>0$, that is, $\int_0^\infty f(ct)\,{d(ct)\over (ct)}\;=\; \int_0^\infty f(t)\,{dt\over t}$. That's a reason to keep the divide-by-$t$ with the $dt$, rather than have $t^{s-1}$ in the integral defining $\Gamma(s)$. "Mellin transform" is just a Fourier transform in different coordinates, etc. Often the terminology is fancier than the underlying mathematics. $\endgroup$ Dec 14, 2013 at 17:32
  • $\begingroup$ @ElementsinSpace I am in somewhat the same position as you were in last year, right now :) Do you, by any chance, have a simple/elementary explanation of those ideas that would make sense to someone who has not studied Haar measures? If so, please let me know, I am thinking about asking a question about these issues. Cheers! (And I realize this post is very old, sorry about that) $\endgroup$
    – user142299
    May 17, 2014 at 23:38

This subject is discussed in the book The Gamma Function by James Bonnar. The $\Pi(z)=z!$ notation is due to Gauss and is sometimes encountered in older literature. The notation $\Gamma(z+1)=z!$ is due to Legendre. Legendre's motivation for the normalization does not appear to be known. Cornelius Lanczos called it "devoid of any rationality" and would instead use $z!$. Legendre's formula does simplify a few formulas, but complicates most others.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .