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.


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 '12 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 '13 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 '13 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 '14 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.