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.
Best Answer
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