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A popular method of proving the formula is to use the infinite product representation of the gamma function. See ProofWiki for example.

However, I'm interested in down-to-earth proof; e.g. using the change of variables. As the formula being connected to the beta function, there could be one-line proof for it.

Could anyone help me?

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Could a down-to-earth proof have one line? –  Dirk Sep 26 '11 at 8:35
With some complex variable theory there's a fairly quick proof. Let f(z) be gamma(z) gamma (1-z) sin(pi*z). Then f(z+1)=f(z), and one wants to show that f is constant. To do this one looks at the growth of f in absolute value as Im(z)-->plus or minus infinity, showing that the singularity at infinity is removable. –  paul Monsky Sep 26 '11 at 13:35
A bit more precisely, since f is entire and f(z+1)=f(z), f=F(exp(pi*2iz) for some entire F. Using the fact, immediate from the integral representation formula for gamma, that gamma-->0 uniformly in bounded vertical strips, we see that F has a removable singularity at infinity. By Liouville, F and f are constant. Letting z-->0 we see that the constant value of f is pi. This seems to have the same flavor as David Speyer's argument. –  paul Monsky Sep 26 '11 at 14:05
Ah, this is nice. I wrote $f=e^g$ to knock down the growth rate and ran into the problem that I could only control $\Re(g)$, not $|g|$. You instead wrote $f(z) = F(e^{2 \pi i z})$, which also knocks down the growth rate but has no such problem. To spell out something which Paul is treating as obvious, "immediate from the integral representation" means "by the Riemman-Lebesgue lemma". –  David Speyer Sep 26 '11 at 17:37
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3 Answers

I suspect the following is a three line proof to the right reader: Set $e^g = \Gamma(1+z) \Gamma(1-z) z \sin(\pi z)$. Then $\Re(g)$ is an even harmonic function with $g(z) = O(|z| \log |z|)$, so $g$ is constant. Plugging in $z=1/2$ evaluates the constant.

The "right reader" is someone who already knows good estimates for $|\Gamma(z)|$, who is familiar with the lemma that a harmonic function where $|g(z)| = o(|z|^k)$ is a polynomial of degree $\leq k-1$, and who knows how to compute $\Gamma(1/2)$. I'll try to edit in proofs of these later today. This answer is CW in case someone else wants to do it.

Estimates for $\Gamma(z)$: For $\Re(z)>0$, we have $|\Gamma(z)| \leq \int_0^{\infty} e^{-t} |t^z| dt/t = \int_0^{\infty} e^{-t} t^{\Re(z)} dt/t = \Gamma(\Re(z))$. By Stirling's formula, this shows that $\log |\Gamma(z)| = O(|z| \log |z|)$. Now using the recursion for the $\Gamma$ function lets us extend this estimate to all of $\mathbb{C}$ (details omitted).

We also have the easy estimate $\log |\sin (\pi z)| = O(|z|)$, on a contour which stays well away from the zeroes of $\sin$: Say a circle of radius $N+1/2$. So $\Re(g) = O(|z| \log |z|)$ for $z$ on a circle of radius $N+1/2$. As $g$ is entire, the maximum modulus principle gives us the same bound everywhere in $\mathbb{C}$.

Harmonic functions with slow growth rate: Let $f$ be a harmonic function on the unit disc. We have the Poisson integral formula:

$$f(x+iy) = \int \frac{1-x^2-y^2}{1-2 (x \cos \phi - y \sin \phi) + x^2+y^2} f(e^{i \phi}) d \phi.$$ Differentiating inside the integral sign, there is some smooth function $K(x,y,\phi)$ such that $$\frac{\partial^{a+b} f}{(\partial x)^a (\partial y)^b}(x+iy) = \int K_{ab}(x,y,\phi) f(e^{i \phi}) d \phi.$$

Now, let $f$ be defined on a circle of radius $R$. Making the appropriate variable changes, $$\frac{\partial^{a+b} f}{(\partial x)^a (\partial y)^b}(x+iy) = \frac{1}{R^{a+b+1}} \int K_{ab}(x/R, y/R, \phi) f(R e^{i \phi}) d \phi.$$ So, if $g$ is entire and $g = o(R^k)$, then every $k$-fold derivative of $g$ is $o(1/R)$. Sending $R$ to infinity, every $k$-fold derivative of $g$ is zero, so we deduce that $g$ is a polynomial of degree $\leq k$.

The Gamma function at $1/2$:

We have $\Gamma(1/2) = \int_0^{\infty} e^{-t} t^{-1/2} dt = \int_0^{\infty} e^{-u^2} (2 du)$. This can be evaluated by a variety of methods.

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Apparently, there is no short proof of the reflection formula (unless you count the one you are alluding to, which does need a fair bit of background). There is, however, at least one down-to-earth one, see


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This is an argument offered by Paul Monsky in the comments. Now that I have the details right, I definitely think it is better than mine.

For $x$ between $1$ and $2$, and $y$ positive, we have $$|\Gamma(x+iy)| = \leq \Gamma(x) = O(1)$$ by the triangle inequality applied to the defining integral. Using the multiplicative recursion for the $\Gamma$ function, we deduce that $\Gamma(x+iy) = O(1)$ for $x \in [0,1]$ and $y \geq 1$. (An earlier version of this proved that, in fact $\Gamma(x+iy) = O(e^{-y})$ in this range, but I don't need it.)

Setting $f(z) = \Gamma(z) \Gamma(1-z) \sin(\pi z)$, for $x \in [0,1]$ we have $$|f(x+iy)| = O(1) O(1) O(e^{\pi y}) = O(e^{(\pi ) y}) \ \mbox{as}\ y \to\infty$$

Since $f(z) = f(z+1)$, we have $f(z) = F(e^{2 \pi i z})$ for some holomorphic function $F$ defined on $\mathbb{C}^*$. The above bounds show that $|F(w)| = O(|w|^{1/2})$ as $w \to \infty$ and similarly $F(w) = O(|w|^{-1/2})$ as $w \to 0$. Since $F$ is an entire function, this shows that $F$ is constant, and any number of arguments can establish the constant.

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Thank you David and Paul. The last argument which involves taking limit & stating constant was new to me. So I researched textbooks and found that chapter 6 of [Stein & Shakarchi] has the satisfactory (1 page) answer. - Both $\Gamma(z)\Gamma(1-z)$ (LHS) and $\pi/\sin(z\pi)$ (RHS) has order 1 pole at $z\in\mathcal{Z}$. - Analytic continuation declares we only have to show (LHS)=(RHS) for $z\in (0,1)$. - (RHS)=$\int^{\infty}_{-\infty}ds\frac{e^{zs}}{1+e^s}=\int^{\infty}_{0}\frac{dt}{‌​t}\frac{t^z}{1+t}$ - (LHS) reduces to the same integral using some tricky (but often used) change of variables. –  juno Sep 28 '11 at 4:53
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