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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    $\begingroup$ Could a down-to-earth proof have one line? $\endgroup$
    – Dirk
    Sep 26, 2011 at 8:35
  • 1
    $\begingroup$ 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. $\endgroup$ Sep 26, 2011 at 13:35
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    $\begingroup$ 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. $\endgroup$ Sep 26, 2011 at 14:05
  • $\begingroup$ 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". $\endgroup$ Sep 26, 2011 at 17:37
  • $\begingroup$ There is a standard residue-based way to compute $\int_0^{\infty} x^{p-1}dx/(1+x)$, which is Beta function. Is it what you need? $\endgroup$ Apr 15, 2016 at 6:41

6 Answers 6


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.


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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    $\begingroup$ 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. $\endgroup$
    – juno
    Sep 28, 2011 at 4:53

It can be shown (from the Beta function) that

\begin{eqnarray} \Gamma(1-x) \Gamma(x) = \mathrm{B}(x, 1 -x) = \int_0^{\infty} \frac{s^{x-1} d s}{s+1} \quad \quad (1) \end{eqnarray} Now we show that

\begin{eqnarray*} \int_0^{\infty} \frac{s^{x-1} d s}{s+1} = \frac{\pi}{\sin \pi x} \end{eqnarray*}

We use the contour shown in Figure below

enter image description here The singularities are a pole at $s=-1$ and a branch point at $s=0$ where the function is multivalued. We can use the positive $x$ axis as a branch cut. The contour encloses the pole, so the integral along the contour $C=C_1 \cup C_2 \cup C_3 \cup C_4$ is given by

\begin{eqnarray*} \int_C \frac{s^{x-1} ds}{s+1} = 2 \pi \mathrm{i} (-1)^{x-1} = 2 \pi \mathrm{i} \, \mathrm{e}^{ \mathrm{i} \pi (x-1)}. \end{eqnarray*} since $(-1)^{x-1}=\mathrm{e}^{- \mathrm{i} \pi (x-1)}$ is the only residue of the integrand. We now evaluate the four individual integrations for the four paths in the figure. Let us call $I_i = \int_{C_i} s^{x-1}/(1+s) ds$. We start with the integrals along the circular paths. For the small circle we can write $s= \epsilon \mathrm{e}^{\mathrm{i} \theta}$ where $\theta \in[ \delta, 2 \pi - \delta]$ with $\epsilon$ the radius of the disk, and $\delta$ the initial angle of integration. We then change variables from $s$ to $\theta$ with $ds = \mathrm{i} \epsilon \mathrm{e}^{\mathrm{i} \theta}$. That is,

\begin{eqnarray*} |I_4| = \left | \int_ {2 \pi - \delta_{\epsilon}}^ {\delta_{\epsilon}} \frac{\mathrm{i} \epsilon^x \mathrm{e}^{\mathrm{i} \theta (x+1) } d \theta } {\epsilon \mathrm{e}^{\mathrm{i} \theta} -1 } \right | \le \int_{\delta_{\epsilon}}^{2 \pi - \delta_{\epsilon}} \frac{|\epsilon^x| d \theta } {1 - \epsilon } = (2 \pi - 2 \delta_{\epsilon} ) \frac{|\epsilon^x| } {1 - \epsilon } \end{eqnarray*} which goes to $0$ as $\epsilon, \delta_{\epsilon} \to 0$, since $0 < x < 1$. Likewise along the big circle $s = R \mathrm{e}^{\mathrm{i} \theta}$ and $ds = \mathrm{i} R \mathrm{e}^{\mathrm{i} \theta} d \theta$, and so

\begin{eqnarray*} |I_2| = \left | \int_{\delta_{R}}^{2 \pi - \delta_{R}} \frac{\mathrm{i} R^x \mathrm{e}^{\mathrm{i} \theta (x+1) } d \theta } {R \mathrm{e}^{\mathrm{i} \theta} -1 } \right | \le \int_{\delta_{R}}^{2 \pi - \delta_{R}} \frac{|R^x| d \theta } {R-1 } = (2 \pi - 2 \delta_{R} ) \frac{R^x } {R-1 } \end{eqnarray*} We use the L'H^{o}pital rule to find that

\begin{eqnarray*} \lim_{R \to \infty} |I_2| = \lim_{R \to \infty} 2 (\pi - \delta_R) \frac{x R^{x-1}}{1} = \lim_{R \to \infty} \frac{x}{R^{1-x}} = 0 \end{eqnarray*} since $1 > 1-x > 0$. We are left with the integrals along $C_1$ and $C_3$. If in the integral along $C_1$ we take the limit as $\epsilon \to 0 , R \to \infty$, is the original integral (1). On the other hand, the integral over $C_3$ has the argument shifted by $2 \pi$ with respect to the original integral. That is,

\begin{eqnarray*} \lim_{\epsilon \to 0, R \to \infty} \int_{C_3} \frac{s^{x-1} ds}{(s+1)^{x+1}} = \int_{\infty}^0 \frac{ \mathrm{e}^{2 \pi \mathrm{i} (x-1)} s^{x-1} ds}{(\mathrm{e}^{2 \pi \mathrm{i}} s + 1)^{x+1}} = -\mathrm{e}^{2 \pi \mathrm{i} ( x-1)} \int_0^{\infty} \frac{s^{x-1} ds }{(s+1)^{x+1}} \end{eqnarray*} Putting all integrals together we find that

\begin{eqnarray*} (1 - \mathrm{e}^{2 \mathrm{i} \pi ( x-1) }) \int_0^{\infty} \frac{s^{x-1} ds}{(s+1)^{x+1}} = 2 \pi \; \mathrm{i} \; \mathrm{e}^{ \mathrm{i} \pi ( x-1)}. \end{eqnarray*} Hence

\begin{eqnarray*} \int_0^{\infty} \frac{s^{x-1} ds}{ (s+1)^{x+1}} = \frac{2 \pi \mathrm{i} \mathrm{e}^{ \mathrm{i} \pi ( x-1)}}{ 1 - \mathrm{e}^{2 \mathrm{i} \pi(x-1)}} = \frac{\pi}{( -\mathrm{e}^{- \pi x} + \mathrm{e}^{\mathrm{i} \pi x})/2 i} = \frac{\pi}{\sin \pi x} \end{eqnarray*}

We then showed that Euler's reflection formula


This can be done without invoking the Beta function explicitly. Directly from the integral definition of $\Gamma(x)$,the product in question is a double integral: $$ \Gamma(x)\Gamma(1-x) =\int^\infty_0\int^\infty_0 (u/v)^x u^{-1} e^{-(u+v)} du\, dv $$ Switching to polar coordinates via $v=x^2=r^2\cos^2\phi$, $u=y^2=r^2\sin^2\phi$, we have $$ \Gamma(x)\Gamma(1-x)= 4\int_0^{\pi/2} (\tan\phi)^{2x-1} \, d\phi \int_0^\infty r e^{-r^2}\, dr = 2\int_0^{\pi/2} (\tan\phi)^{2x-1} d\phi $$ Finally, $\phi= \tan^{-1}\sqrt{s}$, brings us to $$ \Gamma(x)\Gamma(1-x)=\int^\infty_0 {s^{x-1} ds\over 1+s} $$ which can be evaluated straightforwardly by contour integration as shown elsewhere on this page.


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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    $\begingroup$ The link is dead, but there is a working mirror on the Web Archive. $\endgroup$
    – Wojowu
    Feb 2, 2016 at 15:40
  • $\begingroup$ This isn't a one-line proof in any sense (it's actually 4.5 pages), but I was pleased to see that the whole thing can be done with elementary calculus. $\endgroup$
    – sasquires
    May 11, 2018 at 16:54

I have not seen a one line proof of this statement. I believe there is no such a thing. The statement usually takes some work.

Here is my proof: The formula states that \begin{eqnarray} \Gamma(1 -z ) \Gamma(z) = \frac{\pi}{\sin \pi z} \end{eqnarray} We show this formula using contour ingegration. We start with equation for the Beta function in terms of the $\Gamma$ function (second property) with $y=1-x$, and $0 < x < 1$. That is

\begin{eqnarray*} \Gamma(x) \Gamma(1-x) = \mathrm{B}(x, 1-x). \end{eqnarray*} We now show, by using contour integration, that

\begin{eqnarray*} \mathrm{B}(x, 1-x) = \frac{\pi}{\sin \pi x}. \end{eqnarray*} For this, we use a Beta function representation ($\mathrm{B}=\int_0^{\infty} s^{x-1}/(1+s)^{x+y}$) which for $y=1-x$ becomes

\begin{eqnarray*} \mathrm{B}(x, 1-x) = \int_0^{\infty} \frac{s^{x-1} ds}{s+1} \end{eqnarray*} We use contour integration. Let us first make the substitution $s=e^t$, $ ds = e^t dt$, and $t \in (-\infty, \infty)$. So we need to compute

We compute \begin{eqnarray*} \mathrm{B}(x, 1-x) = \int_{-\infty}^{\infty} \frac{\mathrm{e}^{t(x-1)} \mathrm{e}^t dt}{\mathrm{e}^t+1} = \mathrm{B}(x, 1-x) = \int_{-\infty}^{\infty} \frac{\mathrm{e}^{tx} dt}{\mathrm{e}^t+1} \quad , \quad 0 < x < 1. \end{eqnarray*}

Let us consider the contour integral

\begin{eqnarray} I = \int_C f(z) dz, \end{eqnarray} with \begin{eqnarray*} f(z) = \frac{\mathrm{e}^{zx}}{\mathrm{e}^z+1} \quad , \quad 0 < x < 1. \end{eqnarray*} and $C$ is the contour that we need to determine. In the complex plane, the poles of the integrand are the roots of $e^z+1$, that is $\mathrm{e}^z = -1 = \mathrm{e}^{(2k+1) \mathrm{i} \pi}$ so the roots are $z_k= (2k+1) \mathrm{i} \pi$, for $k=0, \pm 1, \pm 2, \cdots$. Then $f(z)$ as an infinite number of poles all lying on the imaginary axis. We will select a contour that has only one pole as shown in the Figure below

enter image description here The contour $C$ can be seen as the union of $C=C_1 \cup C_2 \cup C_3 \cup C_4$, where $C_1$ and $C_3$ are horizontal lines from $-R$ to $R$ with opposite orientation. We want to let $R$ grow to $\infty$. The paths $C_2$ and $C_4$ are vertical lines between $0$ and $2 \pi \mathrm{i}$ with opposite orientations showed in the figure.

From the Residue Theorem we evaluate the integral over $C$. The residue corresponding to the pole $z_0= \pi \mathrm{i}$, is computed using the expression

\begin{eqnarray*} \lim_{z \to z_0} (z-z_0) f(z) = \lim_{z \to z_0} \frac{ (z-z_0) \mathrm{e}^{z x}} {\mathrm{e}^z + 1} = \lim_{z \to z_0} \frac{\mathrm{e}^{zx} + (z-z_0) \mathrm{e}^{zx}}{e^z} = \mathrm{e}^{z_0 (x-1)}. \end{eqnarray*} where we use L'H^{o}pital's rule.

Hence $I = 2 \pi \mathrm{i} \; \mathrm{e}^{\pi i (x-1)}$ since the only residue inside the contour is at $z=\mathrm{i} \pi$ . That is,

\begin{eqnarray*} 2 \pi \; \mathrm{i} \; \mathrm{e}^{ \mathrm{i} \pi (x-1)} =\int_{C_1} f(z) dz + \int_{C_2} f(z) dz + \int_{C_3} f(z) dz + \int_{C_4} f(z) dz . \end{eqnarray*}

We want to find $I_1=\int_{C_1} f(z) dz$, as $R \to \infty$. Let us first find the integral along the vertical path $C_3$.

\begin{eqnarray*} I_3 &=& \int_R^{-R} \frac{\mathrm{e}^{ (t + 2 \pi \mathrm{i} ) x}}{\mathrm{e}^{t+2 \pi \mathrm{i}} + 1 } d t \\ &=& \mathrm{e}^{2 \pi \mathrm{i} x} \int_R^{-R} \frac{\mathrm{e}^{ t x}}{\mathrm{e}^{t+2 \pi \mathrm{i}} + 1 } d t \\ &=& \mathrm{e}^{2 \pi \mathrm{i} x} \int_R^{-R} \frac{\mathrm{e}^{ t x}}{\mathrm{e}^{t} + 1 } d t \\ &=& -\mathrm{e}^{2 \pi \mathrm{i} x} I_1, \end{eqnarray*} where we reversed the sign since $I_1$ is computed from $-R$ to $R$ instead of going in the opposite direction.

The integral $I_2$ along the path $C_2$ is evaluated as follows \begin{eqnarray*} I_2 = \int_0^{2 \pi} \frac{\mathrm{e}^{ (R + \mathrm{i} t ) x}} {\mathrm{e}^{R+\mathrm{i} t} + 1 } d t = \frac{\mathrm{e}^{R x} }{\mathrm{e}^R} \int_0^{2 \pi} \frac{\mathrm{e}^{ (\mathrm{i} t ) x}} {\mathrm{e}^{\mathrm{i} t} + 1/\mathrm{e}^{R} } d t = \mathrm{e}^{R(x-1)} \int_0^{2 \pi} \frac{\mathrm{e}^{ (\mathrm{i} t ) x}} {\mathrm{e}^{\mathrm{i} t} + 1/\mathrm{e}^{R} } d t. \end{eqnarray*} Now, since $0 < x < 1$ (so $x-1 < 0)$, and the last integral is bounded we have that $\lim_{R \to \infty} I_2 = 0$. The same argument applies for the integral $I_4$ along the path $C_4$. We then have that, from $I=I_1+I_2+I_3+I_4$,

\begin{eqnarray*} 2 \pi \mathrm{i} \mathrm{e}^{\mathrm{i} \pi (x-1)} = (1 - \mathrm{e}^{2 \pi \mathrm{i} x}) I_1 \end{eqnarray*} and

\begin{eqnarray*} I_1 = \int_{0}^{\infty} \frac{s^{x-1} ds}{s+1} = \frac{ 2 \pi \mathrm{i} \mathrm{e}^{ \mathrm{i} \pi (x-1)}}{1 - \mathrm{e}^{2 \pi \mathrm{i} x}} = \frac{ \pi}{ \frac{\mathrm{e}^{\mathrm{i} \pi x} - \mathrm{e}^{-\mathrm{i} \pi x}}{2 \mathrm{i}}} = \frac{\pi}{\sin \pi x}. \end{eqnarray*}

We then showed that Euler's reflection formula is correct.

  • $\begingroup$ Doesn't this just shift the problem to being: how do we prove this identity for the beta function? $\endgroup$
    – Yemon Choi
    Apr 15, 2016 at 2:41
  • $\begingroup$ We may integrate without exponential change of variable: join two long horisontal segments $\pm \varepsilon i+[0,R]$ by small semicircle near 0 and large arc (almost circle). $\endgroup$ Apr 15, 2016 at 8:41
  • $\begingroup$ @FedorPetrov HOw do you deal with the branch cuts at s=0? Assume p=1/2 then you have $\int_0^{\infty} ds/\sqrt{s}(s+1}$, then you have a square root branch cut between 0 and $\infty$. Then assume $p=1/3$ then you have two branch cuts, in general you could have as many brunch cuts as you want. Which contour covers them all? $\endgroup$ Apr 15, 2016 at 12:58
  • $\begingroup$ @YemonChoi: The answer is no. I just wanted to add some background. I still proof the reflection formula. Should I add one more line to my proof to state the final formula? $\endgroup$ Apr 15, 2016 at 12:59
  • $\begingroup$ $z^p$ is analytic in the domain ${\mathbb{C}}\setminus [0,+\infty)$, the contour lies in this domain, there is unique residue of $z^p/(1+z)$ inside the contour, in point $-1$. $\endgroup$ Apr 15, 2016 at 13:15

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