TITLE: One-line proof of the Euler's reflection formula
QUESTION [9 upvotes]: 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?

REPLY [3 votes]: 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

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