Identity $\int_{-\infty}^{\infty}\frac{e^{uz}}{1+e^u} \mathrm{d}u=\frac{\pi}{\sin(\pi z)}$ I want to prove the identity 
$$F(z)=\int_{-\infty}^{\infty}\frac{e^{uz}}{1+e^u} \mathrm{d}u=\frac{\pi}{\sin(\pi z)}$$
First of all $F(z)$ defines an analytic function for $0<z<1$. I am little bit confused here because I think the notation is not very good. $\frac{e^{uz}}{1+e^u}$ has the period $2\pi i$ and is bounded in $[0,1]\times S$ where $S=\{u\in\mathbb C: |Im z|\le\pi, |u|\ge r\}$ Ths is a punched range if you draw it.
You have any idea how to compute the integral now?
 A: You could make a substitution and relate it to the beta function.
Or let $ \displaystyle f(s) = \frac{e^{zs}}{1+e^{s}} $ and integrate around a rectangle in the upper half-plane of height $2 \pi$.
Then $$\int_{-\infty}^{\infty} f(x) \ dx + \lim_{R \to \infty}\int^{2 \pi}_{0} f(R+it) i dt  + \int_{\infty}^{-\infty} f(t+2\pi i) \ dt + \lim_{R \to \infty} \int^{2 \pi}_{0} f(-R+it) i   dt$$  $$= 2 \pi i \text{Res} [f(s), s= \pi i] = 2 \pi i \lim_{s \to \pi i} \frac{e^{sz}}{e^{s}} = -2 \pi i e^{\pi i z}$$
The second and fourth integrals vanish and we have
$$ \int_{-\infty}^{\infty} \frac{e^{zx}}{1+e^{s}} - \int_{-\infty}^{\infty} \frac{e^{2 \pi i z}e^{zt}}{1+e^{t}} \ dt = - 2 \pi i e^{z \pi i} $$
$$ \implies \int_{-\infty}^{\infty} \frac{e^{zx}}{1+e^{x}} \ dx = \frac{-2 \pi i e^{\pi i z}}{1-e^{2 \pi i z}} = \frac{2 \pi i}{e^{\pi i z} -e^{\pi i z}} = \frac{\pi}{\sin \pi z}$$
A: Hint. Set $e^u=x^2$, then $e^udu=2xdx$, $e^{uz}=x^{2z}$. The function $x^{z}=\exp(z\log x)$
is analytic in $x$, in $\mathbb C\smallsetminus\{it:t\le 0\}$. And
$$
\int_{-\infty}^\infty \frac{e^{zu}\,du}{1+e^u}=\int_{-\infty}^\infty \frac{x^{2z-1}dx}{1+x^2}.
$$
Then use the countour $c_1\cup c_2\cup c_3\cup c_4$, where
$$
c_1(t)=t,\,\,t\in[\varepsilon,R],\quad c_2(t)=Re^{it}, t\in[0,\pi],\quad c_3(t)=t,\,\,t\in[-R,-\varepsilon], \quad c_4(t)=\varepsilon\,e^{i(\pi-t)}, \,\, t\in [0,\pi].
$$
A: $(0).$ Let $t=e^u.\qquad(1)$. Let $x=\dfrac1{1+t}\qquad(2).$ Recognize the expression of the beta function in the new integral. $\ \quad\ (3).$ Use Euler's reflection formula.

The reason I'm writing this is because all integrals of the form $\displaystyle\int_0^\infty\frac{t^a}{(1+t^b)^c}dt$ can be solved by the steps $(1)$ to $(3)$, by letting $x=\dfrac1{1+t^b}$
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{{\rm F}\pars{z} \equiv
     \int_{-\infty}^{\infty}{\expo{uz} \over 1 + \expo{u}}\,\dd u
    = {\pi \over \sin\pars{\pi z}}:\ {\Large ?}.\qquad z\ \in\ \pars{0,1}}$

Let's $\ds{\quad t \equiv {1 \over 1 + \expo{u}}\ \iff\ \expo{u} = {1 \over t} - 1\,,\ u = \ln\pars{1 - t} - \ln\pars{t}}$. Then
  \begin{align}
\color{#00f}{\large{\rm F}\pars{z}}&=\int_{1}^{0}t\pars{1 - t \over t}^{z}\,
\pars{-\,{1 \over 1 - t} - {1 \over t}}\,\dd t
=\int_{0}^{1}t^{1 - z}\pars{1 - t}^{z - 1}\,\dd t
+ \int_{0}^{1}t^{-z}\pars{1 - t}^{z}\,\dd t
\\[3mm]&={\rm B}\pars{2 - z,z} + {\rm B}\pars{1 - z,z + 1}
={\Gamma\pars{2 - z}\Gamma\pars{z} \over \Gamma\pars{2}}
+ {\Gamma\pars{1 - z}\Gamma\pars{z + 1} \over \Gamma\pars{2}}
\\[3mm]&=\pars{1 - z}\Gamma\pars{1 - z}\Gamma\pars{z}
+\Gamma\pars{1 - z}z\Gamma\pars{z} = \Gamma\pars{z}\Gamma\pars{1 - z}
=\color{#00f}{\large{\pi\over \sin\pars{\pi z}}}
\end{align}

${\rm B}$ and $\Gamma$ are the Beta and Gamma functions, respectively, and we used well known properties of both of them.
A: I really do not know if this could help you or not. The antiderivative is $$\frac{e^{u z} \, _2F_1\left(1,z,z+1,-e^u\right)}{z}$$ which satisfies the equality you want to prove. 
