Formula for $\Gamma (\frac{1}{2} + i t)$ I have been working on the following problem for my complex analysis class involving Euler's Gamma function:
For 
$$\Gamma (s) := \int_0 ^{\infty} t^{s-1} e^{-t} \,dt \ , \ Re(s)>0$$
Show that
$$\left\lvert \Gamma\left(\frac{1}{2} + i t\right)\right\rvert ^2 = \frac{2\pi}{e^{\pi t} + e^{-\pi t}}$$
for $t\in\mathbb{R}$.  I am most of the way there, but have gotten hung up.  So far, I have used the reflection forumla:
$$\Gamma(z) \Gamma (1-z) = \frac{\pi}{\sin (\pi z)}$$
which initially holds only for $Re(z)>0$ but is shown to hold for $z\in\mathbb{C}\setminus \mathbb{Z}_{\le 0}$ by analytic continuation.  It is clear that for any $t \in \mathbb{R}$, $\frac{1}{2} + i t \in \mathbb{C}\setminus \mathbb{Z}_{\le 0}$ , so I apply the reflection formula with $z=\frac{1}{2} + i t$.  A computation using the complex sine function shows that the desired quantity is obtained on the right hand side; namely,
$$\Gamma \left(\frac{1}{2} + it\right)\Gamma \left(1-\left(\frac{1}{2} + it\right)\right) = \frac{2\pi}{e^{\pi t} + e^{-\pi t}}$$
What I am having difficulty with is showing that 
$$\Gamma \left(\frac{1}{2} + it\right)\Gamma \left(1-\left(\frac{1}{2} + it\right)\right) = \left\lvert \Gamma\left(\frac{1}{2} + i t\right)\right\rvert ^2$$
Any guidance would be much appreciated, as always!
 A: Taking the hint from the comment, we can easily verify that $\Gamma(\bar{s}) = \overline{\Gamma(s)}$.  I'll do that at the end.
Then:
$$\begin{align}
\frac{2\pi}{e^{\pi t} + e^{-\pi t}} &= \Gamma\left(\frac{1}{2} + it\right)\Gamma\left(1-\left(\frac{1}{2} + it\right)\right)\\
&= \Gamma\left(\frac{1}{2} + it\right)\Gamma\left(\frac{1}{2} - it\right)\\
&= \Gamma\left(\frac{1}{2} + it\right)\Gamma\left(\overline{\frac{1}{2} + it}\right)\\
&= \Gamma\left(\frac{1}{2} + it\right)\overline{\Gamma\left(\frac{1}{2} + it\right)}\\
&= \left| \Gamma\left(\frac{1}{2} + it\right)\right|^2\\
\end{align}$$
As desired.
Now to show that $\Gamma(\bar{s}) = \overline{\Gamma(s)}$.
$$\begin{align}
\Gamma\left(\overline{a+bi}\right) &= \Gamma\left(a-bi\right) \\
&= \int_0^{\infty} t^{(a-bi)-1} e^{-t} \,dt \\
&= \int_0^{\infty} e^{\ln(t)\left((a-bi)-1\right)} e^{-t} \,dt \\
&= \int_0^{\infty} e^{\ln(t)(a-1)}e^{-\ln(t)bi} e^{-t} \,dt \\
&= \int_0^{\infty} t^{(a-1)}e^{-t}\left(\cos(-\ln(t)b) + i\sin(-\ln(t)b)\right) \,dt \\
&= \int_0^{\infty} t^{(a-1)}e^{-t}\left(\cos(\ln(t)b) - i\sin(\ln(t)b)\right) \,dt \\
&= \int_0^{\infty} t^{(a-1)}e^{-t}\cos(\ln(t)b)\,dt - i\int_0^\infty t^{(a-1)}e^{-t}\sin(\ln(t)b) \,dt \\
&= \left(\overline{\int_0^{\infty} t^{(a-1)}e^{-t}\cos(\ln(t)b)\,dt + i\int_0^\infty t^{(a-1)}e^{-t}\sin(\ln(t)b) \,dt} \right)\\
\vdots\\
&=\overline{\Gamma(a+bi)}
\end{align}$$
(simply follow the same steps backward to complete the conjugate demonstration)
