# 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!

• Taking complex conjugate on $s$ in your defining equation of $\Gamma(s)$ gives you $\Gamma(\bar{s}) = \overline{\Gamma(s)}$. – achille hui Nov 13 '13 at 5:22

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)