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!

  • $\begingroup$ Taking complex conjugate on $s$ in your defining equation of $\Gamma(s)$ gives you $\Gamma(\bar{s}) = \overline{\Gamma(s)}$. $\endgroup$ – 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)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.