# Limiting behavior of gamma function

I am trying to determine whether $\Gamma(x+iy)\rightarrow 0$ as $y\rightarrow\infty$. How should I go about doing it?

I was trying to see if I could get anything from $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}$ but although $|\sin z|\rightarrow\infty$ as $y\rightarrow\infty$, I think it does not follow that $\Gamma(z)\rightarrow 0$. Am I right?

Another approach I was trying was a change of variables by letting $u=\ln t$ so that (for $x>0$,) $\Gamma(x+iy)=\int_{-\infty}^{\infty}e^{xu}e^{-e^{u}}e^{iyu}du$. I have a couple of questions about this. First, is $e^{xu}e^{-e^{u}}$ integrable over the real line? Next, is there something about Fourier transforms that I can use here (perhaps the Riemann-Lebesgue lemma)?

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First a simple estimate: \begin{align} |\Gamma(z)| &=\left|\;\int_0^\infty t^{z-1}e^{-t}\,\mathrm{d}t\;\right|\\ &\le\int_0^\infty t^{\mathrm{Re}(z)-1}e^{-t}\,\mathrm{d}t\\ &=\Gamma(\mathrm{Re}(z))\tag{1} \end{align} As CYC mentions, this estimate can be used to show that $$|\Gamma(z)|=\frac{|\Gamma(z+1)|}{|z|}\le\frac{|\Gamma(\mathrm{Re}(z+1))|}{|z|}\tag{2}$$ Which gives the desired decay.

Polynomial Decay

In fact, for $z=x+iy$, we get \begin{align} |\Gamma(x+iy)| &=\frac{|\Gamma(z+n)|}{|z(z+1)(z+2)\dots(z+n-1)|}\\[6pt] &\le\frac{|\Gamma(x+n)|}{|y|^n}\tag{3} \end{align} which gives decay faster than any power of $1/|y|$.

Exponential Decay

Using estimate $(3)$ and $\frac{\Gamma(x+n)}{n!}\le(n+x)^x$, we get \begin{align} e^{\alpha|y|}|\Gamma(x+iy)| &\le\sum_{n=0}^\infty\frac{\alpha^n|y|^n}{n!}\frac{|\Gamma(x+n)|}{|y|^n}\\ &\le\sum_{n=0}^\infty\alpha^n(n+x)^x\\ &=C(\alpha,x)\tag{4} \end{align} which converges for $\alpha<1$. Thus, for $\alpha<1$ $$|\Gamma(x+iy)|\le C(\alpha,x)e^{-\alpha|y|}\tag{5}$$

Particular Value

Since $\Gamma$ is real on the real axis, $$\Gamma\left(x-iy\right)=\overline{\Gamma\left(x+iy\right)}\tag{6}$$ Therefore, applying the reflection formula for $\Gamma$ yields $$\left|\Gamma\left(\tfrac12+iy\right)\right|^2=\frac{\pi}{\cosh(\pi y)}\tag{7}$$

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It turns out I can use the inequality in the following way: $|\Gamma(z)|=\frac{|\Gamma(z+1)|}{|z|}\leq\frac{\Gamma(\mathrm{Re}(z)+1)}{|z|} \rightarrow 0$. This gives the result for $\mathrm{Re}(z)>0$. Then I can use $\Gamma(z)=\frac{\Gamma(z+1)}{z}$ in an inductive argument to get the result for $\mathrm{Re}(z)\leq 0$. –  user44532 Oct 16 '12 at 1:47
@CYC: of course! very nice. This same argument shows that for fixed $\mathrm{Re}(z)$, $|\Gamma(z)|$ decays faster than any power of $|z|$. –  robjohn Oct 16 '12 at 21:27

Use Stirling approximation for the gamma function and see what you get

$$\Gamma(z+1)\sim\sqrt{2\pi z}\left(\frac{z}{e}\right)^{z}\,.$$

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From your identity $$\Gamma(z)\,\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$$ you may indeed deduce : $$\Gamma(iy)\,\Gamma(-iy)=|\Gamma(iy)|^2=\frac{\pi}{y\,\sinh(\pi y)}$$ observe that $$\Gamma(x+iy)\approx (iy)^x\,\Gamma(iy)\quad\text{for}\ |x|\ll |y|$$ and conclude that you are right !
@robjohn: Well starting with $\Gamma(n+1+iy)=(n+iy)\Gamma(n+iy)$, using $n\ll |y|$ and the regularity of $\Gamma$ for $x\approx n$ to conclude I think... –  Raymond Manzoni Oct 13 '12 at 9:35
I couldn't get the second equation myself ($\Gamma(iy)\,\Gamma(-iy)=|\Gamma(iy)|^2=\frac{\pi}{y\,\sinh(\pi y)}$). How is it done? –  user44532 Oct 13 '12 at 19:05
@ChungYeongChyuan: You may use : $\Gamma(iy)\,\Gamma(1-iy)=\frac{\pi}{\sin(i\pi y)}$ i.e. $-iy\,\Gamma(iy)\,\Gamma(-iy)=\frac{\pi}{-i\sinh(\pi y)}\$ since $\Gamma(1+w)=w\Gamma(w)$ for $w=-iy\$ and $-i\sin(ix)=\sinh(x)$ as seen here (use too $\Gamma(\overline{z})=\overline{\Gamma(z)}$). –  Raymond Manzoni Oct 13 '12 at 20:14