Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question
up vote 3 down vote accepted

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 $$ \left|\Gamma\left(\tfrac12+iy\right)\right|^2=\frac{\pi}{\cosh(\pi y)}\tag{7} $$

share|cite|improve this answer
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

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

share|cite|improve this answer

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 !

share|cite|improve this answer
I know how to prove Gautschi's inequality on the real line, but how does one prove the last estimate you have? Usual convexity arguments appear to fail. – robjohn Oct 13 '12 at 9:32
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.