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)?
 A: 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
$$
\left|\Gamma\left(\tfrac12+iy\right)\right|^2=\frac{\pi}{\cosh(\pi y)}\tag{7}
$$
A: Use Stirling approximation for the gamma function and see what you get
$$ \Gamma(z+1)\sim\sqrt{2\pi z}\left(\frac{z}{e}\right)^{z}\,.$$
A: 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 !
