On the modulus of $\Gamma(z)$ In about two weeks, I'm going to be giving a presentation on the complex-valued Gamma function $\Gamma(z)$. By definition, I know that $$\Gamma(z)= \int_0^\infty e^{-t}t^{z-1}dt.$$  
Now if I let $z=x+iy$, how does the following hold? $$|\Gamma(x+iy)| \leq \Gamma(x).$$  
It might actually be something quite simple, but here's what I attempted: $$|\Gamma(x+iy)|= |\int_0^\infty e^{-t}t^{z-1}dt|$$ $$=|\int_0^\infty e^{-t}t^{x+iy-1}dt|$$ $$=|\int_0^\infty e^{-t}t^{x-1}t^{iy}dt|$$ $$\leq|\int_0^\infty e^{-t}t^{x-1}dt|$$ $$=\int_0^\infty e^{-t}t^{x-1 }dt$$ $$=\Gamma(x),$$
where $t \gg 1$. I hope this was the correct procedure.
 A: As you've noted, for $x>0$,
$$
\begin{align}
|\Gamma(x+iy)|
&=\left|\int_0^\infty e^{-t}\,t^{x+iy-1}\;\mathrm{d}t\right|\\
&=\left|\int_0^\infty e^{-t}\,t^{x-1}\,e^{iy\log(t)}\;\mathrm{d}t\right|\\
&\le\int_0^\infty\left|e^{-t}\,t^{x-1}\,e^{iy\log(t)}\right|\;\mathrm{d}t\\
&=\int_0^\infty e^{-t}\,t^{x-1}\;\mathrm{d}t\\
&=\Gamma(x)\\
&=|\Gamma(x)|\tag{1}
\end{align}
$$
For $-1< x<0$, we can use
$$
\begin{align}
|\Gamma(x+iy)|
&=\frac{1}{|x+iy|}|\Gamma(x+1+iy)|\\
&\le\frac{1}{|x|}|\Gamma(x+1)|\\
&=|\Gamma(x)|\tag{2}
\end{align}
$$
then we can repeat $(2)$ for $-n-1<x<-n$ for $n=1,2,3,\dots$
To finish off, note that $\Gamma(z)$ is not defined (or infinite) for $z$ a non-positive integer.
This yields $|\Gamma(x+iy)|\le|\Gamma(x)|$ for all $x$ where $\Gamma(x)$ is defined (or finite).
A: Added. For $\text{Re}(z)=x>0$ the Gamma integral converges
$$\Gamma(z) = \int_0^\infty e^{-t} t^{z-1} \,{\rm d}t.$$

Apply the following property of moduli of integrals
$$\begin{equation*}
\left\vert \int_{0}^{\infty }w(t)dt\right\vert \leq \int_{0}^{\infty
}\left\vert w(t)\right\vert dt
\end{equation*}$$
to the function $w(t)=e^{-t}t^{x+iy-1}=e^{-t}t^{x-1}t^{iy}$ and observe that 


*

*$\left\vert t^{iy}\right\vert =\left\vert e^{iy\log t}\right\vert =1,$

*$\left\vert e^{-t}t^{x-1}\right\vert =e^{-t}t^{x-1}$ for $t>0$.

