TITLE: Lower bound for $\Gamma(x+i y)$ where $x>0$ involving only the real part
QUESTION [6 upvotes]: I am trying to have some inequalities involving the special Gamma function. I am able to get an upper bound for  $\Gamma(x+i y)$, for $x>0$,
$$
\begin{align}
|\Gamma(x+iy)|
&=\left|\int_0^\infty e^{-t}\,t^{x+iy-1}\;\mathrm{d}t\right|\\
&=\left|\int_0^\infty e^{-t}\,t^{x-1}\,e^{iy\log(t)}\;\mathrm{d}t\right|\\
&\le\int_0^\infty\left|e^{-t}\,t^{x-1}\,e^{iy\log(t)}\right|\;\mathrm{d}t\\
&=\int_0^\infty e^{-t}\,t^{x-1}\;\mathrm{d}t\\
&=\Gamma(x)\\
&=|\Gamma(x)|\tag{1}.
\end{align}
$$
However, I did not succeed to get a lower bound for  $\Gamma(x+i y)$ such that I get rid of the imaginary part. Any help in this direction?
Using the suggestion of @reuns and the Euler's reflection formula, one can shows that 
\begin{align} \left|\Gamma(x+i y)\right| &= \frac{\pi}{\left|\Gamma(1-(x+i y)\right| \left|\sin(\pi (x+i y))\right|} \\
&\ge \frac{\pi}{\left|\Gamma(1-x)\right| \left|\sin(\pi (x+i y))\right|}
 \end{align}
But I can not still get a lower bound for the complex sine term such that I get rid of the imaginary part! Any suggestion?

REPLY [4 votes]: According to the NIST Handbook, edition of 2010, formula 5.6.7, there is the estimate
$$
|\Gamma(x + i y)| \ge \frac{ \Gamma(x)}{\sqrt{\cosh (\pi y)}} 
$$ 
for $x \ge \frac{1}{2}$. By the comment above, this is sharp for $x = \frac{1}{2}$. This seems to say that you can't get rid of the imaginary part.