# Lower bound for $\Gamma(x+i y)$ where $x>0$ involving only the real part

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?

• You won't obtain anything directly from the integral, you need to show the explicit product formula for $\Gamma(s)$, or things like the reflection formula. – reuns Jan 31 '19 at 8:21
• Thank you reuns for your comment. Can you please provide more details or a reference about that. Thanks. – user144209 Jan 31 '19 at 9:55
• $\Gamma(s)\Gamma(1-s) = \pi/\sin(\pi s)$ gives the exact decay on $\Re(s) \in \mathbb{Z}/2$. For other $\Re(s)$ the easiest is to look at the (complex) Stirling approximation, a direct consequence of the product formula en.wikipedia.org/wiki/… – reuns Jan 31 '19 at 10:02

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.
• $|\Gamma|$ goes very fast to zero on any fixed vertical line as the imaginary part goes to infinity (both plus and minus for that matter) so obviously any lower bound must involve the imaginary part... – Conrad Feb 14 '19 at 15:59
• well I can get a better bound from my derivation if I am not worried about the imaginary part using the fact that $|sin(x+iy)| \le sinh(\sqrt{x^2+y^2})$ but my aim is to get rid of the imaginary part. – user144209 Feb 15 '19 at 7:42