# A question on Gamma function

This might be basic but I have difficulty understanding what exactly goes wrong in the following logic:

Consider the Gammma function

$$\Gamma(z) = \int_0^{\infty} t^{z-1} \, e^{-t}\,dt \quad \textrm{Re}(z) >0.$$

Now let us consider $$z=1+iy$$ with $$y\in \mathbb R$$. Then, it seems to me that by triangle inequality there holds: $$|\Gamma(1+iy)| \leq \int_0^{\infty} |t^{iy}\,e^{-t}|\,dt \leq 1.$$ However, the previous inequality is obviously in contradiction with Stirling's asymptotic formula for $$y$$ large. Where is the flaw in this?

• $\Gamma$ is fast decreasing on vertical lines so no contradiction Commented May 5 at 14:57

In fact, Stirling's asymptotic $$\Gamma(s)\sim s^{s-{1\over 2}}\cdot e^{-s}\cdot (1+O(s^{-1}))$$ does apply to complex arguments. The hazard is correctly interpreting the $$s^s$$ that appears, as $$s\to\infty$$ along vertical lines. Namely, for example, $$(1+it)^{1+it} \;=\; e^{(1+it)\log(1+it)}$$ As $$t\to +\infty$$, $$\log(1+it)$$ is essentially $${\pi i\over 2}+\log t$$. Thus, the real part of the exponent is essentially $$-{\pi\over 2}t + \log t$$. Exponentiating gives the vertical exponential decrease...
$$\Gamma(1+iy)=\int_{-\infty}^{\infty}e^{iyx}f(x)dx$$ where $$f(x)=\exp (-e^x+x)$$ is a nice probability density. So $$\Gamma(1+it)\to_{|y |\to \infty}0.$$ More explicitely $$|\Gamma(1+iy)|^2=\frac{\pi y}{\sinh y}$$ from the magic formula $$\Gamma(p)\Gamma(1-p)=\frac{\pi}{\sin \pi p}.$$