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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.

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    Put the absolute value inside the integral (using the triangle inequality $\left|\int f\right| \le \int |f|$), then use $|t^{x+iy}| = t^x$.2012-02-28

2 Answers 2

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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 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).

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    Thanks for the explanation.2012-03-01
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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 $t>0$.