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I am searching for a simple proof of this sentence: For $\sigma$ fixed and $|t|\rightarrow\infty$ $$|\Gamma(\sigma+it)|\sim e^{-\frac{1}{2}\pi |t|}|t|^{\sigma-\frac{1}{2}}\sqrt{2\pi}$$ I am wondering if there is an easy proof.I've searched for other familiar questions but it doesn't seem to be the right answer for me.I would appreciate any suggestion about some bibliography or some specific hints for this proof.

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See the lecture notes by James McKernan.

Rough idea:

Starting from the formula for the second logarithmic derivative of the gamma function: $$ \frac{d}{dz}\left(\frac{\Gamma'(z)}{\Gamma(z)}\right) = \sum_{n=0}^\infty\frac1{(z + n)^2}, $$ and using residue calculus, can be proved that $$ \frac{d}{dz}\left(\frac{\Gamma'(z)}{\Gamma(z)}\right) = \frac1{2z^2} + \int_0^{\infty}\coth\pi v\,\frac{2vz}{(v^2 + z^2)^2}\,dv = \frac1{z} + \frac1{2z^2} + \int_0^{\infty}\frac{4vz}{(v^2 + z^2)^2}\frac{dv}{e^{2\pi v} - 1}. $$ Now, we can integrate in the right half plane: $$ \frac{\Gamma'(z)}{\Gamma(z)} = C + \log z + \frac1{2z} + \int_0^{\infty}\frac{2v}{v^2 + z^2}\frac{dv}{e^{2\pi v} - 1}. $$ Integrating again, $$ \log\Gamma(z) = C' + Cz + (z -1/2)\log z + \frac1\pi\int_0^\infty\frac{z}{(v^2 + z^2)^2}\,\log\frac1{1 - e^{2\pi v}}\,dv $$ (with $C\ne$ previous $C$)

Bounding the integral and using the functional equation $\Gamma(z)\Gamma(1 - z) = \pi/\sin(\pi z)$, the constants can be calculated. This gives $$ \log\Gamma(z) = \frac12\log(2\pi) - z + (z - 1/2)\log z + J(z) $$ with $J(z)$ "small" when $\Re(z)$ is "big". Finally, $$\Gamma(z) = \sqrt{2\pi}z^{z - 1/2}e^{-z}e^{J(z)}.$$

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