Show $\Gamma(z)y^{-z}$ vanishes as $z\to\pm i\infty$

Question

I need to show that $\Gamma(z)y^{-z}$ vanishes as $z\rightarrow\pm i\infty$ if $|\arg(y)|< \frac{\pi}{2}$. Note that $z$ is purely imaginary.

Currently my approach is to use Stirling's Formula to get a representation for this product for large $|z|$. This gives $\sqrt{2\pi}z^{z}e^{-z}y^{-z}$.

I am struggling to make an observation which would cause this this product to vanish for large imaginary $z$. I tried rewriting this as $$\sqrt{2\pi}e^{z \log(z)}e^{-z}e^{-z\log(y)} = \sqrt{2\pi}e^{z\log(z)- z - z\log(y)} = \sqrt{2\pi}e^{z(\log(z) - \log(y) -1)}$$

Am I on the right track here?

ə̷̶̸͇̘̜́̍͗̂̄︣͟ · Accepted Answer · 2020-11-22T10:22:34.280

0

We have $\Gamma(z)y^{-z}=\Gamma(ix)e^{-ix\operatorname{Log}y}$ and $\operatorname{Log}y=\log|y|+i\arg y$ so $$|\Gamma(z)y^{-z}|=|\Gamma(ix)||e^{x\arg y}|<\sqrt{\frac{\pi e^{\pi x}}{x\sinh\pi x}}\to0$$ as $|x|\to\infty$ using Euler's reflection formula.

edited Nov 22 '20 at 10:22

answered Nov 22 '20 at 10:08

ə̷̶̸͇̘̜́̍͗̂̄︣͟

27,005

For the step $|\Gamma(z)y^{-z}| = |\Gamma(ix)|$, how can you make the claim that $y^{-z}$ goes to 1? I was under the impression that $e^{-ixlog(y)}$ has no limit as $x\rightarrow\infty$ – whatwasthat Nov 22 '20 at 10:16
You are correct, will edit – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Nov 22 '20 at 10:18
Is there a method to solve this using Sterling's Formula? – whatwasthat Nov 22 '20 at 10:19
See revised. Stirling's formula is probably too complicated for this problem as you'd have to deal with terms like $z^z$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Nov 22 '20 at 10:26
I see. For $z^{z}$, could you do $e^{zlog(z)}$ then replace z with ix, yielding $e^{ixlog(ix)}$, and as ${x\rightarrow\infty, e^{ixlog(ix)}\rightarrow0}$? – whatwasthat Nov 22 '20 at 10:35
Yes that works too. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Nov 22 '20 at 10:37
Awesome, thank you very much! – whatwasthat Nov 22 '20 at 10:43

Show $\Gamma(z)y^{-z}$ vanishes as $z\to\pm i\infty$

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