1
$\begingroup$

I have found the following asymptotic formula in a book:

$$\lim_{\vert y\vert\rightarrow\infty}\vert \Gamma(x+iy) \vert e^{\frac{\pi}{2}\vert y\vert}\vert y \vert^{\frac{1}{2}-x}= \sqrt{2\pi}. $$

I would like to know if there are even sharper estimates, in particular is it true that: $$\vert \Gamma(x+iy) \vert = \sqrt{2\pi}\cdot e^{-\frac{\pi}{2}\vert y\vert}\vert y \vert^{-\frac{1}{2}+x}\left( 1+O\left( \frac{1}{\vert y \vert}\right) \right) , \quad \vert y \vert\rightarrow \infty$$

Unfortunately I could not find any reference. Do you know some literature where I can find this result stated?

Best wishes

$\endgroup$
2
$\begingroup$

There is an exact solution for $x=\dfrac 12$ as shown in this answer :

$$\tag{1}\left|\Gamma\left(\frac 12+iy\right)\right|=\sqrt{\frac{\pi}{\cosh\left(\pi y\right)}}$$

More general asymptotic results were obtained here like your : $$\tag{2}\left|\Gamma\left(x+iy\right)\right|=\sqrt{2\pi}\,e^{-\pi|y|/2}\,|y|^{x-1/2}(1+r(x,y))$$ with $|r(x,y)|\to 0$ uniformly for $x<K$ as $|y|\to\infty$.

and the more precise (also for $|y|\to\infty$) : \begin{align} |\Gamma(x+iy)|&\sim \sqrt{2\pi}\,\exp\left[\frac{\left(x-\frac 12\right)\;\ln\bigl(x^2+y^2\bigr)}2-y\;\arg(x+iy)-x+\sum\limits_{m=1}^\infty \frac{B_{2m}\;\Re\;{z^{1-2m}}}{2m\,(2m-1)}\right]\\ (3)\qquad&\sim \sqrt{2\pi}\,\bigl(x^2+y^2\bigr)^{\frac x2-\frac 14}\exp\left[-x-y\,\arg(x+iy)+\frac 1{12}\frac x{x^2+y^2}-\frac 1{360}\frac{x^3-3xy^2}{(x^2+y^2)^3}+\cdots\right]\\ \end{align} Addition

We may find more about the error term $r(x,y)$ by dividing the more precise expansion $(3)$ by $(2)$ ex-purged of $(1+r(x,y))$ of course!

For $x> 0\,$ fixed and using the expansion $\;\displaystyle\arg(x+iy)=\arctan\frac yx\sim\frac{\pi}2-\frac xy+\frac{x^3}{3\,y^3}+O\left(\frac 1{y^5}\right)$ we get as $\,y\to +\infty$ : \begin{align} 1+r(x,y)&\sim \frac{\bigl(x^2+y^2\bigr)^{\large{\frac x2-\frac 14}}}{|y|^{x-1/2}}\exp\left(-y\,\arg(x+iy)-(-\pi|y|/2)-x+\frac 1{12}\frac x{x^2+y^2}+O\left(\frac 1{y^4}\right)\right)\\ &\sim \left(1+\frac{x^3-\frac {x^2}2}{2\,y^2}+O\left(\frac 1{y^4}\right)\right)e^{\Large{-y\frac{\pi}2+x-\frac{x^3}{3\,y^2}+\frac{\pi}2|y|-x+\frac x{12\,y^2}+O\left(\frac 1{y^4}\right)}}\\ &\sim \left(1+\frac{x^3-\frac {x^2}2}{2\,y^2}+O\left(\frac 1{y^4}\right)\right)\left(1-\frac{x^3}{3\,y^2}+\frac x{12\,y^2}+O\left(\frac 1{y^4}\right)\right)\\ &\sim 1+\frac{\frac x{12}-\frac{x^3}3+\frac{x^3}2-\frac{x^2}4}{y^2}+O\left(\frac 1{y^4}\right)\\ \end{align} and conclude (with numerical confirmation) that : $$r(x,y)\sim \dfrac{2\,x^3-3\,x^2+x}{12\,y^2}+O\left(\frac 1{y^4}\right)$$ (note that for $x\in\left\{0,\frac 12,1\right\}$ the $\dfrac 1{y^2}$ term disappears and that this formula is valid for any $x\in\mathbb{R}$)

$\endgroup$
  • $\begingroup$ Thank you for your answer. Suppose we fix the real part $x\in\mathbb{R}$. Can there be anything said about the order of $\vert r(x,y) \vert$ with which it converges to $0$, if $\vert y\vert\rightarrow \infty$? $\endgroup$ – Hasti Musti Aug 4 '16 at 13:00
  • $\begingroup$ @HastiMusti: I updated my answer. Cheers, $\endgroup$ – Raymond Manzoni Aug 4 '16 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.