# asymptotics of the Gamma function and remainder

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

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 $x\in\mathbb{R}$)

• 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 $\vert y\vert\rightarrow \infty$? – Hasti Musti Aug 4 '16 at 13:00
• @HastiMusti: I updated my answer. Cheers, – Raymond Manzoni Aug 4 '16 at 17:15