asymptotics of the Gamma function and remainder

Question

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

Answer 1

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 $x\in\mathbb{R}$)

Answer 2

Here is a cleaner way to do it. By $(5.11.8)$, we have that for any fixed real $x$, $$ {\mathop{\rm Re}\nolimits} \log \Gamma (x + {\rm i}y) \sim \tfrac{1}{2}\log (2\pi ) - \tfrac{\pi }{2}\left| y \right| + \left( { - \tfrac{1}{2} + x} \right)\log \left| y \right| + \sum\limits_{k = 1}^\infty {\frac{{( - 1)^{k + 1} B_{2k + 1} (x)}}{{2k(2k + 1)y^{2k} }}} $$ as $|y|\to +\infty$. Here the $B_k(x)$ denote the Bernoulli polynomials. Taking the exponential of each side gives \begin{multline*} \left| {\Gamma (x + {\rm i}y)} \right| \sim \sqrt {2\pi } \,{\rm e}^{ - \frac{\pi }{2}\left| y \right|} \left| y \right|^{ - \frac{1}{2} + x} \!\left( 1 + \frac{{x(2x^2 - 3x + 1)}}{{12y^2 }} \right. \\ \left. + \frac{{x(20x^5 - 132x^4 + 245x^3 - 150x^2 + 5x + 12)}}{{1440y^4 }} + \ldots \right) \end{multline*} as $|y|\to +\infty$, with any fixed real $x$. The coefficient $A_k(x)$ of $y^{-2k}$ in this expansion may be determined via the recurrence relation $$ A_{k + 1} (x) = \frac{1}{{2k + 2}}\sum\limits_{j = 0}^k {\frac{{( - 1)^j B_{2j + 3} (x)A_{k - j} (x)}}{{2j + 3}}} $$ for $k\ge 0$ with $A_0(x)=1$.