Daan Noordenbos

Curious linear asymptote

In this blog post we are going to be considering the function f:R+Rf:\mathbb{R}_+\to\mathbb{R} given by f(x)=Γ(1+1x)xf(x)= \Gamma\left(1+\frac{1}{x}\right)^x. As f(x)f(x) looks remarkably smooth and xf(x)xf(x) is almost linear, we will determine the linear asymptote of xf(x)xf(x). To determine this linear asymptote its slope and intercept have to be calculated. Evidently the slope is given by

limxxf(x)x=limxf(x).\lim\limits_{x\to\infty}\frac{xf(x)}{x}=\lim\limits_{x\to\infty}f(x).

To compute this limit we will use the Taylor expansions of the Gamma function. Famously, Γ(1+x)=1γx+O(x2)\Gamma(1+x)=1-\gamma x + O(x^2), where γ\gamma is the Euler-Mascheroni constant. Now,

limxf(x)=limxΓ(1+1x)x=limx(1γx+O(x2))x=limx(1γx)x=eγ.\begin{align*}\lim\limits_{x\to\infty}f(x)&=\lim\limits_{x\to\infty}\Gamma\left(1+\frac{1}{x}\right)^x\\&=\lim\limits_{x\to\infty}\left(1-\frac{\gamma}{x}+O(x^{-2})\right)^x\\&=\lim\limits_{x\to\infty}\left(1-\frac{\gamma}{x}\right)^x=e^{-\gamma}.\end{align*}

Where we have use of the fact that the term O(x2)O(x^{-2}) is dominated by γx-\frac{\gamma}{x}. The intercept of the linear asymptote is given by the following limit

limx(xΓ(1+1x)xxeγ).\lim\limits_{x\to\infty}\left(x \Gamma\left(1+\frac{1}{x}\right)^x-xe^{-\gamma}\right).

Again a Taylor expansion is used to determine this limit. The expansion in question is the following

logΓ(1+x)=γx+k=2(1)kζ(k)kxk,\log\Gamma(1+x)=-\gamma x+\sum\limits_{k=2}^{\infty}\frac{(-1)^k\zeta(k)}{k}x^k,

where ζ(s)=k=1ks\zeta(s)=\sum\limits_{k=1}^{\infty}k^{-s}. This Taylor expansion yields that, logΓ(1+x)=γx+π212x2+O(x3)\log\Gamma(1+x)=-\gamma x + \frac{\pi^2}{12}x^2+O(x^3). Now,

limx(xΓ(1+1x)xxeγ)=limxx(Γ(1+1x)xeγ)=limx0Γ(1+x)1xeγx=limx0ddx(Γ(1+x)1xeγ)=limx0(Γ(1+x)Γ(1+x)1x1x2logΓ(1+x))Γ(1+x)1x=limx01x2(xddxlogΓ(1+x)logΓ(1+x))Γ(1+x)1x=limx01x2(γx+π26x2+xO(x2)+γxπ212x2+O(x3))Γ(1+x)1x=limx01x2(π212x2+O(x3))Γ(1+x)1x=π212eγ\begin{align*}\lim\limits_{x\to\infty}\left(x \Gamma\left(1+\frac{1}{x}\right)^x-xe^{-\gamma}\right)&=\lim\limits_{x\to\infty}x \left(\Gamma\left(1+\frac{1}{x}\right)^x-e^{-\gamma}\right)\\&=\lim\limits_{x\to0}\frac{\Gamma\left(1+x\right)^\frac{1}{x}-e^{-\gamma}}{x}\\&=\lim\limits_{x\to0}\frac{\mathrm{d}}{\mathrm{d}x}\left(\Gamma\left(1+x\right)^\frac{1}{x}-e^{-\gamma}\right)\\&=\lim\limits_{x\to0}\left(\frac{\Gamma'(1+x)}{\Gamma(1+x)}\frac{1}{x}-\frac{1}{x^2}\log\Gamma(1+x)\right)\Gamma(1+x)^{\frac{1}{x}}\\&=\lim\limits_{x\to0}\frac{1}{x^2}\left(x\frac{\mathrm{d}}{\mathrm{d}x}\log\Gamma(1+x)-\log\Gamma(1+x)\right)\Gamma(1+x)^{\frac{1}{x}}\\&=\lim\limits_{x\to0}\frac{1}{x^2}\left(-\gamma x+\frac{\pi^2}{6}x^2+xO(x^2)+\gamma x - \frac{\pi^2}{12}x^2+O(x^3)\right)\Gamma(1+x)^{\frac{1}{x}}\\&=\lim\limits_{x\to0}\frac{1}{x^2}\left(\frac{\pi^2}{12}x^2+O(x^3)\right)\Gamma(1+x)^{\frac{1}{x}}\\&=\frac{\pi^2}{12}e^{-\gamma}\end{align*}

The linear asymptote is therefore π212eγ+eγx\frac{\pi^2}{12}e^{-\gamma}+e^{-\gamma}x, and Γ(1+1x)xπ2+12x12xeγ\Gamma\left(1+\frac{1}{x}\right)^x\approx \frac{\pi^{2}+12x}{12x}e^{-\gamma} for large xx. This approximation is within one procent of the true value for x1.7844x\ge1.7844.