In this blog post we are going to be considering the function f:R+→R given by f(x)=Γ(1+x1)x. As f(x) looks remarkably smooth and xf(x) is almost linear, we will determine the linear asymptote of xf(x). To determine this linear asymptote its slope and intercept have to be calculated. Evidently the slope is given by
x→∞limxxf(x)=x→∞limf(x).
To compute this limit we will use the Taylor expansions of the Gamma function. Famously, Γ(1+x)=1−γx+O(x2), where γ is the Euler-Mascheroni constant. Now,
x→∞limf(x)=x→∞limΓ(1+x1)x=x→∞lim(1−xγ+O(x−2))x=x→∞lim(1−xγ)x=e−γ.
Where we have use of the fact that the term O(x−2) is dominated by −xγ. The intercept of the linear asymptote is given by the following limit
x→∞lim(xΓ(1+x1)x−xe−γ).
Again a Taylor expansion is used to determine this limit. The expansion in question is the following
logΓ(1+x)=−γx+k=2∑∞k(−1)kζ(k)xk,
where ζ(s)=k=1∑∞k−s. This Taylor expansion yields that, logΓ(1+x)=−γx+12π2x2+O(x3). Now,
x→∞lim(xΓ(1+x1)x−xe−γ)=x→∞limx(Γ(1+x1)x−e−γ)=x→0limxΓ(1+x)x1−e−γ=x→0limdxd(Γ(1+x)x1−e−γ)=x→0lim(Γ(1+x)Γ′(1+x)x1−x21logΓ(1+x))Γ(1+x)x1=x→0limx21(xdxdlogΓ(1+x)−logΓ(1+x))Γ(1+x)x1=x→0limx21(−γx+6π2x2+xO(x2)+γx−12π2x2+O(x3))Γ(1+x)x1=x→0limx21(12π2x2+O(x3))Γ(1+x)x1=12π2e−γ
The linear asymptote is therefore 12π2e−γ+e−γx, and Γ(1+x1)x≈12xπ2+12xe−γ for large x. This approximation is within one procent of the true value for x≥1.7844.