Question 10.36: Show that Γ(z + 1) ∼ √2πz z^z e^z {1 + 1/12z + 1/288z² - 139...
Show that \Gamma(z+1) \sim \sqrt{2 \pi z} z^{z} e^{-z}\left\{1+\frac{1}{12 z}+\frac{1}{288 z^{2}}-\frac{139}{51,840 z^{3}}+\cdots\right\}.
We have \Gamma(z+1)=\int_{0}^{\infty} \tau^{z} e^{-\tau} d \tau. By letting \tau=z t, this becomes
\Gamma(z+1)=z^{z+1} \int\limits_{0}^{\infty} t^{z} e^{-z t} d t=z^{z+1} \int\limits_{0}^{\infty} e^{z(\ln t-t)} d t (1)
which has the form (10.37), page 330,
I(z)=\int_{C}e^{z F(t)}\,d t (10.37)
where F(t)=\ln t-t.
F^{\prime}(t)=0 when t=1. Letting t=1+w, we find, using Problem 6.23, page 185, or otherwise, the Taylor series
\begin{aligned} F(t)=\ln t-t & =\ln (1+w)-(1+w)=\left(w-\frac{w^{2}}{2}+\frac{w^{3}}{3}-\frac{w^{4}}{4}+\cdots\right)-1-w \\ & =-1-\frac{w^{2}}{2}+\frac{w^{3}}{3}-\frac{w^{4}}{4}+\cdots=-1-\frac{(t-1)^{2}}{2}+\frac{(t-1)^{3}}{3}-\frac{(t-1)^{4}}{4}+\cdots \end{aligned}
Hence from (1),
\begin{aligned} \Gamma(z+1) & =z^{z+1} e^{-z} \int\limits_{0}^{\infty} e^{-z(t-1)^{2} / 2} e^{z(t-1)^{3} / 3-z(t-1)^{4} / 4+\cdots} d t \\ & =z^{z+1} e^{-z} \int\limits_{-1}^{\infty} e^{-z w^{2} / 2} e^{z w^{2} / 3-z w^{4} / 4+\cdots} d w & (2) \end{aligned}
Letting w=\sqrt{2 / z} v, this becomes
\Gamma(z+1)=\sqrt{2} z^{z+1 / 2} e^{-z} \int\limits_{-\sqrt{z / 2}}^{\infty} e^{-v^{2}} e^{(2 / 3) \sqrt{2} z^{-1 / 2} v^{3}-z^{-1} v^{4}+\cdots} d v (3)
For large values of z, the lower limit can be replaced by -\infty, and on expanding the exponential, we have
\Gamma(z+1) \sim \sqrt{2} z^{z+1 / 2} e^{-z} \int\limits_{-\infty}^{\infty} e^{-v^{2}}\left\{1+\left(\frac{2}{3} \sqrt{2} z^{-1 / 2} v^{3}-z^{-1} v^{4}\right)+\cdots\right\} d v (4)
or
\Gamma(z+1) \sim \sqrt{2 \pi z} z^{z} e^{-z}\left\{1+\frac{1}{12 z}+\frac{1}{288 z^{2}}-\frac{139}{51,840 z^{3}}+\cdots\right\} (5)
Although we have proceeded above in a formal manner, the analysis can be justified rigorously.
Another Method. Given
F(t)=-1-\frac{(t-1)^{2}}{2}+\frac{(t-1)^{3}}{3}-\frac{(t-1)^{4}}{4}+\cdots=-1-u^{2}
Then
u^{2}=\frac{(t-1)^{2}}{2}-\frac{(t-1)^{3}}{3}+\cdots
and by reversion of series or by using the fact that F(t)=\ln t-t, we find
\frac{d t}{d u}=b_{0}+b_{1} u+b_{2} u^{2}+\cdots=\sqrt{2}+\frac{\sqrt{2}}{6} u^{2}+\frac{\sqrt{2}}{216} u^{4}+\cdots
Then, from (10.41), page 330,
I(z)\sim\sqrt{{\frac{\pi}{z}}}\,e^{z F(t_{0})}\biggl\{b_{0}+{\frac{1}{2}}{\frac{b_{2}}{z}}+{\frac{1\cdot3}{2\cdot2}}{\frac{b_{4}}{z^{2}}}+{\frac{1\cdot3\cdot5}{2\cdot2\cdot2}}{\frac{b_{6}}{z^{3}}}+\cdot\cdot\cdot\biggr\} (10.41)
we find
\Gamma(z+1) \sim \sqrt{\frac{\pi}{z}} z^{z+1} e^{z(\ln 1-1)}\left\{\sqrt{2}+\frac{1}{2}\left(\frac{\sqrt{2}}{6}\right) \frac{1}{z}+\frac{1 \cdot 3}{2 \cdot 2}\left(\frac{\sqrt{2}}{216}\right) \frac{1}{z^{2}}+\cdots\right\}
or
\Gamma(z+1) \sim \sqrt{2 \pi z} z^{z} e^{-z}\left\{1+\frac{1}{12 z}+\frac{1}{288 z^{2}}+\cdots\right\}
Note that since F^{\prime \prime}(1)=-1, we find on using (10.42), page 330,
I(z)\sim\sqrt{\frac{-2\pi}{z F^{\prime\prime}(t_{0})}}\,e^{z F(t_{0})} (10.42)
\Gamma(z+1) \sim \sqrt{2 \pi z} z^{z} e^{-z}
which is the first term. For many purposes this first term provides sufficient accuracy.