Notes on Whittaker & Watson, Chapter XII, part 2

Whittaker & Watson, A Course of Modern Analysis. Chapter XII, The Gamma Function.

12.2 Euler’s integral expression $\int_0^\infty e^{-t}t^{z-1}dt$ for the Gamma function.

The integral expression is analytic for $\Re(z)>0$. To prove that it agrees with $\Gamma(z)$, consider $$ \begin{align*} \Pi(z,n) &= \int_0^n\left(1-\frac{t}{n}\right)^nt^{z-1}dt\newline & = n^z\int_0^1(1-\tau)^n\tau^{z-1}d\tau &\tau\equiv t/n \newline & = \frac{1\cdot2\cdots n}{z(z+1)\cdots(z+n)} & \text{repeated integration by parts} \end{align*} $$ the limit of which $\lim_{n\to\infty}\Pi(z,n)=\Gamma(z)$ by Euler’s infinite product formula.

On the other hand, $\lim_{n\to\infty}\Pi(z,n)=\int_0^\infty e^{-t}t^{z-1}dt$, which establishes that $\Gamma(z) = \int_0^\infty e^{-t}t^{z-1}dt$ for $\Re(z)>0$. The interchange of the limit and integral is justified by bounding the difference $0\le e^{-t}-\left(1-\frac{t}{n}\right)^n\le n^{-1}t^2e^{-t}$.

[Alternatively, if $\Pi(z,n)=\int_{[0,\infty]}(1-t/n)^nt^{z-1}\mathbf{1}_{[0,n]}dt$ is defined by the Lebesgue integration, the monotone or the dominated convergence theorem applies since $(1-t/n)^n\mathbf{1}_{[0,n]} \nearrow e^{-t}$.]

12.21 Analytic continuation of the $\Gamma$ integral.

Consider the function $\Gamma_2$ defined on $-k-1<\Re(z)<-k$, $k$ being positive integer, by the integral $$ \begin{align*} \Gamma_2(z) & = \int_0^\infty t^{z-1}\left(e^{-t}-1+t-\frac{t^2}{2!}+\cdots+(-1)^k\frac{t^k}{k!}\right)dt\newline & =\left.\frac{t^z}{z}\left(e^{-t}-1+t-\frac{t^2}{2!}+\cdots+(-1)^k\frac{t^k}{k!}\right)\right|_0^\infty & \text{vanishes when $\Re(z)<-k$}\newline & + \frac{1}{z}\int_0^\infty t^{z}\left(e^{-t}-1+t-\frac{t^2}{2!}+\cdots+(-1)^{(k-1)}\frac{t^(k-1)}{(k-1)!}\right)dt\newline & = \frac{1}{z}\Gamma_2(z+1). \end{align*} $$ When $k=0$, i.e. $-1<\Re(z)<0$, $\Gamma_2(z)=\frac{1}{z}\Gamma(z+1)$ showing that $\Gamma_2$ agrees with $\Gamma$ in $-1<\Re(z)<0$, thus is the analytic continuation of the integral expression for $z$ with negtive real parts.

12.22 Henkel’s expression of $\Gamma$ as a contour integral.

Consider the integral on the Henkel contour $D$ such that $\int_D = \int_\rho^\delta+\int_{|t|=\delta}+\int_\delta^\rho$ with $\rho\ge\delta>0$ $$ \int_D(-t)^{z-1}e^{-t}dt = -2i\sin\pi z\int_\delta^\rho t^{z-1}e^{-t}dt + i\delta^z\int_{-\pi}^\pi \exp\big(iz\theta+\delta(\cos\theta+i\sin\theta)\big)d\theta. $$ In the limit $\delta\to 0, \rho\to\infty$, the contour $D\to C$, we have $$ \int_C(-t)^{z-1}e^{-t}dt = -2i\sin\pi z\int_0^\infty t^{z-1}e^{-t}dt, $$ or $$ \Gamma(z) = -\frac{1}{2i\sin\pi z}\int_C(-t)^{z-1}e^{-t}dt $$ which extends to $\mathbb{C}$ except for all integer values $z=0, \pm 1, \pm 2\cdots$.

Using the reflection formula (§12.14) $$ \frac{1}{\Gamma(z)} = \frac{i}{2\pi}\int_C(-t)^{-z}e^{-t}dt. $$ 12.3 Gauss' expression for the logarithmic derivateof the Gamma function as an infinite integral (digamma function).

First, an expression of $\gamma$ (from §12.1 example 2 and §12.2 example 4) $$ \begin{align*} \gamma &= \lim_{n\to\infty}\int_0^1\frac{1-t^n}{1-t}dt-\log n = \lim_{n\to\infty}\int_0^1\frac{1-(1-t)^n}{t}dt-\log n \newline & = \lim_{n\to\infty}\int_0^n 1-\left(1-\frac{t}{n}\right)^n\frac{dt}{t}- \int_1^n \frac{dt}{t} \newline & = \lim_{n\to\infty}\int_0^1\frac{1-(1-\frac{t}{n})^n}{t}dt - \int_1^n\frac{(1-\frac{t}{n})^n}{t}dt & \text{12.1 example 2}\newline & = \int_0^1 (1-e^{-t})\frac{dt}{t} - \int_1^\infty \frac{e^{-t}dt}{t} & \text{12.2 example 4}\newline & = \lim_{\delta\to 0}\int_\delta^1 \frac{dt}{t} - \int_\delta^\infty \frac{e^{-t}dt}{t} = \lim_{\delta\to 0}\int_{1-e^{-\delta}}^1 \frac{dt}{t} - \int_\delta^\infty \frac{e^{-t}dt}{t} & \log t\Big|_{1-e^{-\delta}}^\delta\to 0\newline & = \lim_{\delta\to 0}\int_\delta^\infty\frac{e^{-t}}{1-e^{-t}}dt- \int_\delta^\infty \frac{e^{-t}dt}{t} = \int_0^\infty\left(\frac{1}{1-e^{-t}}-\frac{1}{t}\right)e^{-t}dt & t\rightarrow 1-e^{-t}. \end{align*} $$ Then the derivative of $\log\Gamma$, i.e. the digamma $\psi$ can be obtained by this expression of $\gamma$ and rewritting $\frac{1}{z+m}=\int_0^\infty e^{-(m+z)t}dt$: $$ \begin{align*} \psi(z)=\frac{\Gamma'(z)}{\Gamma(z)} &= -\gamma-\frac{1}{z}+\lim_{n\to\infty}\sum_{m=1}^n\left(\frac{1}{m}-\frac{1}{z+m}\right)\newline &=-\gamma+\lim_{n\to\infty}\left(\sum_{m=1}^n\int_0^\infty e^{-mt}dt- \sum_{m=0}^n\int_0^\infty e^{-(m+z)t}dt\right) \newline &=\int_0^\infty\left(\frac{e^{-t}}{t}-\frac{e^{-zt}}{1-e^{-t}}\right)dt -\lim_{n\to\infty}\int_0^\infty\frac{1-e^{-zt}}{1-e^{-t}}e^{-(n+1)t}dt. \end{align*} $$ The last term vanishes as $n\to\infty$. Thus we have $$ \psi(z)=\int_0^\infty\left(\frac{e^{-t}}{t}-\frac{e^{-zt}}{1-e^{-t}}\right)dt. $$ A result due to Dirichlet can be obtained by the substitution $t=\log(1+x)$ $$ \psi(z) = \int_0^\infty\left(e^{-x}-\frac{1}{(1+x)^z}\right)\frac{dx}{x}. $$

12.31 Binet’s first expression for $\log\Gamma(z)$ in terms of a infinite integral.

When $\Re(z)>0$, using Gauss’s result for $z+1$ and extracting the diverging terms $\log z = \int_0^\infty (e^{-t}-e^{-tz})\frac{dt}{t}$ (by a contour around $\delta, \rho, \rho z, \delta z$ with $\delta\to 0$ and $\rho\to\infty$) and $\frac{1}{2z}=\int_0^\infty \frac{e^{-tz}}{2}dt$, we have $$ \begin{align*} \frac{d}{dz}\log\Gamma(z+1) &=\frac{1}{2z}+\log z-\int_0^\infty\left(\frac{1}{2}-\frac{1}{t}+\frac{1}{e^t-1}\right)e^{-tz}dt. \end{align*} $$ The logarithm of the Gamma function can then be obtained by integrating in $z$ from 1 to $z$, $$ \log\Gamma(z) = \log\Gamma(z+1) - \log z = \left(z-\frac{1}{2}\right)\log z-z+1+\int_0^\infty\left(\frac{1}{2}-\frac{1}{t}+\frac{1}{e^t-1}\right)\frac{e^{-tz}-e^{-t}}{t}dt, $$ and since $\left(\frac{1}{2}-\frac{1}{t}+\frac{1}{e^t-1}\right)/t\to\frac{1}{12}$ is continuous as $t\to 0$, the integrals $\int_0^\infty\left(\frac{1}{2}-\frac{1}{t}+\frac{1}{e^t-1}\right)\frac{e^{-tz}}{t}dt$ and $\int_0^\infty\left(\frac{1}{2}-\frac{1}{t}+\frac{1}{e^t-1}\right)\frac{e^{-t}}{t}dt$ can be evaluated individually. The first integral evaluates to $\frac{1}{2}+\frac{1}{2}\log\frac{1}{2}$ when $z=\frac{1}{2}$, and relates to the second integral via $\log\Gamma(\frac{1}{2})=\frac{1}{2}\log\pi$ by the result above. Thus the second integral evaluates to $1-\frac{1}{2}\log 2\pi$, and we have $$ \log\Gamma(z)=\left(z-\frac{1}{2}\right)\log z-z+\frac{1}{2}\log 2\pi+\int_0^\infty\left(\frac{1}{2}-\frac{1}{t}+\frac{1}{e^t-1}\right)\frac{e^{-tz}}{t}dt. $$ Since $\left(\frac{1}{2}-\frac{1}{t}+\frac{1}{e^t-1}\right)\frac{1}{t}$ is bounded for $t>0$, the integral term above is of $O(\Re(z)^{-1})$. Thus $\log\Gamma(z) \sim \left(z-\frac{1}{2}\right)\log z-z+\frac{1}{2}\log 2\pi$ for large $\Re(z)$.

12.32 Binet’s second expression for $\log\Gamma(z)$ in terms of a infinite integral.

Applying the Abel-Plana summation formula to $\frac{d^2}{dz^2}\log\Gamma(z)=\sum_{n=0}^\infty\frac{1}{(z+n)^2}$, we have $$ \frac{d^2}{dz^2}\log\Gamma(z)=\frac{1}{2z^2}+\frac{1}{z}+\int_0^\infty\frac{4tz}{(z^2+t^2)^2(e^{2\pi t}-1)}dt $$ which after being integrated twice gives $$ \log\Gamma(z)=\left(z-\frac{1}{2}\right)\log z+(C-1)z+C'+2\int_0^\infty \frac{\arctan(t/z)}{e^{2\pi t}-1}dt. $$ in which $C$ and $C'$ are constants of integration. When $z\to+\infty$ on the real axis, the integral is of $O(z^{-1})$. Comparing this with the result in 12.31, we can determine the constants $C$ and $C'$, and $$ \log\Gamma(z)=\left(z-\frac{1}{2}\right)\log z-z+\frac{1}{2}\log 2\pi+2\int_0^\infty \frac{\arctan(t/z)}{e^{2\pi t}-1}dt. $$ 12.33. The asymptotic expansion of the logarithm of the Gamma function (Stirling’s series).

Let $\phi(z)=2\int_0^\infty \frac{\arctan(t/z)}{e^{2\pi t}-1}dt$ be the integral in Binet’s second formula. Using the Maclaurin series of $\arctan$, and the integral form of Bernoulli number that $B_n=4n\int_0^\infty\frac{t^{2n-1}}{e^{2\pi t}-1}$ (§7.2), we have $$ \phi(z)=\sum_{r=1}^n\frac{(-1)^{r-1}B_r}{2r(2r-1)z^{2r-1}}+\frac{(-1)^n\cdot2}{z^{2n-1}}\int_0^\infty\frac{dt}{e^{2\pi}-1}\int_0^t\frac{u^{2n}}{u^2+z^2}du $$ in which $\int_0^t\frac{u^{2n}}{u^2+z^2}du$ is the remainder of the Maclaurin series. When $-\frac{\pi}{2}+\Delta\le\arg z\le\frac{\pi}{2}-\Delta$, the double integral is asymptotically $O(|z|^{-2})$ and consequently the last term is of $O(|z|^{2n+1})$. Therefore $$ \log\Gamma(z)\sim\left(z-\frac{1}{2}\right)\log z-z+\frac{1}{2}\log 2\pi+\sum_{r=1}^\infty\frac{(-1)^{r-1}B_r}{2r(2r-1)z^{2r-1}} $$ which is known as Stirling’s series.