Notes on Whittaker & Watson, Chapter XIV, part 2
Whittaker & Watson, A Course of Modern Analysis. Chapter XIV, The Hypergeometric Function. 14.5 Barnes' contour integral for the hypergeometric function. Consider the integral $$ \frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^sds $$ with $\arg(-z)<\pi$, on the path from ${-i\infty}$ to ${i\infty}$ such that the poles of $\Gamma(a+s)$ and $\Gamma(b+s)$ at $s=-a-n,-b-n, n=1,2,\dots$ are to the left of the path, and the poles of $\Gamma(-s)$ at $s=1,2,\dots$ are to the right of the path. This path exists when $a,b$ are not negative integers.
Notes on Whittaker & Watson, Chapter XIV, part 1
Whittaker & Watson, A Course of Modern Analysis. Chapter XIV, The Hypergeometric Function. 14.1 The hypergeometric series $$ 1+\frac{a\cdot b}{1\cdot c}z + \frac{a(a+1)b(b+1)}{1\cdot2\cdot c(c+1)}z^2 + \frac{a(a+1)(a+2)b(b+1)(b+2)}{1\cdot2\cdot3\cdot c(c+1)(c+2)}z^3+\cdots $$ defines an analytic function $F(a,b;c;z)$ for $|z|<1$. Some examples: $$ \begin{gather*} (1+z)^n=F(-n,\beta;\beta;-z),\newline \log(1+z)=zF(1,1;2;-z),\newline e^z=\lim_{\beta\to\infty}F(1,\beta;1;z/\beta). \end{gather*} $$ 14.11 The values of $F(a,b;c;1)$ when $\Re(c-a-b)>0$. Consider the series $$ \begin{align*} & c\big(c-1-(2c-a-b-1)x\big)F(a,b;c;x) + (c-a)(c-b)xF(a,b;c+1;x) \newline ={}& c(c-1)(1-x)F(a,b;c-1;x) \newline ={}& c(c-1)\left(1+\sum_{n=1}^\infty(u_n-u_{n-1})x^n\right) \qquad\qquad F(a,b;c-1;x)=\sum_{n=0}^{\infty}u_nx^n \end{align*} $$ in which, the RHS tends to 0 as $x\to1$.
Notes on Whittaker & Watson, Chapter XIII, part 3
Whittaker & Watson, A Course of Modern Analysis. Chapter XIII, The Zeta Function of Riemann. 13.5 Inequalities satisfied by $\zeta(s,a)$ when $\sigma > 0$. $$ \zeta(s,a)=\sum_{n=0}^N(n+a)^{-s} - \frac1{(1-s)(N+a)^{s-1}} - \sum_{n=N}^\infty f_n(s), $$ where $$ \begin{align*} f_n(s) &= \frac1{1-s}\left(\frac{1}{(n+a+1)^{s-1}}-\frac{1}{(n+a)^{s-1}}\right) - \frac{1}{(n+a+1)^s} = s\int_{n}^{n+1}\frac{u-n}{(u+a)^{s+1}}du. \end{align*} $$ The series $\sum_{n=N}^\infty f_n(s)$ is absolutely and locally uniformly convergent since $|f_n(s)|\le|s|(n+a)^{-\sigma-1}$ for $\sigma>0$. Hence the series is convergent to an analytic function by Weierstrass’s theorem.
Notes on Whittaker & Watson, Chapter XIII, part 2
Whittaker & Watson, A Course of Modern Analysis. Chapter XIII, The Zeta Function of Riemann. 13.2 Hermite’s formula for $\zeta(s,a)$. Applying the Abel-Plana formula to the summation $\zeta(s,a)=\sum_{n=0}^{\infty}(n+a)^{-s}$, with the summand function $\phi(z)=(a+z)^{-s}$, and $z=x+iy$, we have for $\sigma>1$ $$ \zeta(s,a)=\lim_{x_2\to\infty}\frac{1}{2}\alpha^{-s} + \int_{x_1=0}^{x_2}(a+x)^{-s}dx + \frac1i\int_{0}^{\infty}\frac{\phi(x_2+iy)-\phi(x_1+iy) -\phi(x_2+iy)+\phi(x_1+iy)}{e^{2\pi y}-1}dy. $$ With $\arg(a+z)=\arctan\frac{y}{a+x}$, consider the function $$ q(z) = \frac{\phi(z)-\phi(\bar z)}{2i}= - \left((a+x)^2+y^2\right)^{-\frac12s}\sin\left(s\arctan\frac{y}{a+x}\right). $$ By the inequalities that $|\sin z|=|\sinh iz|\le \big|\sinh|z|\big|$, $\arctan\frac{y}{a+x}\le \frac{|y|}{a+x}$, and the fact that $\arctan \le \frac{\pi}{2}$, we have $$ \begin{align*} |q(z)|&\le \left((a+x)^2+y^2\right)^{\frac12-\frac12\sigma}\left|y^{-1}\right|\sinh \frac{\pi|s|}2, & \frac{|y|}{\left((a+x)^2+y^2\right)^{\frac12}}\le1\newline |q(z)|&\le \left((a+x)^2+y^2\right)^{-\frac12\sigma}\sinh \frac{y|s|}{a+x}.
Notes Whittaker & Watson, Chapter XIII, part 1
Whittaker & Watson, A Course of Modern Analysis. Chapter XIII, The Zeta Function of Riemann. 13.1 The definition of Zeta-function. The series definition $$ \zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s} $$ is uniformly convergent and analytic for $s=\sigma+it$ when $\sigma\ge 1+\delta$ with $\delta>0$. 13.11 The generalized Zeta-function (Hurwitz) $$ \zeta(s,a)=\sum_{n=0}^\infty\frac{1}{(a+n)^s} $$ 13.12 The expression of $\zeta(s,a)$ as an infinite integral. By the integral $(n+a)^{-s}\Gamma(s) = \int_0^\infty x^{s-1}e^{-(n+a)x}dx$, for $\sigma>0$ and $\arg x=0$, we have $$ \begin{align*} \Gamma(s)\zeta(s,a) &=\lim_{N\to\infty}\sum_{n=0}^N \int_0^{\infty}x^{s-1}e^{-(n+a)x}dx\newline &=\lim_{N\to\infty}\int_0^{\infty}\frac{x^{s-1}e^{-ax}}{1-e^{-x}}dx - \int_0^{\infty}\frac{x^{s-1}e^{-(N+a+1)x}}{1-e^{-x}}dx\newline &=\lim_{N\to\infty}\int_0^{\infty}\frac{x^{s-1}e^{-ax}}{1-e^{-x}}dx -(N+a)^{1-\sigma}\Gamma(\sigma-1) \end{align*} $$ in which the last term tends to 0 when $\sigma\ge 1+\delta$.
Notes on Whittaker & Watson, Chapter XII, part 3
Whittaker & Watson, A Course of Modern Analysis. Chapter XII, The Gamma Function. 12.4 The Beta function in its integral form defined by the Eulerian integral of the first kind: $$ B(p,q) = \int_0^1x^{p-1}(1-x)^{q-1}dx $$ is analytic when $\Re(p),\Re(q)>0$. It is symmetric so that $B(p,q)=B(q,p)$. By integration by parts $B(p,q+1)=\frac{q}{p}B(p+1,q)$. 12.41 To establish that $$ B(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}, $$ consider for $m,n$ such that $\Re(2m-1), \Re(2n-1) >0$ $$ \begin{align*} \Gamma(m)\Gamma(n) & = \int_0^\infty e^{-x}x^{m-1}dx\int_0^\infty e^{-y}x^{n-1}dy\newline & = 4\lim_{R\to\infty}\int_0^R\int_0^R e^{-(x^2+y^2)}x^{2m-1} e^{-y^2}x^{2n-1}dx\ dy & x\to x^2,y\to y^2\newline & = 4\lim_{R\to\infty}\int_0^Re^{-r^2}r^{2(m+n-1)}dr^2\int_0^{\frac{\pi}{2}}\cos^{2m-1}\theta\sin^{2n-1}\theta\ d\theta & \iint_{\substack{(x,y)\in[0,1]^2\newline\ x^2+y^2>R^2}}\to0\newline &=\Gamma(m+n)B(m,n).
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.
Notes on Whittaker & Watson, Chapter XII, part 1
Whittaker & Watson, A Course of Modern Analysis. Chapter XII, The Gamma Function. 12.1 The Weierstrass product definition of $1/\Gamma(z)$ $$ \frac{1}{\Gamma(z)}=ze^{\gamma z}\prod_{n=1}^\infty\left(1+\frac{z}{n}\right)e^{-\frac{z}{n}} $$ in which $\gamma=\lim_{m\to\infty}\frac{1}{1}+\frac{1}{2}+\cdots\frac{1}{m}-\log m$ is the Euler-Mascheroni constant. The product is an entire function, i.e. analytic for all $z\in\mathbb{C}$. 12.11 Substituting the Euler-Mascheroni constant by its definition, we have Euler’s formula of $\Gamma(z)$ $$ \begin{align*} \Gamma(z) &= \frac{1}{z}\prod_{n=1}^\infty\left(1+\frac{1}{n}\right)^{-z}\left(1+\frac{z}{n}\right)^{-1}, & z\neq0,-1,-2,\cdots.\newline &=\lim_{n\to\infty}\frac{1\cdot2\cdot\cdots\cdot(n-1)}{z(z+1)\cdots(z+n-1)}n^z. \end{align*} $$ 12.12 The recurrence relation $\Gamma(z+1)=z\Gamma(z)$ and $\Gamma(n)=(n-1)!
Notes on Whittaker & Watson, Chapter XI, part 3
Whittaker & Watson, A Course of Modern Analysis. Chapter XI, Integral Equations. 11.3 Introducing Fredholm’s integral equation of the first kind $$ f(x)=\lambda\int_a^bK(x,\xi)\phi(\xi)d\xi $$ with $\phi(\cdot)$ unknown, and the integral equations with variable upper limits $$ f(x)=\lambda\int_a^xK(x,\xi)\phi(\xi)d\xi\newline \phi(x) = f(x)+\lambda\int_a^xK(x,\xi)\phi(\xi)d\xi. $$ 11.31. The integral equation of the first kind with variable upper limit is also known as Volterra’s equation. When $\frac{\partial K}{\partial x}$ exists and is continuous, we can differentiate the equation to get $$ f'(x) = \lambda K(x,x)\phi(x) + \lambda\int_a^x\frac{\partial K}{\partial x}\phi(\xi)d\xi $$ which is an equation of the second kind.
Notes on Whittaker & Watson, Chapter XI, part 2
Whittaker & Watson, A Course of Modern Analysis. Chapter XI, Integral Equations. 11.21 Verification of the solution to Fredholm’s equation of the second kind from 11.2. By Hadamard’s inequality (11.11) series expansion of $D(\lambda)$ converges absolutely for all $\lambda$, and $D(x,y; \lambda)-\lambda K(x,y)$ is uniformly convergent function of $x,y$. Factorizing $K(x, y)$ from $D(x,y; \lambda)$ $$ \begin{align*} D(x,y;\lambda) = \lambda D(\lambda)K(x,y) & - \lambda^2 \int_a^b\begin{vmatrix} 0 & K(x,\xi_1)\newline K(\xi_1,y) & K(\xi_1,\xi_1) \end{vmatrix} d\xi_1 \newline & + \frac{\lambda^3}{2!