Notes on Whittaker & Watson, Chapter XVII, part 3
Whittaker & Watson, A Course of Modern Analysis. Chapter XVII, Bessel Functions. 17.6 The second solution of Bessel’s equation when the order is an integer. The Bessel functions $J_n$ and $J_{-n}$ are two solutions to Bessel’s equation, which are not linearly independent when $n$ is an integer. The Bessel function of the second kind is defined by $$ \begin{gather*} Y_n(z)=\frac{J_n(z)\cos n\pi- J_{-n}(z)}{\sin n\pi} & \text{or}\\ \mathsf{Y}_n=2\pi e^{n\pi i}\frac{J_n(z)\cos n\pi- J_{-n}(z)}{\sin 2n\pi} \end{gather*} $$ where $Y_n$ is due to Weber and Schläfli, and $\mathsf{Y}_n$ is introduced by Hankel.
Notes on Whittaker & Watson, Chapter XVII, part 2
Whittaker & Watson, A Course of Modern Analysis. Chapter XVII, Bessel Functions. 17.3 Hankel’s contour integral for $J_n(z)$. The integral on a figure of eight contour around $t=\pm1$, with the starting point $A$ on the right of $t=1$ and $\arg(t-1)=\arg(t+1)=0$ at $t=A$ $$ \begin{align*}& z^n\int_A^{1+,-1-}(t^2-1)^{n-\frac12}\cos(zt),dt \\ = {} & \sum_{r=0}^\infty\frac{(-1)^rz^{n+2r}}{(2r)!}\int_A^{1+,-1-}t^{2r}(t^2-1)^{n-\frac12}dt \\ = {} & \sum_{r=0}^\infty\frac{(-1)^rz^{n+2r}}{(2r)!}(-4i)\sin{\left(n-\frac12\right)\pi}\int_0^1t^{2r}(t^2-1)^{n-\frac12}dt\\ ={}&\sum_{r=0}^\infty\frac{(-1)^rz^{n+2r}}{(2r)!}\frac{2i\sin{\left(n-\frac12\right)\pi}\Gamma(r+\frac12)\Gamma(n+\frac12)}{\Gamma(n+r+1)}\\ ={}&2^{n+1}i\sin{\left(n-\frac12\right)\pi}\Gamma\left(r+\frac12\right)\Gamma\left(n+\frac12\right)J_n(z) \end{align*} $$ gives us the contour integral of the Bessel function $J_n(z)$ $$ J_n(z)=\frac{\Gamma(\frac12-n)(\frac12z)^n}{2\pi i\Gamma(\frac12)}\int_A^{1+,-1-}(t^2-1)^{n-\frac12}\cos(zt),dt.
Notes on Whittaker & Watson, Chapter XVII, part 1
Whittaker & Watson, A Course of Modern Analysis. Chapter XVII, Bessel Functions. 17.1 The Bessel coefficients. The Bessel functions are confluent hypergeometric functions with one regular and one irregular singularity. Particular cases of Bessels functions for integer $n$ are the coefficients $J_n(z)$ of $t^n$ in the Laurent series of the generating function $$ e^{\frac12z\left(t-\frac1t\right)} $$ which can be calculated as $$ \begin{align*} J_n(z)&=\frac{1}{2\pi i}\int^{(0+)}u^{-n-1}e^{\frac12z\left(u-\frac1u\right)}du \\ &=\frac{1}{2\pi i}\left(\frac z2\right)^n\int^{(0+)}t^{-n-1}e^{t-\frac{z^2}{4t}}du & u=\frac{2t}{z} \\ &= \frac{1}{2\pi i}\sum_{r=0}^{\infty}\frac{(-1)^r}{r!
Notes on Whittaker & Watson, Chapter XVI, part 3
Whittaker & Watson, A Course of Modern Analysis. Chapter XVI, The Confluent Hypergeometric Function. 16.5 The parabolic cylinder functions. Weber’s equation. The equation $$ \frac{d^2w}{dz^2}+\left(2k-\frac14z^2\right)w=0 $$ is satisfied by $w=z^{-\frac12}W_{k,-\frac14}(\frac12z^2)$, and the parabolic cylinder function $$ D_n(z)=2^{\frac12n+\frac14}z^{-\frac12}W_{\frac12n+\frac14,-\frac14}\left(\frac12z^2\right) $$ satisfies Weber’s equation $$ \frac{d^2D_n(z)}{dz^2}+\left(n+\frac12-\frac14z^2\right)D_n(z)=0. $$ Expressed in $M_{k,m}$, we have $$ D_n(z)=\frac{\Gamma(\frac12)2^{\frac12n+\frac14}z^{-\frac12}}{\Gamma(\frac12-\frac12n)}M_{\frac12n+\frac14,-\frac14}\left(\frac12z^2\right) + \frac{\Gamma(-\frac12)2^{\frac12n+\frac14}z^{-\frac12}}{\Gamma(-\frac12n)}M_{\frac12n+\frac14,\frac14}\left(\frac12z^2\right) $$ which is one-valued throughout $z\in\mathbb{C}$. We have the asymptotic expansion of $D_n(z)$ for $\arg(z)<\frac34\pi$ $$ D_n(z)\sim e^{-\frac14z^2}z^n\left(1-\frac{n(n-1)}{2z^2}+\frac{n(n-1)(n-2)(n-3)}{2\cdot4z^4}-\cdots\right).
Notes on Whittaker & Watson, Chapter XVI, part 2
Whittaker & Watson, A Course of Modern Analysis. Chapter XVI, The Confluent Hypergeometric Function. 16.2 Expression of various functions by functions of the type $W_{k,m}(z)$. 1º The error function $$ \operatorname{Erfc}(x)=\int_x^\infty e^{-t^2}dt =\frac12x^{-\frac12}e^{-\frac12x^2}W_{-\frac14,\frac14}(x^2). $$ 2º The Incomplete Gamma function $$ \gamma(n,x)=\int_0^xt^{n-1}e^{-t}dt=\Gamma(n)-x^{\frac12(n-1)e^{-\frac12x}}W_{\frac12(n-1),\frac12n}(x). $$ 3º The Logarithmic-integral function $$ \operatorname{li}(z)=\int_0^{z}\frac{dt}{\log t}=-(-\log z)^{-\frac12}z^{\frac12}W_{-\frac12,0}(-\log z). $$ 16.3 The asymptotic expansion of $W_{k,m}(z)$, when $|z$| is large. We substitute $(1+\frac tz)^{\lambda}$ in the definition of $W_{k,m}(z)$ by its the expansion $$ \left(1+\frac tz\right)^\lambda = 1+ \frac{\lambda t}z+\cdots+\frac{\lambda(\lambda-1)\cdots(\lambda-n+1)}{n!
Notes on Whittaker & Watson, Chapter XVI, part 1
Whittaker & Watson, A Course of Modern Analysis. Chapter XVI, The Confluent Hypergeometric Function. 16.1 The confluence of two singularities of Riemann’s equation. Riemann’s $P$-function $$ P\begin{Bmatrix}0 & \infty & c\\ \frac12+m & -c & c-k & z\\ \frac12-m & 0 & k\end{Bmatrix} $$ in the limit $c\to\infty$ is the solution to the differential equation $$ \frac{d^2u}{dz^2}+\frac{du}{dz}+\left(\frac{k}{z}+\frac{\frac14-m^2}{z^2}\right)u=0. $$ Substituting $u=e^{-\frac12z}W_{k,m}$, we have the equation for $W$ $$ \frac{d^2W}{dz^2}+\left(-\frac14+\frac{k}{z}+\frac{\frac14-m^2}{z^2}\right)W=0 $$ which has a regular singularity at $z=0$ and an irregular singularity at $z=\infty$.
Notes on Whittaker & Watson, Chapter XV, part 3
Whittaker & Watson, A Course of Modern Analysis. Chapter XV, Legendre Functions. 15.5 Ferrer’s associated Legendre functions $P_n^m(z)$ and $Q_n^m(z)$ are defined as $$ \begin{gather*} P_n^m(z) = (1-z^2)^{\frac12m}\frac{d^m}{dz^m}P_n(z)\\ Q_n^m(z) = (1-z^2)^{\frac12m}\frac{d^m}{dz^m}Q_n(z) \end{gather*} $$ for positive integer $m$, $z\in(-1,1)$, and $n\in\mathbb{C}$, which satisfy the differential equation $$ (1-z^2)\frac{d^2w}{dz^2}-2\frac{dw}{dz}+\left(n(n+1)-\frac{m^2}{1-z^2}\right)w=0 $$ derived from Legendre’s equation. 15.51 The integral properties of the associated Legendre functions (orthogonality). For positive integers $n,r>m$ $$ \int_{-1}^1P_n^m(z)P_r^m(z)dz=\frac{2\delta_{rn}}{2n+1}\frac{(n+m)!}{(n-m)!}, $$ which is obtained for $n\ne r$ by multiplying the differential equation for $P_n^m(z)$ by $P_r^m(z)$ and the equation for $P_r^m(z)$ by $P_n^m(z)$; and for $n=r$ through the recurrence relation that $\int_1^1\left(P_n^{m+1}(z)\right)^2dz = (n-m)(n+m+1)\int_1^1\left(P_n^m(z)\right)^2dz$.
Notes on Whittaker & Watson, Chapter XV, part 2
Whittaker & Watson, A Course of Modern Analysis. Chapter XV, Legendre Functions. 15.3 Legendre functions of the second kind. In solving Legendre’s differential equation (§15.2), we instead consider the branch cut connecting $t=-1,1$ and the contour $D$ which is an ellipse around the cut. Thus, we have the solution on the $z$-plane cut on real axis from 1 to $-\infty$ $$ Q_n(z)=\frac{1}{4i\sin n\pi}\int_D\frac{(t^2-1)^n}{2^n(z-t)^{n+1}}dt $$ for $n\notin\mathbb{Z}$ and under the condition that $\Re(n+1)>0$.
Notes on Whittaker & Watson, Chapter XV, part 1
Whittaker & Watson, A Course of Modern Analysis. Chapter XV, Legendre Functions. 15.1 Definition of Legendre polynomials. Expanding the generating function $(1-2zh+h^2)^{-\frac12} = P_0(z)+hP_1(z)+h^2P_2(z)+\cdots$, we have the Legendre polynomials $$ P_n(z)=\sum_{r=0}^{\lfloor\frac n2\rfloor}\frac{(-1)^r(2n-2r)!}{2^nr!(n-r)!(n-2r)!}z^{n-2r}. $$ 15.11 Rodrigues' formula for the Legendre polynomials. By differentiating $(z^2-1)^n$, we have $$ P_n(z)=\frac{1}{2^nn!}\frac{d^n}{dz^n}(z^2-1)^n. $$ 15.12 Schläfli’s integral for $P_n(z)$. Applying Cauchy’s integral formula to the result above, we have $$ P_n(z)=\frac{1}{2\pi i}\int_C\frac{(t^2-1)^n}{2^n(t-z)^{n+1}}dt $$ on a counterclockwise contour $C$ around $z$ in the $t$-plane.
Notes on Whittaker & Watson, Chapter XIV, part 3
Whittaker & Watson, A Course of Modern Analysis. Chapter XIV, The Hypergeometric Function. 14.6 Solution of Riemann’s equation by a contour integral. With the substitution of the dependent variable $u$ by $$ u=(z-a)^\alpha(z-b)^\beta(z-c)^\gamma I, $$ Riemann’s equation becomes $$ Q(z)\frac{d^2I}{dz^2} + \Big[(\lambda-2)Q'(z) + R(z)\Big]\frac{dI}{dz} +\left(\frac12\left(\lambda-1\right)(\lambda-2)Q''(z)+(\lambda-1)R'(z)\right)=0, $$ where $$ \begin{cases}\lambda=1-\alpha-\beta-\gamma=\alpha'+\beta'+\gamma',\\ Q(z)=(z-a)(z-b)(z-c),\\ R(z)=\sum_{\text{cycl}}(\alpha'+\beta+\gamma)(z-b)(z-c). \end{cases} $$ Assuming the solution in the form $$ I=\int_C(t-a)^{\alpha'+\beta+\gamma-1}(t-b)^{\alpha+\beta'+\gamma-1}(t-c)^{\alpha+\beta+\gamma'-1}(z-t)^{-\alpha-\beta-\gamma}dt $$ the equation becomes $\int_C\frac{\partial V}{\partial t}dt=0$ with $V=(t-a)^{\alpha'+\beta+\gamma}(t-b)^{\alpha+\beta'+\gamma}(t-c)^{\alpha+\beta+\gamma'}(t-z)^{-(1+\alpha+\beta+\gamma)}$. Note that the integrand of $I$ and $V$ differ by a factor of $(t-a)(t-b)(t-c)(z-t)^{-1}$ which is single valued.