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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 $$ \left(1+\frac tz\right)^\lambda = 1+ \frac{\lambda t}z+\cdots+\frac{\lambda(\lambda-1)\cdots(\lambda-n+1)}{n!}\frac{t^n}{z^n}+R_n(t,z) $$ where the remainder $$ R_n(t,z)=\frac{\lambda(\lambda-1)\cdots(\lambda-n+1)}{n!}\left(1+\frac{t}{z}\right)^\lambda\int_0^{\frac tz}u^n(1-u)^{-\lambda-1}du $$ is bounded for $|\arg(z)|\le\pi-\alpha$ and $|z|>1$, $$ |R_n(t,z)|<\left|\frac{\lambda(\lambda-1)\cdots(\lambda-n+1)}{n!}\right|\left(\frac{1+t}{\sin\alpha}\right)^{|\lambda|}\left(\frac tz\right)^{n+1}(1+t)^{|\lambda|}(n+1)^{-1}, $$ in which the term $|1+\frac tz|$ is controlled by $1\le|1+\frac tz|\le1+t$ for $\Re(z)\ge 0$ and $|1+\frac tz|\le\frac{1}{\sin\alpha}$ for $\Re(z)\le0$. Thus the integral of the remainder $R_n(t,z)$ evaluates to $O(z^{-n-1})$. We have $$ W_{k,m}\sim e^{-\frac12z}z^k\left(1+\sum_{n=1}^\infty\frac{\left(m^2-(k-\frac12)^2\right)\left(m^2-(k-\frac32)^2\right)\cdots\left(m^2-(k-n+\frac12)^2\right)}{n!z^n}\right). $$ 16.31 The second solution of the equation for $W_{k,m}(z)$.
The equation for $W_{k,m}$ is invariant under the transform $k\to-k,z\to-z$. Thus with $\arg(-z)<\pi$, $W_{-k,m}(-z)$ is also a solution, linearly independent of $W_{k,m}$, as the ratio $\frac{W_{k,m}(z)}{W_{-k,m}(z)}$ is not constant.
16.4 Contour integrals of Barnes' type for $W_{k,m}(z)$. $$ W_{k,m}(z)=\frac{e^{-\frac12z}z^k}{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(s)\Gamma(-s-k-m+\frac12)\Gamma(-s-k+m+\frac12)}{\Gamma(-k-m+\frac12)\Gamma(-k+m+\frac12)}z^sds $$ where $\arg(z)\le\frac32\pi-\alpha<\frac32\pi$, and the contour is chosen such that the poles of $\Gamma(s)$ and those of $\Gamma(-s-k-m+\frac12)\Gamma(-s-k+m+\frac12)$ are on the opposite sides of the contour.
Consider the integral on the rectangle $\pm\xi i,-N-\frac12\pm\xi i$, with $\xi\gg1$ and $N$ chosen sufficiently large such that the poles of $\Gamma(-s-k-m+\frac12)\Gamma(-s-k+m+\frac12)$ are to the right of $\Re(s)=-N-\frac12$. The integrals $\int_{-\xi i}^{-N-\frac12-\xi i}$ and $\int_{\xi i}^{-N-\frac12+\xi i}$ vanishes as $\xi\to\infty$, and the integral $\int_{-N-\frac12-i\infty}^{-N-\frac12+i\infty}=O(|z|^{-N-\frac12})$. By calculating the residues at $s=-N,-N+1\dots,0$, the sum of residues evaluates to the asymptotic expansion of $W_{k,m}$.
16.41 Relations between $W_{k,m}(z)$ and $M_{k,\pm m}(z)$.
When $2m$ is not integer $$ \begin{align*}W_{k,m}(z)=\frac{\Gamma(-2m)}{\Gamma(\frac12-m-k)}M_{k,m}(z)+\frac{\Gamma(2m)}{\Gamma(\frac12+m-k)}M_{k,-m}(z) & & |\arg(z)|<\frac32\pi\\ W_{-k,m}(-z)=\frac{\Gamma(-2m)}{\Gamma(\frac12-m+k)}M_{-k,m}(-z)+\frac{\Gamma(2m)}{\Gamma(\frac12+m+k)}M_{-k,-m}(-z), & & |\arg(-z)|<\frac32\pi \end{align*} $$ and conversely with $-\frac12\pi<\arg(z)<\frac32\pi$ and $-\frac32\pi<\arg(-z)<\frac12\pi$ $$ M_{k,m}(z)=\frac{\Gamma(2m+1)}{\Gamma(\frac12+m-k)}e^{k\pi i}W_{-k,m}(-z)+\frac{\Gamma(2m+1)}{\Gamma(\frac12+m+k)}e^{(\frac12+m+k)\pi i}W_{k,m}(z). $$