# What Arthur Erdélyi left out —

## Confluent Hypegeometric Functions

\begin{align*} U(a,b,z)& = \frac{1}{\Gamma\left(a\right)}\int_{0}^{\infty}e^{-zt}t^{a-1}\left(1+t\right)^{b-a-1}\mathrm{d}t= \frac{z^{1-b}}{\Gamma\left(1+a-b\right)}\int_{0}^{\infty}e^{-zt}\frac{t^{a-b}}{\left(1+t\right)^{a}}\mathrm{d}t \\ & = z^{1-b}U\left(1+a-b,2-b,z\right)= \frac{\Gamma(1-b)}{\Gamma(a-b+1)}M(a,b,z)+\frac{\Gamma(b-1)}{\Gamma(a)}z^{1-b}M(a-b+1,2-b,z) \end{align*} in terms of Laguerre polynomials $$M(a,b,z)=\frac{\binom{b-1}{a}}{\binom{b-1+\alpha}{a}}\sum_{k=0}(-1)^{k}L_{k}^{(-\alpha-k)}(z)\frac{\binom{-a}{k}}{\binom{-b-\alpha}{k}}$$ \begin{align*} U(a,b,z)&= \frac{(1+\alpha-b)!}{(1+\alpha-b+a)!}\sum_{k=0}L_{k}^{(\alpha)}(z)\frac{\binom{-a}{k}}{\binom{b-a-\alpha-2}{k}} = z^{1-b}\frac{(b-\alpha-1)!}{(a-\alpha)!}\sum_{k=0}L_{k}^{(-\alpha)}(z)\frac{\binom{b-a-1}{k}}{\binom{\alpha-a-1}{k}} \\ &= e^{-z}\frac{2^{b-a}\sqrt{\pi}}{\Gamma\left(a+\tfrac{1}{2}\right)}\sum_{i=0}L_{2i}^{\left(a+b-2\right)}\left(2z\right)\frac{\binom{-\frac{1}{2}}{i}}{\binom{-\frac{1}{2}-a}{i}} = e^{-z}z^{1-b}\frac{2^{1-a}\sqrt{\pi}}{\Gamma\left(a-b+\tfrac{3}{2}\right)}\sum_{i=0}L_{2i}^{\left(a-2b+1\right)}\left(2z\right)\frac{\binom{-\frac{1}{2}}{i}}{\binom{b-a-\frac{3}{2}}{i}} \end{align*}

## (Lower and Upper) Incomplete Gamma Function

\begin{align*} \frac{\Gamma(s,z)}{z^s e^{-z}} &= -\frac{1}{s}\sum_{i=0}\frac{\binom{i-s}{i+1}}{L_{i}^{-s}\left(-z\right)L_{i+1}^{-s}\left(-z\right)}= \frac{1}{z}-\frac{1}{z}\sum_{i=0}\frac{\binom{i+1-s}{i+1}}{L_{i}^{1-s}\left(-z\right)L_{i+1}^{1-s}\left(-z\right)} \xleftarrow[n\to\infty]{} \sum_{\lambda\colon L_{n}^{-s}(\lambda)=0}\frac{\binom{n-s}{n}}{L_n^{1-s}(\lambda)^2 }\frac{1}{\lambda(z+\lambda)}\\ \Gamma(s,z) &= e^{-2z}\frac{\sqrt{\pi}}{\Gamma\left(\frac{3}{2}-s\right)}\sum_{j=0}L_{2j}^{\left(-2s\right)}\left(2z\right)\frac{\binom{-\frac{1}{2}}{j}}{\binom{s-\frac{3}{2}}{j}} = e^{-2z}\left(2z\right)^{s}\sum_{j=0}\frac{L_{2j}^{\left(s\right)}\left(2z\right)}{j+\frac{1}{2}} \\ \gamma (s,z) &= \frac{2\sqrt{\pi}}{s}z^{s-\frac{1}{2}}e^{-\frac{z}{2}}\sum_{n=0}\frac{\binom{s-1}{n}}{\binom{-s-1}{n}}\left(n+\frac{1}{2}\right)I_{n+\frac{1}{2}}\left(\frac{z}{2}\right) = \frac{1}{2}4^{s}\sqrt{z}e^{-\frac{z}{2}}\Gamma\left(s-\frac{1}{2}\right)\sum_{n=0}\frac{(s-1)\binom{1-2s}{n}}{(s+n-1)(s+n)}\left(s+n-\frac{1}{2}\right)I_{s+n-\frac{1}{2}}\left(\frac{z}{2}\right) \end{align*}

## Riemann's Zeta function

\begin{equation*} s\left(s-1\right)\frac{\zeta\left(s\right)\Gamma\left(\frac{s}{2}\right)}{\pi^{\frac{s}{2}}}= \left(1-s\right)\alpha^{s}+s\alpha^{s-1} -s\left(1-s\right)\sum_{k=1}\frac{\Gamma\left(\frac{s}{2},\alpha^{2}k^{2}\pi\right)}{\left(k^{2}\pi\right)^{\frac{s}{2}}}+\frac{\Gamma\left(\frac{1-s}{2},\frac{k^{2}\pi}{\alpha^{2}}\right)}{\left(k^{2}\pi\right)^{\frac{1-s}{2}}} \end{equation*} with $\alpha$ a free parameter.

## Borel Summations of Rapidly Divergent Series

Divergent summation of Riemann's (and Hurwitz's) Zeta function ($B_k$ are Bernoulli polynomials): $\sum_{k=0}\frac{B_{k}(x)}{z^{k}}\frac{\binom{1-s}{k}}{s-1}=z^{s-1}\zeta(s,x+z)$ Gamma Function$\sum_{k=0}^{\infty}\binom{s}{k}k!\,z^k=\int_0^\infty e^{-t}\mathcal{B}y(tz)\, dt=\int_{0}^{\infty}e^{-t}(1+tz)^{s}\, dt=z^{s}e^{\frac{1}{z}}\Gamma\left(s+1,\frac{1}{z}\right)$ as S. Ramanujan already noted in his letter to G. H. Hardy for $s=-1$ and $z=1$.
Hypergeometric Function$U(a,b,z)=\frac{z^{c-a}}{\Gamma(c)}\int_{0}^{\infty}e^{-tz}\cdot t^{c-1}\cdot_{2}F_{1}(a,a-b+1;c;-t)dt=z^{-a}\cdot\,_{2}F_{0}\left(a,a-b+1;;-\frac{1}{z}\right)$

## Laguerre Polynomials

Generating series \begin{align*} &\frac{e^{-\frac{tz}{1-t}}}{(1-t)^{n+1+\alpha}}L_{n}^{(\alpha)}\left(\frac{z}{1-t}\right) = \sum_{k=n}\binom{k}{n}t^{k-n}L_{k}^{(\alpha)}(z) \quad\text{ and }\quad \frac{e^{t z}}{(1-t)^{n-\alpha}}L_n^{(\alpha-n)}(z(1-t))=\sum_{k=n}(-1)^{k-n}\binom{k}{n}t^{k-n}L_k^{(\alpha-k)}(z),\\ &\frac{e^{-\frac{tz}{1-t}}}{(1-t)^{n+1+\alpha}}L_{n}^{(\alpha)}\left(\frac{tz}{1-t}\right) = \sum_{k=0}\binom{k+\alpha+n}{n}t^{k}L_{k}^{(\alpha)}(z),\\ &e^{t}L_{n}^{(\alpha)}(z-t) = \sum_{k=0}\frac{t^{k}}{k!}L_{n}^{(\alpha+k)}(z),\\ &\frac{e^{-z\sqrt{1+4t}}}{\sqrt{1+4t}}=e^{-z}\sum_{n=0}t^n L_n^{(-2n-1)}(2z). \end{align*}

## Gegenbauer Polynomials

Generating series (cf. \eqref{ref2}): $$\label{ref1} \sum_{n=0}^{\infty}\frac{C_{n}^{(\alpha)}(x)}{\binom{2\alpha+n-1}{n}}\frac{t^{n}}{n!}=\Gamma\left(\alpha+\frac{1}{2}\right)e^{tx}\frac{J_{\alpha-\frac{1}{2}}\left(t\sqrt{1-x^{2}}\right)}{\left(\frac{1}{2}t\sqrt{1-x^{2}}\right)^{\alpha-\frac{1}{2}}}$$

## Bessel Functions

Generating series (cf. \eqref{ref1}): $$\label{ref2} \frac{J_{s}\left(tx\right)}{\left(\frac{1}{2}tx\right)^{s}}=\frac{e^{-x\sqrt{1-t^{2}}}}{\Gamma\left(s+1\right)}\sum_{n=0}C_{n}^{s+\frac{1}{2}}\left(\sqrt{1-t^{2}}\right)\frac{x^{n}}{n!\binom{2s+n}{n}}$$ Modified Bessel functions (hyperbolic Bessel functions): \begin{align*} I_{n+\frac{1}{2}}\left(z\right) &= \frac{n!e^{-z}}{\sqrt{\pi}\left(2z\right)^{n+\frac{1}{2}}}\sum_{k=2n+1}\binom{k-n-1}{n}\frac{\left(2z\right)^{k}}{k!} \\ &= \frac{n!}{\sqrt{\pi}\left(2z\right)^{n+\frac{1}{2}}}\left(e^{z}\cdot L_{n}^{\left(-2n-1\right)}\left(-2z\right)-e^{-z}\cdot L_{n}^{\left(-2n-1\right)}\left(2z\right)\right) = \frac{1}{\pi}\left(K_{n+\frac{1}{2}}\left(-z\right)i-\left(-1\right)^{n}K_{n+\frac{1}{2}}\left(z\right)\right) \end{align*}

## Fourier Transforms

Definition: $\mathcal F(f)(\xi)= \int_{-\infty}^\infty e^{-2\pi i \xi z} f(z)$ is unitary (FourierTransform[..., z, $\xi$, FourierParameters->{0,-2 $\pi$}]).
Eigenfunctions (for $a=\sqrt{2\pi}$; Hermite polynomials) \begin{equation*} \mathcal F\left(e^{-\frac{a^2 z^2}2} H_n(a z)\right)(\xi)= \frac{\sqrt{2\pi}(-i)^n}{a}\cdot e^{-\frac{2\pi^2\xi^2}{a^2}} H_n\left(\frac{2\pi\xi}a\right) \end{equation*} \begin{equation*} \mathcal F\left(e^{-\frac{a^2 z^2}2} L_n^{(b)}\left(\frac {a^2 z^2}2\right)\right)(\xi)= \frac{(-1)^n\sqrt{2\pi}}{a} e^{-\frac{2\pi^2\xi^2}{a^2}} \cdot L_n^{\left(-b-n-\frac 1 2\right)}\left(\frac {2\pi^2\xi^2}{a^2}\right) \end{equation*} \begin{equation*} \mathcal F\left(e^{-a^2 z^2} z^n\right)(\xi)= \left(-\frac i{2 a}\right)^n \frac{\sqrt \pi} a e^{-\frac{\pi^2\xi^2}{a^2}} H_n\left(\frac {\pi\xi}{a}\right) \end{equation*} Compact support: \begin{equation*} \mathcal F\left(\frac{J_{s+n} (z)}{\left(\frac z 2\right)^s}\right)(\xi)= \frac{i^n 2\sqrt\pi C_n^s(2\pi\xi)}{\Gamma\left(s+\frac12\right)\binom{-2s}{n}} \cdot \ (1-4\pi^2\xi^2)^{s-\frac12} \operatorname{rect}( \pi \xi ) \end{equation*} Schwartz distribution: \begin{equation*} \mathcal F\left(\frac{e^{-z^2 \pi}}{(z^2\pi)^{n+1}}\right)(k)= \frac{\sqrt\pi L_n^{\frac 1 2}(-k^2\pi)}{\binom{-\frac 1 2}{n+1}(n+1)!} \sqrt{k^2\pi}\left(1- \mathrm{Erfc}\left(\sqrt{k^2\pi}\right)\right) + \frac{e^{-k^2\pi}}{n!} \left(\frac{L_n^{\frac 1 2}(-k^2\pi)}{\binom{-\frac 1 2}{n+1}(n+1)} - \frac{1}{n+\frac 1 2} \sum_{j=1}^{\frac{n+1}{2}} \frac{L_{n+1-2j}^{\frac 1 2+2j}(-k^2\pi)}{\binom{-\frac 1 2-j}{n-j}}\right) \end{equation*} \begin{equation*} \int_{-\infty}^\infty e^{s x}e^{-a x^2}H_n(x)dx= e^{s^2/(4a)}\sqrt{\frac{\pi}a}\sqrt{1-\frac{1}{a}}^n H_n \left(\frac{s}{2\sqrt{a(a-1)}}\right) \end{equation*} \begin{equation*} \int_{-\infty}^\infty e^{i s x}e^{-a x^2}H_n(x)dx= e^{-s^2/(4a)}\sqrt{\frac{\pi}a} \left(i\frac{\sqrt{a(1-a)}}a\right)^n H_n\left(\frac{s}{2\sqrt{a(1-a)}}\right) \end{equation*}

### Poisson summation formula

\begin{equation*} \sum_{z\in \mathbb Z} f((z-z_0)T)e^{2\pi i \xi_0 z} = \frac 1 T \sum_{\xi\in\mathbb Z}\mathcal{F}(f)\left(\frac{\xi-\xi_0}{T}\right)e^{-2\pi i z_0 (\xi-\xi_0)} \end{equation*} Regularization of poles \begin{align*} \sum_{z=1}\frac{e^{-z^2\pi x^2}}{(z^2\pi x^2)^{n+1}}&= \frac{1}{2x \binom{-\frac 1 2}{n+1}(n+1)!}+\frac{1}{2(n+1)!}\sum_{j=0}^{n+1}(-1)^{n-j}\frac{\binom{n+1}{j}}{j!\binom{-\frac 1 2}{j}}B_{2j}\left(\frac\pi{x^2}\right)^j \\ & +\frac 1 x \sum_{k=1} \frac{e^{-\frac{k^2\pi}{x^2}}}{n!}\left(\frac{L_n^{\frac 1 2}(-\frac{k^2\pi}{x^2})}{\binom{-\frac 1 2}{n+1}(n+1)} -\frac 1 {n+\frac 1 2}\sum_{j=1}^{\frac{n+1}2}\frac{L_{n+1-2j}^{\frac 1 2+2j}(-\frac{k^2\pi}{x^2})}{\binom{-\frac 1 2-j}{n-j}}\right) -\sqrt\pi \frac{L_n^{\frac 1 2}(-\frac{k^2\pi}{x^2})}{\binom{-\frac 1 2}{n+1}(n+1)!}\sqrt{\frac{k^2\pi}{x^2}} \mathrm{Erfc}\left(\sqrt{\frac{k^2\pi}{x^2}}\right) \end{align*}

Corrections are welcome.