Evaluation the Elsasser function:$\text E(y,u)=\int_{-\frac12}^\frac12e^{\frac{2\pi uy\sinh(2\pi y )}{\cos(2\pi x)-\cosh(2\pi y)}}dx$ from MathWorld

Solution 1:

Transforming the integral as \begin{align} \operatorname E(y,u)&=\int_{-\frac12}^\frac12 \exp\left( \frac{2\pi uy\sinh(2\pi y)}{\cos(2\pi x)-\cosh(2\pi y)} \right)\,dx\\ &=\frac{1}{\pi}\int_0^\pi \exp\left( \frac{\alpha}{\beta\cos t-1} \right)\,dt\\ &=\frac{e^{-\alpha}}{\pi}\int_0^\pi \exp\left( \frac{\alpha \beta\cos t}{\beta\cos t-1} \right)\,dt \end{align} where \begin{align} \alpha&=2\pi uy\tanh( 2\pi y)\\ \beta&=\frac{1}{\cosh 2\pi y}, \end{align} we can use the generating function of the Laguerre polynomials \begin{equation} (1-z)^{-1}\exp\left(\frac{xz}{z-1}\right)=\sum_{n=0}^{\infty}L_{n}\left(x\right)z^{n} \end{equation} valid for $|z|<1$. Here, taking $x=\alpha,z=\beta\cos t$, we have ($|\beta\cos t|<1$) \begin{equation} \exp\left( \frac{\alpha \beta\cos t}{\beta\cos t-1} \right)=\sum_{n=0}^{\infty}L_{n}\left(\alpha\right)\beta^{n}(1-\beta\cos t)\cos^nt \end{equation} Then by changing the order of integration and summation, one obtains \begin{align} \operatorname E(y,u)&=\frac{e^{-\alpha}}{\pi}\sum_{n=0}^{\infty}L_{n}\left(\alpha\right)\beta^{n}\int_0^\pi (1-\beta\cos t)\cos^nt\,dt\\ &=\frac{e^{-\alpha}}{\pi}\left[\sum_{n=0}^{\infty}L_{n}\left(\alpha\right)\beta^{n}\int_0^\pi \cos^nt\,dt-\sum_{n=0}^{\infty}L_{n}\left(\alpha\right)\beta^{n+1}\int_0^\pi \cos^{n+1}t\,dt\right] \end{align} For $p=0,1,2,\cdots$, the trigonometrics integrals \begin{equation} \int_0^\pi\cos^pt\,dt=\begin{cases} 0&\text{ if }n=2p+1\\ \pi\frac{(2p-1)!!}{(2p)!!}&\text{ if }n=2p \end{cases} \end{equation} Denoting \begin{equation} a_p=\frac{(2p-1)!!}{(2p)!!}=2^{-2p}\binom{2p}{p} \end{equation} \begin{align} \operatorname E(y,u)&=e^{-\alpha}\left[\sum_{p=0}^{\infty}L_{2p}\left(\alpha\right)\beta^{2p}a_p-\sum_{p=0}^{\infty}L_{2p+1}\left(\alpha\right)\beta^{2p+2}a_{p+1}\right] \end{align} Finally, \begin{equation} \operatorname E(y,u)=e^{-\alpha}\sum_{p=0}^{\infty}\beta^{2p}\left[\frac{(2p-1)!!}{(2p)!!}L_{2p}\left(\alpha\right)-\frac{(2p+1)!!}{(2p+2)!!}\beta^2L_{2p+1}\left(\alpha\right)\right] \end{equation} or \begin{equation} \operatorname E(y,u)=e^{-\alpha}\sum_{p=0}^{\infty}\binom{2p}{p}\left(\frac{\beta}{2} \right)^{2p}\left[L_{2p}\left(\alpha\right)-\frac{2p+1}{2(p+1)}\beta^2L_{2p+1}\left(\alpha\right)\right] \end{equation} which seems to be numerically correct.

Edit: A Laplace transform method

Alternatively, one can use a Laplace transform method. From the expression \begin{equation} \operatorname E(y,u)=\frac{1}{\pi}\int_0^\pi \exp\left( \frac{\alpha}{\beta\cos t-1} \right)\,dt \end{equation} taking the Laplace transform wrt $\alpha$ under the integral leads to \begin{equation} \mathcal{L}\left[\operatorname E,\alpha\to p\right]=\frac1\pi\int_0^\pi\frac{\beta\cos t}{p\beta\cos t-p-1}\,dt \end{equation} with $p>0$, and thus \begin{equation} \mathcal{L}\left[\operatorname E,\alpha\to p\right]=\frac{1}{p}-\frac{1}{p\sqrt{(p+1)^2}-\beta^2p^2} \end{equation} The inverse transform can be expressed as \begin{equation} \operatorname E(y,u)=1-\frac{1}{\sqrt{1-\beta^2}}\int_0^\alpha I_0\left( \frac{\beta s}{1-\beta^2} \right)e^{-\frac{s}{1-\beta^2}}\,ds \end{equation} or \begin{equation} \operatorname E(y,u)=1-\sqrt{1-\beta^2}\int_0^{\frac{\alpha}{1-\beta^2}} I_0\left(\beta s \right)e^{-s}\,ds \end{equation} $I_0(.)$ is a modified Bessel function. From the multiplication theorem \begin{equation} I_{\nu}\left(\lambda z\right)=\lambda^{\nu}\sum_{k=0}^{\infty}% \frac{(\lambda^{2}-1)^{k}(\frac{1}{2}z)^{k}}{k!}I_{\nu+ k}\left(z% \right), \end{equation} with$\nu=0$ \begin{align} \operatorname E(y,u)&=1-\sqrt{1-\beta^2}\int_0^{\frac{\alpha}{1-\beta^2}}\sum_{k=0}^\infty\frac{(\beta^2-1)^k}{2^{k}k!}s^kI_k(s)e^{-s}\,ds\\ &=1-\sum_{k=0}^\infty(-1)^k\frac{(1-\beta^2)^{k+1/2}}{2^{k}k!}\int_0^{\frac{\alpha}{1-\beta^2}}s^kI_k(s)e^{-s}\,ds \end{align} From the tabulated integral (DLMF \begin{equation} \int e^{\pm z}z^{\nu}I_{\nu}\left(z\right)\mathrm{d}z=\frac{e^{\pm z }z^{\nu+1}}{2\nu+1}\left(I_{\nu}\left(z\right)\mp I_{\nu+1} \left(z\right)\right) \end{equation} we obtain \begin{equation} \operatorname E(y,u)=1-\sum_{k=0}^\infty(-1)^k\frac{(1-\beta^2)^{k+1/2}}{2^{k}k!} \frac{e^{- \frac{\alpha}{1-\beta^2} }\left( \frac{\alpha}{1-\beta^2} \right)^{k+1}}{2k+1}\left(I_{k}\left(\frac{\alpha}{1-\beta^2}\right)+ I_{k+1} \left(\frac{\alpha}{1-\beta^2}\right)\right) \end{equation} with $\gamma=\frac{\alpha}{1-\beta^2}=2\pi uy\coth2\pi y$, \begin{equation} \operatorname E(y,u)=1-2\pi uy\sum_{k=0}^\infty\frac{(-1)^k}{2^{k}k!} \frac{(2\pi uy\tanh(2\pi y))^{k}}{2k+1}\left(I_{k}\left(\gamma\right)+ I_{k+1} \left(\gamma\right)\right)e^{- \gamma } \end{equation} which seems numerically valid.

Solution 2:

Thanks to @Paul Enta’s answer here is a new decomposition formula for the Kampé de Fériet function found on Wolfram Mathworld among other sources with this photo:

Using the Bessel Sum Representation:

$$\begin{equation} \operatorname E(y,u)=1-2\pi uy\sum_{k=0}^\infty\frac{(-1)^k}{2^{k}k!} \frac{(2\pi uy\tanh(2\pi y))^{k}}{2k+1}\left(I_{k}\left(\gamma\right)+ I_{k+1} \left(\gamma\right)\right)e^{- \gamma }, \gamma=2\pi uy\coth2\pi y \end{equation} $$

Remember that $\alpha=2\pi uy\tanh( 2\pi y)$ and use the Bessel function definition and Pochhammer symbol:

$$\frac{e^\gamma(1-\text E(y,u))}{2\pi uy}= \sum_{m,n\ge0}\left(-\frac{\alpha}{2}\right)^m \frac{\left(\frac12\right)_m}{\left(\frac32\right)_m \Gamma(n+m+1)n!}\left(\frac \gamma 2\right)^{2n+m} +\sum_{m,n\ge0}\left(-\frac{\alpha}{2}\right)^m \frac{\left(\frac12\right)_m}{\left(\frac32\right)_m \Gamma(n+m+2)n!}\left(\frac \gamma 2\right)^{2n+m+1}$$

With the Kampé de Fériet function defined above as:

$$\text F^{p,r,u}_{q,s,v}\left(^{a_1,…,a_p;c_1,…,c_r;f_1,…,f_u}_{b_1,…,b_q;d_1,…,d_s;g_1,…,g_v}\ x,y\right)\mathop=^\text{def}\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{\prod\limits_{j=1}^p(a_j)_{m+n} \prod\limits_{j=1}^r(c_j)_m \prod\limits_{j=1}^u (f_j)_n x^my^n}{\prod\limits_{j=1}^q (b_j)_{m+n} \prod\limits_{j=1}^s(d_j)_m \prod\limits_{j=1}^v(g_j)_n m!n!}$$

Therefore:

$$\frac{e^\gamma(1-\text E(y,u))}{2\pi uy} =\text F^{0,1,0}_{1,1,0}\left(^{\ \ ;\frac 12;}_{1;\frac 32;}\ \frac{\gamma \alpha}{4},\frac{\gamma ^2}4\right)+\frac \gamma 2 \text F^{0,1,0}_{1,1,0}\left(^{\ \ ;\frac 12;}_{2;\frac 32;}\ \frac{\gamma \alpha}{4},\frac{\gamma ^2}4\right) $$

and:

$$\text E(y,u)=1-\frac{2\pi y}{e^{2\pi uy\coth(2\pi y)}}\left[\text F^{0,1,0}_{1,1,0}\left(^{\ \ ;\frac 12;}_{1;\frac 32;}\ -(\pi uy)^2,(\pi uy \coth(2\pi y))^2\right)+\pi uy\coth(2\pi y) \text F^{0,1,0}_{1,1,0}\left(^{\ \ ;\frac 12;}_{2;\frac 32;}\ -(\pi uy)^2,(\pi uy \coth(2\pi y))^2\right)\right]$$

Unfortunately, $\text F^{0,1,0}_{1,1,0}$ seems to have no Horn function form to simplify the already simple Kampé de Fériet representation. Please correct me and give me feedback!