Integral representation for reverse Bessel polynomials

Question

$\DeclareMathOperator ZZ$Series solutions to partly invert $e^x(x^2+a)$ and $e^xx(x+a)$ exist when more general quadratic-exponential equations are reduced. Using the reverse Bessel polynomials $p_n(x)$:

$$e^xx(x+a)=b\implies x=-\sum_{n=1}^\infty\frac{(-2 ba^{-2})^n}{nn!}p_n\left(\frac{a n}2\right)\tag1$$

The inverse Z transform, like here, lets us find an integral representation by looking at the generating function:

$$\begin{align}\Z_x(p_n(x))=\sum_{n=0}^\infty\frac{p_n(x)}{n!}w^n=e^{(1-\sqrt{1-2w})x}\\ \implies p_n(x)=\Z^{-1}\left(e^{(1-\sqrt{1-2w})x} \right)=\frac{n!}{2\pi i }\oint\exp((1-\sqrt{1-2w})x)w^{-n-1}dw\\\mathop=^?\frac{n!}{2\pi}\int_{-\pi}^\pi \exp\left(\left(1-\sqrt{1-2e^{it}})x-i t n\right)\right)dt\tag2\end{align}$$

This would allow an integral representation of $(1)$, but $(2)$ does not numerically match $p_n(x)$.

What is an actual integral representation for $p_n(x)$?

Answer 1

To evaluate \begin{equation} x=-\sum_{n=1}^\infty\frac{(-2 ba^{-2})^n}{nn!}p_n\left(\frac{a n}2\right) \end{equation} one can use the expression for the reverse Bessel polynomials $p_n(x)$ in term of the modified Bessel function eq. (2): \begin{align} p_n(z)&=z^ny_{n-1}\left( \frac1z \right)\\ y_n(z)&=\sqrt{\frac{2}{\pi z}}e^{1/z}K_{-n-1/2}\left( \frac 1z \right) \end{align} with $u=-2a^{-2}b$ and $X_n=an/2$ we have then \begin{equation} x=-\sqrt{\frac{2}{\pi}}\sum_{n=1}^\infty\frac{u^n}{nn!}X_n^{n+1/2}e^{X_n}K_{-n+1/2}(X_n) \end{equation} The integral representation (DLMF) \begin{equation} K_{\nu}\left(sz\right)=\frac{\Gamma\left(\nu+\frac{1}{2}\right)(2z)^{\nu}}{\pi% ^{\frac{1}{2}}s^{\nu}}\int_{0}^{\infty}\frac{\cos\left(st\right)\,\mathrm{d}t} {(t^{2}+z^{2})^{\nu+\frac{1}{2}}} \end{equation} is valid when $\nu>-1/2,s>0,\Re z>0$.

As $K_{-n+1/2}(X_n)=K_{n-1/2}(X_n)$, taking $z=1,s=X_n$ and after simplifications it comes \begin{equation} x=-\frac{a}{2\pi}\sum_{n=1}^\infty \frac{(2ue^{a/2})^n}{n}\int_0^\infty\frac{\cos(nat/2)}{(1+t^2)^n}\,dt \end{equation} and, by swaping summation and integration, \begin{equation} x=-\frac{a}{2\pi}\int_0^\infty\,dt\sum_{n=1}^\infty \frac1n \left( \frac{2ue^{a/2}}{1+t^2}\right)^n\cos(nat/2) \end{equation} Obtained series corresponds to the generating function of the Tchebyshev polynomials \begin{equation} -\ln\left(1-2xz+z^{2}\right)=2\sum_{n=1}^{\infty}\frac{T_{n}\left(x\right)}{n}% z^{n} \end{equation} which is valid when $|z|<1$. With \begin{equation} M=\frac{4b}{a^2}e^{a/2} \end{equation} one obtains \begin{equation} x=\frac{a}{4\pi}\int_0^\infty\ln\left( 1+2M\frac{\cos(at/2)}{1+t^2}+\frac{M^2}{(1+t^2)^2} \right)\,dt \end{equation} This integral converges for any values of $M$ which makes the result correct for any values of the parameters $a\ne0$ and $b$. It can be numerically checked that this value of $x$ verifies \begin{equation} xe^x\left( x+a \right)=b \end{equation}