Consider the single variable Bell polynomials $\phi_{n}(x)$ given by: $$\phi_{n}(x)=e^{-x}\sum_{k=0}^{\infty}\frac{k^{n}x^{k}}{k!}$$ I am looking for a set of functions $\tilde{\phi}_{n}(x)$ such that, for some inner product, the pair $\phi_{n}(x)$, and $\tilde{\phi}_{m}(x)$ is orthogonal. Boyadzhiev proved a semi-orthogonality property for the polynomials $\;\phi_{n}(x)$: $$\int_{-\infty}^{0}\phi_{n}(x)\phi_{m}(x)\frac{e^{2x}}{x}dx=(-1)^{n}\frac{2^{n+m}-1}{n+m}B_{n+m}$$ $B_{k}$ being the kth Bernoulli number. But i wish for complete orthogonality ! i have tried the following :
It's easy to check the validity of $$\int_{0}^{\infty}e^{-x}\phi_{n}(-x)x^{s-1}dx=(-s)^{n}\Gamma(s)\;\;\;\;(\Re(s)>0)$$
By Parseval's theorem for the Mellin transform, we have: $$\int_{0}^{\infty}\phi_{n}(-x)\tilde{\phi}_{m}(x)e^{-x}\frac{dx}{x}=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}(-s)^{n}\Gamma(s)\Phi_{m}(-s)ds$$ Where : $$\Phi_{m}(s)=\int_{0}^{\infty}\tilde{\phi}_{m}(x)x^{s-1}dx$$ And $\sigma$ lies in the common domain of analycity of $\Gamma(s)\;$ and $\; \Phi_{m}(s)$. But that is the farthest I could go trying to get (weighted) Kronecker delta from the integral !!
EDIT 1
Using the generating function of $\phi_{n}(x)$ : $$\sum_{n=0}^{\infty}\frac{\phi_{n}(x)}{n!}t^{n}=\exp\left[x\left(e^{t}-1 \right ) \right ]$$ We have: $$\frac{\phi_{m}(x)}{m!}=\frac{1}{2\pi}\int_{0}^{2\pi}\exp\left[x\left(e^{e^{it}}-1 \right ) \right ]e^{-imt}dt$$ Now, for a suitable choice of the domain of integration $I$, we put: $$\int_{I}\frac{\phi_{m}(x)\tilde{\phi}_{n}(x)}{m!}dx=\frac{1}{2\pi}\int_{I}\int_{0}^{2\pi}\tilde{\phi}_{n}(x)\exp\left[x\left(e^{e^{it}}-1 \right ) \right ]e^{-imt}dtdx$$ And we get orthogonality by requiring : $$\int_{I}\tilde{\phi}_{n}(x)\exp\left[x\left(e^{e^{it}}-1 \right ) \right ]dx=e^{int}$$ Or, by a suitable choice of the branch cut of $\log$: $$\int_{I}\tilde{\phi}_{n}(x)\exp\left[x\left(z-1 \right ) \right ]dx=(\log z)^{n}$$
EDIT 2 We use the integral representation of the gamma function to obtain: $$\int_{0}^{\infty}x^{a}e^{-sx}dx=\frac{\Gamma(a+1)}{s^{a+1}}\;\;\;\;\Re(s)>0$$ Thus: $$\int_{0}^{\infty}\left(\log x \right )^{n}e^{-sx}dx=\lim_{a\rightarrow 0}\frac{d^{n}}{da^{n}}\frac{\Gamma(a+1)}{s^{a+1}}=\frac{1}{s}\sum_{k=0}^{n}a_{k}(\log s)^{k}$$ Therefore, there exist nth order polynomials in $\log x$, that we'll denote by $f_{n}(x)$, such that:
$$\int_{0}^{\infty}f_{n}(x)e^{-sx}dx=\frac{(\log s)^{n}}{s}$$ Now we put $s=e^{z}$, and obtain: $$\int_{0}^{\infty}e^{-x}f_{n}(x)e^{-x(e^{z}-1)}dx=e^{-z}z^{n}$$ Or: $$\sum_{m=0}^{\infty}\int_{0}^{\infty} e^{-x}f_{n}(x)\frac{\phi_{m}(-x)}{m!}z^{m}dx=e^{-z}z^{n}$$ This is the closest i got !
We use the integral representation of the gamma function to obtain: $$\int_{0}^{\infty}x^{a}e^{-sx}dx=\frac{\Gamma(a+1)}{s^{a+1}}\;\;\;\;\Re(s)>0$$ Thus: $$\int_{0}^{\infty}\left(\log x \right )^{n}e^{-sx}dx=\lim_{a\rightarrow 0}\frac{d^{n}}{da^{n}}\frac{\Gamma(a+1)}{s^{a+1}}=\frac{1}{s}\sum_{k=0}^{n}a_{k}(\log s)^{k}$$ Therefore, there exist nth order polynomials in $\log x$, that we'll denote by $\tilde{\phi}_{n}(x)$, such that:
$$\int_{0}^{\infty}\tilde{\phi}_{n}(x)e^{-sx}dx=\frac{(\log s)^{n}}{s}$$ Now we put $s=e^{z}$, and obtain: $$\int_{0}^{\infty}e^{-x}\tilde{\phi}_{n}(x)e^{-x(e^{z}-1)}dx=e^{-z}z^{n}$$ Or: $$\sum_{m=0}^{\infty}\int_{0}^{\infty} e^{-x}\tilde{\phi}_{n}(x)\frac{\phi_{m}(-x)}{m!}z^{m}dx=e^{-z}z^{n}$$ Comparing the two series, we get the quasi-orthogonality: $$\int_{0}^{\infty}e^{-x}\tilde{\phi}_{n}(x)\frac{\phi_{m}(-x)}{m!}dx=\left\{\begin{matrix} 0 &, &m<n \\ 1& , & m=n\\ \frac{(-1)^{m-n}}{(m-n)!} &, & m>n \end{matrix}\right.$$ Now, i suspect that the following holds: $$\int_{0}^{\infty}e^{-x}\frac{d}{dx}\left[\tilde{\phi}_{n}(x) \right ]\phi_{m}(-x)dx=m!\delta_{nm}$$ Couldn't prove it, but experimentally it holds for a couple of examples. it's worth mentioning that, by Bromwich integral formula, $\tilde{\phi}_{n}(x)$ are given by: $$\tilde{\phi}_{n}(x)=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{\left(\log s \right )^{n}}{s}e^{sx}ds$$ Couldn't obtain a closed form expression. But i guess the Hankel integral representation of the reciprocal gamma function can be utilized to obtain such an expression.