functions orthogonal to the exponential Bell polynomials 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 !
 A: 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.
