17
$\begingroup$

One can show by computation the following for $b>1$ $$\sum_{n=0}^\infty\frac{n^ne^{-b n}}{\Gamma(n+1)}=\frac{1}{1+W_{\color{blue}{0}}(-e^{-b})},\tag{1}$$ (here one assumes that the term with $n=0$ is understood as the limit $\lim_{n\to 0}$ and is equal to $1$) and $$\int_{0}^\infty\frac{x^xe^{-b x}}{\Gamma(x+1)}dx=\boldsymbol{\color{red}{-}}\frac{1}{1+W_{\color{red}{-1}}(-e^{-b})}.\tag{2}$$

$W_0$ and $W_{-1}$ are different branches of the Lambert W function. One can see that this formulas look similar. I considered them in the hope of obtaining a function for which sum equals integral: $$ \sum_{n=0}^\infty f(n)=\int_0^\infty f(x) dx. $$ $(1)$ is the consequence of Lagrange inversion and the integral arises in the probability distribution theory, namely the Kadell-Ressel pdf (see also this MSE post).

Question 1. Can anybody explain the symmetry between $(1)$ and $(2)$ without resorting to direct calculation?

Question 2. Is it possible to alter $(1)$ and $(2)$ to obtain a nice function for which sum equals integral?

If $b=1$ then there is the Knuth series $$ \sum_{n=1}^\infty\left(\frac{n^ne^{-n}}{\Gamma(n+1)}-\frac1{\sqrt{2\pi n}}\right)=-\frac23-\frac1{\sqrt{2\pi}}\zeta(1/2),\tag{3} $$ and the "Knuth integral" $$ \int_0^\infty\left(\frac{x^xe^{-x}}{\Gamma(x+1)}-\frac1{\sqrt{2\pi x}}\right)dx=-\frac13.\tag{4} $$ Again we see there is a discrepancy.

Question 3. Is it possible to modify the term $\frac1{\sqrt{2\pi x}}$ in $(3)$ and $(4)$ so that the series and the integral agree?

Edit. Of course by mounting some additional terms and parameters one can come up with a formula that technically answers question 2 or 3. What is meant as nice in question 2 might be difficult to formulate explicitly. It is best illustrated by formulas in this MSE post.

$\endgroup$
  • $\begingroup$ Doesn't the Euler-McLaurin formula implies that sums cannot equal integrals exactly, when the sumand is identical to the integrand? $\endgroup$ – Alecos Papadopoulos Oct 18 '17 at 17:22
  • 1
    $\begingroup$ @AlecosPapadopoulos, there are a lot of examples demonstrating that the difference can vanish. $\endgroup$ – Nemo Oct 18 '17 at 17:24
  • $\begingroup$ @Nemo would you mind to provide some examples or even a resource? maybe someone can spot a pattern $\endgroup$ – tired Oct 18 '17 at 19:45
  • $\begingroup$ @Nemo Great initiative, but something's wrong with the link. Can you clearly separate the link from surrounding text? Thanks! $\endgroup$ – Alecos Papadopoulos Oct 18 '17 at 20:20
  • 2
    $\begingroup$ Your post shows an interest in "a function for which sum equals integral." But this cannot be what you mean; you must mean you are looking for a continuous (or even nicer) function having that properly. Otherwise there are trivial examples (which I know you know): $f(x)=2^{-x}$ when $x$ is a positive integer, say, and zero otherwise. $\endgroup$ – symplectomorphic Oct 19 '17 at 6:18
5
$\begingroup$

Question 2. Is it possible to alter (1) and (2) to obtain a function for which sum equals integral?

A simpler form for $z\in[0,\mathrm{e}^{-1})$:

\begin{align} \sum_{n=0}^\infty \frac{(z\,n)^n}{\Gamma(n+1)} &= \frac1{1+\operatorname{W}_{0}(-z)} \tag{1}\label{1} ,\\ \int_0^\infty \frac{(z\,x)^x}{\Gamma(x+1)}\,dx &=-\frac1{1+\operatorname{W}_{-1}(-z)} \tag{2}\label{2} . \end{align}

For some $u\in\mathbb{R}$ consider \begin{align} \sum_{n=0}^\infty \frac{u}{(n+1)^2} &=\frac{u\pi^2}6 \tag{3}\label{3} ,\\ \int_0^\infty \frac{u}{(x+1)^2}\,dx&=u \tag{4}\label{4} . \end{align}

Let's add \eqref{3} and \eqref{4} to \eqref{1} and \eqref{2}, respectively:

\begin{align} \sum_{n=0}^\infty \left( \frac{(z\,n)^n}{\Gamma(n+1)} +\frac{u}{(n+1)^2} \right) &= \frac1{1+\operatorname{W}_{0}(-z)} +\frac{u\pi^2}6 \tag{5}\label{5} ,\\ \int_0^\infty \left( \frac{(z\,x)^x}{\Gamma(x+1)} +\frac{u}{(x+1)^2} \right) \,dx &=-\frac1{1+\operatorname{W}_{-1}(-z)} +u \tag{6}\label{6} . \end{align}

From the right hand sides of \eqref{5} and \eqref{6} for any $z\in[0,\mathrm{e}^{-1})$ we have

\begin{align} u&= -6\frac{2+\operatorname{W_0}(-z)+\operatorname{W_{-1}}(-z)}{(\pi^2-6)(1+\operatorname{W_0}(-z))(1+\operatorname{W_{-1}}(-z))} \end{align}

such that the pair $(z,u)$ satisfies \eqref{5}=\eqref{6}.

For example,

\begin{align} z&=\tfrac12\ln2 ,\quad\operatorname{W_0(-z)}=-\ln2,\quad\operatorname{W_{-1}(-z)}=-2\ln2 ,\\ &\sum_{n=0}^\infty \left( \frac{(n\ln2)^n}{2^n\Gamma(n+1)} - \frac{6(2-3\ln2)}{ (\pi^2-6)(1-\ln2)(1-2\ln2)(n+1)^2 } \right) \\ =& \int_{0}^\infty \left( \frac{(x\ln2)^x}{2^x\Gamma(x+1)} - \frac{6(2-3\ln2)}{ (\pi^2-6)(1-\ln2)(1-2\ln2)(x+1)^2 } \right) \\ =& \frac{\pi^2(\ln2-1)+6(2\ln2-1)}{ (\pi^2-6)(\ln2-1)(2\ln2-1) } \approx 1.549536 . \end{align}

Edit

Similarly,

\begin{align} &\sum_{n=0}^\infty 2^{-n} \left( \frac{(n\ln2)^n}{\Gamma(n+1)} + \frac{\ln2\,(3\ln2-2)}{ (\ln2-1)(2\ln2-1)^2 } \right) \\ =& \int_{0}^\infty 2^{-x} \left( \frac{(x\ln2)^x}{\Gamma(x+1)} + \frac{\ln2\,(3\ln2-2)}{ (\ln2-1)(2\ln2-1)^2 } \right) \\ =& \frac{2(\ln2)^2-1}{ (\ln2-1) (2\ln2-1)^2 } \approx 0.8537740 . \end{align}

$\endgroup$
  • 1
    $\begingroup$ Apologies for not clarifying the question earlier. $\endgroup$ – Nemo Oct 19 '17 at 6:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.