Difficulty finding Expectation of a special function

Question

I have a special function given as:

$${\rm f}\left(r\right) ={1 \over \beta\lambda}\,2^{r/\beta} \exp\left({\left[2^{r/\beta} - 1\right]K \over \lambda}\right)$$

I should find the Expectation of the random variable $r$. Mathematica was not able to solve the associated Integral function. So it returns:

$$ \int_0^{\infty}\left\{{1 \over \beta\lambda}\,2^{r/\beta} \exp\left({\left[2^{r/\beta} - 1\right]K \over \lambda}\right)\right\}\ r\,{\rm d}r $$

Does anyone recognize how I can reduce this function so I can solve it further?

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_0^{\infty}\braces{{1 \over \beta\lambda}\,2^{r/\beta} \exp\pars{{\bracks{2^{r/\beta} - 1}K \over \lambda}}}\ r\,\dd r:\ {\large ?}}$ and we'll assume $\ds{{K \over \lambda} < 0}$.

Let's $\ds{t \equiv -\,{\bracks{2^{r/\beta} - 1}K \over \lambda}}$ such that $\ds{r = {\beta\ln\pars{1 - \lambda t/K} \over \ln\pars{2}}}$ and $\ds{\dd r = {-\lambda/K \over \ln\pars{2}\pars{1 - \lambda t/K}}\,\dd t}$:

\begin{align} &\int_0^{\infty}\braces{{1 \over \beta\lambda}\,2^{r/\beta} \exp\pars{{\bracks{2^{r/\beta} - 1}K \over \lambda}}}\ r\,\dd r \\[3mm]&={1 \over \beta\lambda}\int_{0}^{\infty}\pars{1 - {\lambda \over K}\,t}\expo{-t}\, {{\beta\ln\pars{1 - \lambda t/K} \over \ln\pars{2}}}\, {-\lambda/K \over \ln\pars{2}\pars{1 - \lambda t/K}}\,\dd t \\[3mm]&=-\,{1 \over K\ln^{2}\pars{2}}\int_{0}^{\infty}\ln\pars{1 + \mu t}\expo{-t} \,\dd t\qquad\mbox{where}\qquad\mu \equiv -\,{\lambda \over K} > 0. \end{align}

Then, \begin{align} &\int_0^{\infty}\braces{{1 \over \beta\lambda}\,2^{r/\beta} \exp\pars{{\bracks{2^{r/\beta} - 1}K \over \lambda}}}\ r\,\dd r =-\,{1 \over K\ln^{2}\pars{2}}\int_{0}^{\infty}\expo{-t}\, {\mu \over 1 + \mu t}\,\dd t \\[3mm]&=-\,{\expo{1/\mu} \over K\ln^{2}\pars{2}}\int_{1/\mu}^{\infty} {\expo{-t} \over t}\,\dd t ={\expo{1/\mu} \over K\ln^{2}\pars{2}}\,{\rm E_{i}}\pars{-\,{1 \over \mu}} \end{align} where $\ds{{\rm E_{i}}\pars{z}}$ is the Exponential Integral Function.

$$\color{#00f}{\large% \int_0^{\infty}\braces{{1 \over \beta\lambda}\,2^{r/\beta} \exp\pars{{\bracks{2^{r/\beta} - 1}K \over \lambda}}}\ r\,\dd r ={\expo{-K/\lambda} \over K\ln^{2}\pars{2}}\, {\rm E_{i}}\pars{K \over \lambda}}\,,\quad {K \over \lambda} < 0 $$

Answer 2

The integral is not convergent if $K$ and $\lambda$ are positive constants.

If the upper boundary of the integral is not infinity, but $R$, the result is :

convenient in case of $K>0$ and $\lambda >0$ and finite $R$.