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I need to evaluate the integral:

$$\int_0^{\infty} x p^xe^{\Large-\frac{x}{a}}\ dx$$

for $0<p<1$. Unfortunately I do not know where to begin. I tried integration by parts but got nowhere so I would appreciate some help.

Thank you.

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$$\int_0^{\infty} e^{-x/a} x p^x dx = \int_0^{\infty} e^{-x/a} x e^{x \log p} dx = \int_0^{\infty} x e^{x(\log p -1/a)} dx = \frac{1}{(\log p -1/a)^2} $$

This is using this bit of integration by parts and assuming $\log p < 1/a$

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Write $p^x$ as $e^{x\ln p}$, then \begin{align} \int_0^{\infty} e^{\Large-\frac{x}{a}} x p^x\ dx&=\int_0^{\infty} e^{\Large-\frac{x}{a}} x e^{\large x\ln p}\ dx\\ &=\int_0^\infty xe^{\Large-x\left(\frac{1}{a}-\ln p\right)}\ dx \end{align} Let $u=x\left(\frac{1}{a}-\ln p\right)$, then \begin{align} \int_0^\infty xe^{\Large-x\left(\frac{1}{a}-\ln p\right)}\ dx&=\frac{1}{\left(\frac{1}{a}-\ln p\right)^2}\int_0^\infty ue^{-u}\ du\\ &=\frac{a^2}{(1-a\ln p)^2}\Gamma(2)\\ &=\frac{a^2}{(1-a\ln p)^2} \end{align} where $\displaystyle\int_0^\infty ue^{-u}\ du$ can be evaluated by using integration by parts or using gamma function.

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  • $\begingroup$ Very helpful, thank you. $\endgroup$ – JohnK Jun 2 '14 at 15:05
  • $\begingroup$ Your substitution misses an extra factor, examples show my answer is correct. $\endgroup$ – John Fernley Jun 2 '14 at 15:14
  • $\begingroup$ @JohnFernley I know that and I fixed it but you no need to downvote my answer. You at least can comment to ask me to fix it ┻━┻ ︵ヽ(`▭´)ノ︵ ┻━┻ $\endgroup$ – Anastasiya-Romanova 秀 Jun 2 '14 at 15:18
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Your integral is equivalent to:

$$\int_{0}^{\infty}xe^{bx}\mathrm{d}x,$$

where $b=\dfrac{-1}{a}+\log p$.

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align}&\color{#c00000}{\int_{0}^{\infty}x\,p^{x}\expo{-x/a}\,\dd x} =p\,\partiald{}{p}\int_{0}^{\infty}p^{x}\expo{-x/a}\,\dd x =p\,\partiald{}{p}\int_{0}^{\infty}\pars{p\expo{-1/a}}^{x}\,\dd x \\[3mm]&=p\,\partiald{}{p}\bracks{{1 \over \ln\pars{p\expo{-1/a}}}\int_{0}^{\infty} \partiald{\pars{p\expo{-1/a}}^{x}}{x}\,\dd x} =p\,\partiald{}{p}\bracks{-1 \over \ln\pars{p\expo{-1/a}}} \\[3mm]&={1 \over \ln^{2}\pars{p\expo{-1/a}}} \end{align}

$$\color{#66f}{\large\int_{0}^{\infty}x\,p^{x}\expo{-x/a}\,\dd x ={1 \over \bracks{\ln\pars{p} - 1/a}^{2}}}\,,\qquad 0 <\verts{p\expo{-1/a}} < 1 $$

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  • $\begingroup$ VERY nice! Thank you. $\endgroup$ – JohnK Jul 2 '14 at 20:18
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    $\begingroup$ @JohnK You're welcome. Thanks. $\endgroup$ – Felix Marin Jul 2 '14 at 20:21

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