I have to calculate analytically this integral: $$ {\rm J}\left(q\right) = \int_{0}^{\infty}{{\rm d}x \over x^{q}\left({\rm e}^{kx}-1\right)} $$ where $-1\le q\le N$ with: $N\in\mathbb{N}$ and $q\in\mathbb{N}$, $k\le 5\times10^{-5}$ I didn't find anything on the Gradshteyn Ryzhik and Mathematica isn't able to integrate it. Is it possible to make some approximation because the little value of $k$? Thanks in advance.
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3 Answers
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \color{#0000ff}{\large{\rm J}\left(q\right)} &= \int_{0}^{\infty}{{\rm d}x \over x^{q}\left({\rm e}^{kx}-1\right)} = k^{q - 1}\int_{0}^{\infty}{\expo{-x} \over x^{q}\pars{1 - \expo{-x}}}\,\dd x = k^{q - 1}\int_{0}^{\infty}{\expo{-x} \over x^{q}} \sum_{\ell = 0}^{\infty}\expo{-\ell x}\,\dd x \\[3mm]&=k^{q - 1}\sum_{\ell = 0}^{\infty}\int_{0}^{\infty}x^{-q}\expo{-\pars{\ell + 1}x} \,\dd x =k^{q - 1}\sum_{\ell = 0}^{\infty}{1 \over \pars{\ell + 1}^{1 - q}} \int_{0}^{\infty}x^{-q}\expo{-x}\,\dd x \\[3mm]&= \color{#0000ff}{\large k^{q - 1}\zeta\pars{1 - q}\Gamma\pars{1 - q}} \end{align} The series converge when $1 - q > 1\quad\imp\quad \color{#0000ff}{\large q < 0 }$
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It isn't bounded for $q\ge 0$.
For $q<0$, it equals $$ \zeta(1-q)\Gamma(1-q)k^q, $$ I believe.
answered Oct 23, 2013 at 14:34
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$\begingroup$ $k^{q - 1}$ instead of $k^{q}$. $\endgroup$ Commented Jan 18, 2014 at 8:35
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Mathematica is perfectly capable of evaluating this:
ConditionalExpression[k^(-1 + q) Gamma[1 - q] PolyLog[1 - q, 1], q < 0 && k > 0]
Since the integrate has an $x^{-q - 1}$ singularity at the origin, the integral diverges for $q \geq 0.$
answered Oct 23, 2013 at 17:09