I encountered this function in Statistical Mechanics. $$f(x) =\int_{0}^{\infty}\frac{u^2}{1+\frac{e^{u^2}}{x}}du$$ For $x=0$, we define its value to be zero. I wanted to see it's asymptotic behavior in the limit x tending to $\infty$. Can we express the asymptotic behavior of this function, in the limit x tending to $\infty$, in terms of other known mathematical functions (if possible; in elementary functions)? For some physical reasons (which are irrelevant here), $x$ belongs to $[0,\infty)$.


In a first step, we perform the substitution $v=e^{u^2}$. we obtain $$ f(x) =\int_{0}^{\infty}\frac{u^2}{1+\frac{e^{u^2}}{x}}du = \int_1^\infty \frac{(\ln v)^{1/2}}{2 v(1+v/x)} dv \;.$$

Next, we take a look at the function $$ f_0(x) = \int_1^x \frac{(\ln v)^{1/2}}{2 v } dv =\frac13 (\ln x)^{3/2}$$ The intuition for this is that the integral for $f(x)$ is dominated for $v \lesssim x$. In this region the integral is approximately given by the expression above.

Indeed, we have $$f(x) - f_0(x) = -\int_1^x \frac{(\ln v)^{1/2}}{2(x+v)} dv+ \int_x^\infty \frac{(\ln v)^{1/2}}{2 v(1+v/x)} dv;$$ with the estimates $$\int_1^x \frac{(\ln v)^{1/2}}{2(x+v)} dv < \int_1^x \frac{(\ln x)^{1/2}}{2x} dv <\frac{(\ln x)^{1/2}}{2} $$ and $$\int_x^\infty \frac{(\ln v)^{1/2}}{2 v(1+v/x)} dv < \frac{x}{2}\int_x^\infty \frac{(\ln v)^{1/2}}{ v^2} dv = \frac12 (\ln x)^{1/2} + \frac{x}{4} \int_1^\infty \frac{1}{v^2(\ln v)^{1/2}} dv \\< \frac12 (\ln x)^{1/2} +\frac{x}{4 (\ln x)^{1/2}} \int_1^\infty \frac{1}{v^2} dv < \frac12 (\ln x)^{1/2} + \frac{1}{4 (\ln x)^{1/2}}\;. $$

With this, we have established that $$f(x) = \frac{1}{3} (\ln x)^{3/2} + O(\ln^{1/2} x).$$


1. (Not so illuminating) analytic expression. Assume for a moment that $0 < x < 1$. Then

\begin{align*} f(x) &= \int_{0}^{\infty} \frac{xu^2e^{-u^2}}{1 + xe^{-u^2}} \, du = \sum_{n=1}^{\infty} (-1)^{n-1} x^n \int_{0}^{\infty} u^2 e^{-nu^2} \, du \\ &= -\frac{\sqrt{\pi}}{4} \sum_{n=1}^{\infty} \frac{(-x)^n}{n^{3/2}} = -\frac{\sqrt{\pi}}{4} \operatorname{Li}_{3/2}(-x). \end{align*}

The last function is analytic outside $(-\infty, -1]$, and hence this identity extends to all of $x \geq 0$ by the principle of analytic continuation. But this is not so useful when investigating the asymptotic bahavior of $f(x)$.

2. Asymptotic expansion. Write $\alpha = \log x$ and make the substitution $u = \sqrt{\alpha(v+1)}$. Then

\begin{align*} f(x) &= \frac{\alpha^{3/2}}{2} \int_{-1}^{\infty} \frac{\sqrt{1+v}}{1 + e^{\alpha v}} \, dv \\ &= \frac{\alpha^{3/2}}{2} \left( \int_{0}^{1} \sqrt{1-v} \, dv - \int_{0}^{1} \frac{\sqrt{1-v}}{1 + e^{\alpha v}} \, dv + \int_{0}^{\infty} \frac{\sqrt{1+v}}{1 + e^{\alpha v}} \, dv \right). \end{align*}

This easily yields the following asymptotics

$$ f(x) = \frac{1}{3}(\log x)^{3/2} + \mathcal{O}\left( (\log x)^{1/2} \right). $$

For a better resolution, recall that the polylogarithm is defined as $\operatorname{Li}_s(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^s}$ for $|z| < 1$. Then

$$ \frac{1}{1 + e^{\alpha v}} = -\operatorname{Li}_0(-e^{-\alpha v}), \qquad \frac{d}{dv} \operatorname{Li}_{s+1}(-e^{-\alpha v}) = - \alpha \operatorname{Li}_s(-e^{-\alpha v}) $$

and hence

\begin{align*} \int_{0}^{\infty} \frac{\sqrt{1+v}}{1 + e^{\alpha v}} \, dv &= -\int_{0}^{\infty} (1+v)^{1/2} \operatorname{Li}_0(-e^{-\alpha v}) \, dv \\ &= -\frac{\operatorname{Li}_1(-1)}{\alpha} - \frac{1}{2\alpha} \int_{0}^{\infty} \frac{\operatorname{Li}_1(-e^{-\alpha v})}{(1+v)^{1/2}} \, dv \\ &= -\frac{\operatorname{Li}_1(-1)}{\alpha} - \frac{\operatorname{Li}_2(-1)}{2\alpha^2} + \frac{1}{4\alpha^2} \int_{0}^{\infty} \frac{\operatorname{Li}_2(-e^{-\alpha v})}{(1+v)^{3/2}} \, dv \end{align*}

and, in principle, the same argument can be applied to extract an asymptotic expansion up to any fixed order. Similarly,

\begin{align*} \int_{0}^{1} \frac{\sqrt{1-v}}{1 + e^{\alpha v}} \, dv &= -\int_{0}^{1} (1-v)^{1/2} \operatorname{Li}_0(-e^{-\alpha v}) \, dv \\ &= -\frac{\operatorname{Li}_1(-1)}{\alpha} + \frac{1}{2\alpha} \int_{0}^{1} \frac{\operatorname{Li}_1(-e^{-\alpha v})}{(1-v)^{1/2}} \, dv \\ &= -\frac{\operatorname{Li}_1(-1)}{\alpha} + \frac{\operatorname{Li}_2(-1) - \operatorname{Li}_2(e^{-\alpha})}{2\alpha^2} \\ &\qquad + \frac{1}{4\alpha^2} \int_{0}^{1} \frac{\operatorname{Li}_2(-e^{-\alpha v}) - \operatorname{Li}_2(-e^{-\alpha})}{(1-v)^{3/2}} \, dv \end{align*}

and so on. Using the results above, we obtain a better asymptotics

$$ f(x) = \frac{1}{3} (\log x)^{3/2} + \frac{\pi^2}{24} \frac{1}{(\log x)^{1/2}} + \mathcal{O}\left( \frac{1}{(\log x)^{3/2}} \right). $$

  • $\begingroup$ Indeed, the last term must be $\displaystyle\mathcal{O}\left(1 \over \left(\log x\right)^{\color{red}{5}/2}\right)$. $\endgroup$ – Felix Marin Apr 9 '18 at 22:57
  • $\begingroup$ @FelixMarin, Thank you for your answer. Indeed I was quite sure that there is a well-known asymptotic expansion of polylogarithms of fractional orders, but was lazy enough not to look for it and just to stick to chunky computations. $\endgroup$ – Sangchul Lee Apr 9 '18 at 23:01
  • $\begingroup$ It's always useful anyway. $\endgroup$ – Felix Marin Apr 9 '18 at 23:05

We have a linear upper bound: $$f=\int_0^\infty\frac{xu^2e^{-u^2}}{1+xe^{-u^2}}du\le x\int_0^\infty u^2e^{-u^2} du.$$But we also have a $O(1)$ lower bound: $$f\ge\frac{x}{1+xe^{-1}}\int_0^1u^2e^{-u^2}du=\frac{\int_0^1u^2e^{-u^2}du}{1/x+e^{-1}}.$$


$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{f}\pars{x} & \equiv \int_{0}^{\infty}{u^{2} \over 1 + \expo{u^{2}}/x}\,\dd u \,\,\,\stackrel{u^{2}\ \mapsto\ u}{=}\,\,\, {1 \over 2}\int_{0}^{\infty}{u^{1/2} \over \expo{u}/x + 1}\,\dd u = -\,{1 \over 2}\,\Gamma\pars{3 \over 2}\mrm{Li}_{3/2}\pars{-x} \\[5mm] & = -\,{1 \over 4}\,\root{\pi}\,\mrm{Li}_{3/2}\pars{-x}\label{1}\tag{1} \end{align}

See the Polylogarithm Integral Representation.

By using the Polylogarithm Asymptotic Expansion, \eqref{1} becomes ( $\ds{B_{n}}$ is a Bernoulli Number )

\begin{align} \mrm{f}\pars{x} & = -\,{1 \over 4}\,\root{\pi}\sum_{k = 0}^{\infty}\pars{-1}^{k}\pars{1 - 2^{1 - 2k}}\pars{2\pi}^{2k}\, {B_{2k} \over \pars{2k}!}\,{\ln^{3/2 - 2k}\pars{x} \over \Gamma\pars{5/2 - 2k}} \\[5mm] & = \bbx{{1 \over 3}\,\ln^{3/2}\pars{x} + {\pi^{2} \over 24}\,\ln^{-1/2}\pars{x} + {7\pi^{4} \over 1920}\,\ln^{-5/2}\pars{x} + \mrm{O}\pars{\ln^{-9/2}\pars{x}}\ \mbox{as}\ x\ \to\ \infty} \end{align}


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.