Asymptotic expansion of an integral

Question

I came up with a simpler example which illustrates the technical difficulty I have encountered in my work.

Consider an integral depending on parameter $\epsilon$:

\begin{equation} \int\limits^{\infty}_{1 + \frac{2 \epsilon^{2}}{R^{2} - \epsilon^{2}}} \frac{1}{t^{2}-1} \frac{1}{\sqrt{R^{2} - \epsilon^{2}\left(t+1\right)/\left(t-1\right)\,}}\,{\rm d}t. \end{equation}

I am interested in the behaviour of this integral $\left(~\mbox{evaluated at}\ t = \infty~\right)$ as $\epsilon \to 0$.

This integral can be performed analytically, and it is straightforward to see that the end result diverges logarithmically as $\epsilon \to 0$. My question is: how to extract this logarithmic divergence without performing full integration ?. Naive expansion of the integrand in Taylor series in $\epsilon$ never leads to a logarithmic divergence.

In my real problem the integral cannot be evaluated analytically, but I am interested only in the logarithmic divergence. Is there a systematic way to extract it?

Thank you in advance!

Yegor

Answer 1

First make the change of variables $x = \epsilon^2\frac{t+1}{t-1}$ to get

$$ I(\epsilon) = \int_{\frac{R^2+\epsilon^2}{R^2-\epsilon^2}}^{\infty} \frac{dt}{(t^2-1)\sqrt{R^2-\epsilon^2 \frac{t+1}{t-1}}} = \frac{1}{2} \int_{\epsilon^2}^{R^2} \frac{dx}{x\sqrt{R^2-x}}. $$

Fix $0 < \delta < R^2$ and split the integral into the two pieces

$$ I(\epsilon) = I_1(\epsilon) + I_2 = \frac{1}{2} \int_{\epsilon^2}^{\delta} \frac{dx}{x\sqrt{R^2-x}} + \frac{1}{2} \int_{\delta}^{R^2} \frac{dx}{x\sqrt{R^2-x}}, $$

each of which is finite for $\epsilon > 0$. We see here that the singularity comes from $I_1(\epsilon)$, whose largest contribution comes from a neighborhood of $x = \epsilon^2 \approx 0$. Informally,

$$ I_1(\epsilon) = \frac{1}{2} \int_{\epsilon^2}^{\delta} \frac{dx}{x\sqrt{R^2-x}} \approx \frac{1}{2} \int_{\epsilon^2}^{\delta} \frac{dx}{x\sqrt{R^2}} \approx - \frac{\log(\epsilon^2)}{2\sqrt{R^2}}. $$

To make this precise, let's write

$$ \frac{1}{\sqrt{R^2 - x}} = \frac{1}{\sqrt{R^2 - x}} - \frac{1}{\sqrt{R^2}} + \frac{1}{\sqrt{R^2}}, $$

so that

$$ I_1(\epsilon) = \frac{1}{2\sqrt{R^2}} \int_{\epsilon^2}^{\delta} \frac{dx}{x} + \frac{1}{2} \int_{\epsilon^2}^{\delta} \frac{1}{x} \left(\frac{1}{\sqrt{R^2 - x}} - \frac{1}{\sqrt{R^2}}\right)dx. $$

The integrand in the integral on the right is bounded by a constant for all $x \in [0,\delta]$ (note that $\lim_{x\to 0}$ exists), and the integral on the left is

$$ \frac{\log(\delta/\epsilon^2)}{2\sqrt{R^2}} = -\frac{\log \epsilon}{|R|} + O(1). $$

Putting all this together, we conclude that

$$ \int_{\frac{R^2+\epsilon^2}{R^2-\epsilon^2}}^{\infty} \frac{dt}{(t^2-1)\sqrt{R^2-\epsilon^2 \frac{t+1}{t-1}}} = -\frac{\log \epsilon}{|R|} + O(1) $$ as $\epsilon \to 0$.

In fact we know what the constant in the $O(1)$ term is; it's just the finite terms we threw out along the way. Indeed,

$$ I(\epsilon) = -\frac{\log \epsilon}{|R|} + C + o(1), $$

where

$$ C = \frac{\log \delta}{2|R|} + \frac{1}{2} \int_{0}^{\delta} \frac{1}{x} \left(\frac{1}{\sqrt{R^2 - x}} - \frac{1}{\sqrt{R^2}}\right)dx + \frac{1}{2} \int_{\delta}^{R^2} \frac{dx}{x\sqrt{R^2-x}}. $$

This constant depends on $R$ but not on $\delta$, so we may take the limit as $\delta \to R^2$ to obtain

$$ \begin{align} C &= \frac{\log |R|}{|R|} + \frac{1}{2} \int_{0}^{R^2} \frac{1}{x} \left(\frac{1}{\sqrt{R^2 - x}} - \frac{1}{\sqrt{R^2}}\right)dx \\ &= \frac{\log(2|R|)}{|R|}. \end{align} $$

Thus

$$ \int_{\frac{R^2+\epsilon^2}{R^2-\epsilon^2}}^{\infty} \frac{dt}{(t^2-1)\sqrt{R^2-\epsilon^2 \frac{t+1}{t-1}}} = -\frac{\log \epsilon}{|R|} + \frac{\log(2|R|)}{|R|} + o(1) $$ as $\epsilon \to 0$.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow} \newcommand{\ol}[1]{\overline{#1}}$ $\ds{% {\cal J} \equiv\!\!\!\!\! \int\limits^{\infty}_{1 + \frac{2 \epsilon^{2}}{R^{2} - \epsilon^{2}}}\!\!\!\!\! {\dd t \over \pars{t^{2}-1}\sqrt{R^{2} - \epsilon^{2}\left(t+1\right)/\left(t-1\right)\,}} = {1 \over \verts{R}} \int\limits^{\infty}_{1 + \mu^{2} \over 1 - \mu^{2}}\!\!\! {\dd t \over \pars{t^{2}-1}\sqrt{1 - \mu^{2}\left(t+1\right)/\left(t-1\right)\,}} }$

where $\ds{\mu \equiv \epsilon/R}$. \begin{align} {\cal J} &= \int_{1 + \mu^{2} \over 1 - \mu^{2}}^{\infty} {\dd t \over \root{t - 1}\pars{t + 1}\root{\pars{1 - \mu^{2}}t - 1 - \mu^{2}}} \\[3mm]&= \int^{1 - \mu^{2} \over 1 + \mu^{2}}_{0} {\dd t/t^{2} \over \root{1/t - 1}\pars{1/t + 1}\root{\pars{1 - \mu^{2}}/t - 1 - \mu^{2}}} \\[3mm]&= \int^{1 - \mu^{2} \over 1 + \mu^{2}}_{0} {\dd t \over \root{1 - t}\pars{t + 1}\root{\pars{1 - \mu^{2}} - \pars{1 + \mu^{2}}t}} \\[3mm]&= -\,{1 \over 2}\int^{1 - \mu^{2} \over 1 + \mu^{2}}_{0}{\dd t \over t - 1} + \overbrace{\int^{1 - \mu^{2} \over 1 + \mu^{2}}_{0}\bracks{% {1 \over \root{1 - t}\pars{t + 1}\root{\pars{1 - \mu^{2}} - \pars{1 + \mu^{2}}t}} - {1 \over 2\pars{1 - t}}}\,\dd t}^{\ds{\equiv\ {\cal K}\pars{\mu}}} \\[3mm]&= -\,{1 \over 2}\,\ln\pars{1 - {1 - \mu^{2} \over 1 + \mu^{2}}} + {\cal K}\pars{\mu} = -\ln\pars{\verts{\epsilon \over R}} + {1 \over 2}\ln\pars{1 + \bracks{\epsilon \over R}^{2}} + {\cal K}\pars{\epsilon \over R} \end{align}

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$$\bbox[15px,border:1px dotted navy]{\ds{ {\cal J} \sim -\ln\pars{\verts{\epsilon \over R}} \quad\mbox{when}\quad \epsilon \gtrsim 0}} $$

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The 'next contribution' is finite and it's given by $$ {\cal K}\pars{0} = {1 \over 2}\int_{0}^{1}{\dd t \over t + 1} = {1 \over 2}\,\ln\pars{2} $$