I want to understand the asymptotic behavior of an integral of the form

$$ I_f(\epsilon) = \int_0^1 \frac{\log(1/x)}{\sqrt{x}\sqrt{x+\epsilon}} f(x) dx $$ as $\epsilon \to 0^+$ for a generic function $f(x)$ which is smooth on $[0,1]$.

I suspect that the asymptotic behavior should be something like: $$ I_f(\epsilon) = c_1 \log^2(1/\epsilon)+c_2 \log(1/\epsilon) + c_3 + o(1) $$ where the $c_k$ are some $f$ dependent constants. Can someone show me how to compute the expansion?

Ideally, I'd like to compute the expansion up to terms which vanish as $\epsilon \to 0^+$. Any help would be greatly appreciated.

  • $\begingroup$ Those constants are surprisingly not very simple. With the help of Mathematica to evaluate the integral explicitly when $f(x) \equiv 1$ and then calculate the resulting asymptotic I get $$\begin{align} I_f(\epsilon) = &\left(\frac{1}{2}(\log \epsilon)^2 - (\log 4)(\log\epsilon) + \frac{\pi^2}{3} + 2(\log 2)^2\right)f(0) \\&\qquad + \int_0^1 \log(1/x) \frac{f(x)-f(0)}{x}\,dx + o(1). \end{align}$$ When I have some time I'll work on a way of showing this analytically if someone else hasn't yet. $\endgroup$ Nov 24, 2013 at 18:36

1 Answer 1


After looking at Antonio's comment, I thought a little bit more about how to find the expansion of my integral. I don't know if it's in poor taste to answer one's own question, but it was interesting enough, to me anyway, that I thought it was worth posting.

First, as Antonio points out it is sufficient to only consider $f(x)=1$. To expand $I_1(\epsilon)$ let $x=\epsilon e^{-t}$ and define $L:= \log(1/\epsilon)$, after the dust clears the integral becomes $$ I_1(\epsilon) = \int_{-L}^\infty \frac{ L+t }{\sqrt{1+e^{t}}} dt. $$
The singular behavior comes from the contribution to this integral near $t=-\infty$. We can subtract off the limiting divergent behavior to write \begin{align*} I_1(\epsilon) &= \frac{1}{2}L^2 + \int_{-L}^0 \frac{ L+t }{\sqrt{1+e^{t}}} - (L+t) dt + \int_0^\infty \frac{ L+t }{\sqrt{1+e^{t}}} dt \\ &= \frac{1}{2}L^2 + L \int_{-\infty}^{\infty} \left( \frac{1}{\sqrt{1+e^{t}}} - \chi_{\mathbb{R}_-}(t) \right) dt + \int_{-\infty}^\infty \left( \frac{t}{\sqrt{1+e^{t}}} - t \chi_{\mathbb{R}_-}(t) \right) dt \\ &\qquad- \int_{-\infty}^{-L} \frac{ L+t }{\sqrt{1+e^{t}}} - (L+t) dt. \end{align*} Here $\chi_{\mathbb{R}_-}(t)$ is the indicator function of the negative half line: $$ \chi_{\mathbb{R}_-}(t) = \begin{cases} 1 & x<0 \\ 0 & x>0. \end{cases} $$ Each of the integrals is convergent. The first two are $\epsilon$ independent so that the formula really is an asymptotic series, and they can be computed numerically or using Mathematica (I simplified further using dilogarithm identities which Mathematica didn't use). The last integral is vanishingly small as can be verified by expanding the denominator in powers of $e^t$ and integrating term by term. The result is \begin{align*} I_1(\epsilon) &= \frac{1}{2} \log(1/\epsilon)^2 + 2\log(2) \log(1/\epsilon) + \frac{\pi^2}{3} + 2 \log^2 (2) + \mathcal{O}({\epsilon \log \epsilon}) \\ &= \frac{1}{2} \log(4/\epsilon)^2 + \frac{\pi^2}{3} + \mathcal{O}({\epsilon \log \epsilon}). \end{align*}

For general $f(x)$ using Antonio's comment the expansion is given by $$ I_f(\epsilon) = f(0) \left[\frac{1}{2} \log(4/\epsilon)^2 + \frac{\pi^2}{3} \right] + \int_0^1 \log(1/x) \frac{ f(x) - f(0) }{x} dx + o(1). $$

  • $\begingroup$ +1, nice! And answering your own questions here is definitely encouraged, when possible. :) $\endgroup$ Nov 25, 2013 at 18:52

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