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The integral

$$I = \int_{-\infty}^\infty \frac{e^{-\varepsilon x^2}} { \sqrt{1+x^2} } dx$$

is convergent for $\varepsilon > 0$ and can even be given in terms of the Bessel function $K_0$. As $\varepsilon \to 0$ it is divergent and $I \sim -\log \varepsilon$. What would be the simplest way to derive the above leading term in an asymptotic $\varepsilon \to 0$ expansion directly in terms of the above integral?

Clearly, if one uses the exact result in terms of $K_0$ and then for instance uses the second order differential equation satisfied by $K_0$ it is quite simple to derive the $\log\varepsilon$ form of the divergence. But I'm looking for a way to derive this directly from the above integral. The reason is that I have a much more complicated integral to analyze which can not be given in a closed form but the above simple integral captures its difficulty so would like to understand this one first.

Another related question: the integral

$$J = \int_0^{2\pi} e^{-\sin(x)^2/\varepsilon^2}dx $$

can also be evaluated exactly in terms of the Bessel function $I_0$ which result will imply $J \sim \varepsilon$ as $\epsilon\to 0$. But again, directly from the integral what is the simplest way to see this leading term in the $\varepsilon\to 0$ expansion?

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up vote 2 down vote accepted

Differentiate $I(\varepsilon):=2\int_0^\infty e^{-\varepsilon x^2}(1+x^2)^{-1/2}dx$ with respect to $\varepsilon$ under the sign of integral; change variable putting $u:=\varepsilon x^2$. We get $$I'(\varepsilon) = -\frac{1}{\varepsilon} \int_0^\infty e^{-u}\sqrt{\frac{u}{u+\varepsilon}}du= -\frac{1}{\varepsilon}\big(1+o(1)\big)\, ,$$ by the dominated convergence theorem, and integrating $$I(\varepsilon)=-\log(\varepsilon)\big(1+o(1)\big),\,\, \mathrm{as }\, \varepsilon\to0\, .$$

For the latter integral, change variable with $x:=\varepsilon u $, so

$$J(\varepsilon):=2\int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}}e^{-{\sin(x)^2}/{\varepsilon^2}}dx=2\varepsilon\int_\mathbb{R} e^{-\sin(\varepsilon u)^2/\varepsilon^2}\chi_{ [-\frac{\pi}{2\varepsilon},\, +\frac{\pi}{2\varepsilon} ] }(u) du\, .$$ The integrand converges pointwise to $e^{-u^2}$ as $\varepsilon\to0$, and it is dominated by $e^{-u^2/4}$ for any $u\in\mathbb{R}$ (just because $\sin(x)\ge x/2\ge 0$ for any $0\le x \le \pi/2$ ). By the dominated convergence theorem,

$$J(\varepsilon)=2\varepsilon\int_\mathbb{R} e^{-u^2}du\big(1+o(1)\big)=2\sqrt{\pi}\varepsilon\big(1+o(1)\big),\, \, \mathrm{as }\, \varepsilon\to0\, .$$

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Thanks, I didn't see your answer before adding the second part of the question. For the first part it's definitely what I was looking for, thanks a lot! – Daniel Nov 15 '12 at 21:48

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