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?


1 Answer 1


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,

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

  • $\begingroup$ 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! $\endgroup$
    – Daniel
    Nov 15, 2012 at 21:48

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.