Large $a$ asymptotics of $\int_0^{\pi/4} \exp(-a(x^2-\frac{x^4}{3}))$

Question

I'm looking for a way to prove that $\displaystyle \int_0^{\pi/4} \exp(-a(x^2-\frac{x^4}{3}))dx=\int_0^{\pi/4} \exp(-ax^2)dx+o\left(\int_0^{\pi/4} \exp(-ax^2)dx\right)$

as $a$ goes to $\infty$

I already know the substitution $u=ax^2$ yields $\displaystyle \int_0^{\pi/4} \exp(-ax^2)dx=\sqrt{\frac{\pi}{a}}+o_{a\to\infty}\left(\sqrt{\frac{\pi}{a}}\right)$, but a similar trick cannot be performed to tackle $\int_0^{\pi/4} \exp(-a(x^2-\frac{x^4}{3}))dx\\$.

I attempted to expand the exponential: $\exp(-a(x^2-\frac{x^4}{3}))=\exp(-ax^2)\exp(a\frac{x^4}{3})$, but $\exp(a\frac{x^4}{3})\to \infty$ quite fast.

Answer 1

You want to look for the maximum of $-a(x^2 - x^4/3)$; this is the point from which the main contribution of the integral will come as $a \to \infty$. Let $\phi(x) = x^2 - x^4/3$. We're now looking for the minimum of $\phi$. $$\phi^\prime(x) = 2x - 3x^3/2, \\ \phi^\prime(x) = 0 \iff x \in \{ 0, \pm 2/ \sqrt{3} \}.$$ The only two in the range are $0$ and $2/\sqrt{3}$. By knowledge of quartics, we know that the minimum will occur at one of $0$ or the end points, ie $0$ or $\pi/4$. $$\phi(0) = 0, \ \phi \left( \frac{1}{4}\pi \right) \simeq 0.5 > 0, $$ thus the minimum of $\phi$ is at $0$. Taylor (/MacLaurin) expand: $$\phi(x) - \phi(0) \simeq x \phi^\prime(0) + \frac{1}{2}x^2 \phi^\prime(0) = x^2,$$ for $0 < x < \epsilon$, for some suitably small (but not too small) $\epsilon$, since $\phi^\prime(x) = 2 - 9x/2,$ so $\phi^\prime(0) = 2$. So we now have that $$\int_0^{\pi/4} \exp(-a(x^2 - x^4/3)) dx \sim \int_0^\epsilon \exp(-ax^2) dx \sim \int_0^R \exp(-ax^2) dx $$ for any $R > \epsilon$, in particular for $R = \frac{1}{4}\pi$, since $\psi(x) = -x^2$ is a decreasing function in $x$, so the asymptotically dominant part of the integrand is $x=0$.

Hopefully this helps! =D

All the best, Sam :)

---

Note that you could replace the upper limit $\pi/4$ by $R$ for any $R$ such that $\phi(R) > 0$ - specifically, $\phi(r) > 0 \ \forall \ 0<r<R$, but we know that $0$ is a local minimiser of the (negative) quartic, so is a minimiser over $[0,R']$ where $R'>0$ and $\phi(R')=0$.

Also, $\phi(y) = 0$ with $y > 0$ if and only if $y = \sqrt{3}$, so you can replace $R'$ by $\sqrt{3}$ above, but the above is slightly more general. :)

Answer 2

I would recommend you to consult a textbook about asymptotic analysis (Erdelyi old ,but good for this problem, or Bleistein/Handelsman). For your problem the Laplace method is suited. It gives asymptotic approximations for integrals $$ \int_a^b h(x) \exp(\lambda f(x)) dx$$ as $\lambda \to\infty$. Depending on the position of the maxima of $f(x)$, in the interior or on the boundary you get different expansions.