I have the integral $$I(x)=\int_0^{\pi/2}\exp(-xt^3\cos t)dt$$ and I want to derive the first two terms in the asymptotic expansion for $x\rightarrow \infty$, which should give me $$\frac{1}{3x^{1/3}}\Gamma(1/3)+\left(\frac{1}{6}+\frac{8}{\pi^3} \right)\frac{1}{x}+\dots$$

Using the Laplace Method as stated here, I get only one term so my idea was to use partial integration, which gives me $$\pi/2-\int_0^{\pi/2} \left[ \exp(-xt^3\cos t)\,\, t\,\,(xt^3\sin(t)-3xt^2\cos(t)) \right] dt$$

  • $\begingroup$ Did you try Laplace's method again with your result? What did you get? $\endgroup$ Feb 11, 2014 at 19:16
  • $\begingroup$ The problem here is if you take a look at the term $t^3\cos(t)$, differentiate it and set it equal to zero, you get a lot of numerical values and zero itself, but zero doesnt fit the the recquirement of Laplace Methods $\endgroup$
    – Alexander
    Feb 11, 2014 at 19:22
  • $\begingroup$ Here a technique. $\endgroup$ Feb 11, 2014 at 19:27
  • $\begingroup$ For an explanation on how to obtain an arbitrary number of terms with mathematica check out this question on mathematica.SE $\endgroup$
    – glS
    May 6, 2015 at 7:46

1 Answer 1


Well, there are two contributions because the exponent is zero at $t=0$ and $t=\pi/2$. Let's consider $t=0$ first. In the immediate neighborhood of $t=0$ (we'll get to what that means in a bit), the exponent behaves as $x t^3$ so that as $x\to\infty$, we have

$$I(x) \sim \int_0^{\infty} dt \, e^{-x t^3} = \frac{\Gamma\left ( \frac{4}{3}\right )}{x^{1/3}} = \frac{\Gamma\left ( \frac{1}{3}\right )}{3 x^{1/3}} \quad (x\to\infty)$$

What is the size of the neighborhood? Well, we want $0 \lt x t^3 \lt \epsilon$ for some small $\epsilon$, so we have $0 \lt t \lt (\epsilon/x)^{1/3}$.

We also have a contribution in a neighborhood near $t=\pi/2$; we Taylor expand and get that

$$t^3 \cos{t} = -\frac{\pi^3}{8} \left ( t-\frac{\pi}{2}\right ) + O\left [ \left ( t-\frac{\pi}{2}\right )^2\right ]$$

Really, we are only interested in positive values of this expansion, as the exponent is positive through the integration region. If we look at only an immediate neighborhood near $t=\pi/2$, but with $t \lt \pi/2$, then we may approximate the contribution to the integral there as

$$\int_0^{\infty} dy \, e^{-\pi^3 x y/8} = \frac{8}{\pi^3 x}$$

It doesn't make sense to simply add these two terms together and declare them the leading behavior of $I(x)$ until we investigate the next leading behavior of the contribution at $x=0$. Note that the next contribution in the exponent is $x t^5/2$; within the interval of interest, this is $O(x^{-2/3})$, so we may Taylor expand this exponential term separately. The result is the following integral for the next contribution at $t=0$:

$$\frac12 x\int_0^{\infty} dt \;t^5 \, e^{-x t^3} = \frac1{6 x}$$

Note that this is $O(1/x)$ as is the leading contribution from $t=\pi/2$, so we may add these. The stated result follows.


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