I am trying to verify the following asymptotic approximation as $x \rightarrow \infty$: $$\int^{1}_{0}t^{-\frac{1}{2}} \cos(t) e^{-xt^{\frac{1}{2}}} \, dt \sim \frac{2}{x}$$

This method is such that $$\int^\beta_\alpha g(t)e^{xh(t)} \, dt \sim g(a) \Big(\frac{-2\pi}{xh''(a)}\Big)^{\frac{1}{2}}e^{xh(a)}.$$

Where $t=a$ is a maximum of $h(t)$ for $\alpha <t<\beta$, note that also $g(a) \neq 0$.

Now I am advised to first let $t = \tau^2.$

So with this substitution our integral becomes : $$2 \int^1_0 \cos(\tau^2)e^{-x\tau} \, d\tau$$

For which I find

$$g(\tau) = \cos(\tau^2)$$

$$h(\tau)= -\tau,\, h'(\tau)= -1,\, h''(\tau)=0$$

But these values would lead to me dividing by zero so I think I misunderstood something... if any one can spot my misunderstanding it would be greatly appreciated!

  • 1
    $\begingroup$ Note that proper notation is $\cos(t)$ rather than $cos(t)$, and that is coded as \cos(t). That not only prevents italicization but also results in proper spacing in things like $a\cos b$ and $a\cos(b).$ In include both of those latter examples to show the context-dependent nature of the spacing: you see that there is less space to the right of $\cos$ in $a\cos(b)$ than in $a\cos b. \qquad$ $\endgroup$ – Michael Hardy Feb 12 '18 at 17:53
  • $\begingroup$ Thank you I will keep this in mind on future questions :) $\endgroup$ – Evan Feb 12 '18 at 17:58

You've missed the fact that the version of the Laplace method you've stated,

$$\int^\beta_\alpha g(t)e^{xh(t)} \, dt \sim g(a) \Big(\frac{-2\pi}{xh''(a)}\Big)^{\frac{1}{2}}e^{xh(a)},$$ where $t=a$ is a maximum of $h(t)$ for $\alpha <t<\beta$, note that also $g(a) \neq 0$

assumes that the maximum of $h(t)$ satisfies $\alpha < t < \beta$, i.e. it is neither at $t = \alpha$ nor at $t = \beta$.

But for the integral you ended up with,

$$2\int^1_0 \cos(\tau^2)e^{-x\tau} \, d\tau = 2\int_0^1 g(\tau) e^{x h(\tau)} \,d\tau,$$

the maximum of $h(\tau) = -\tau$ over the interval $[0,1]$ is located at $\tau=0$. So you can't apply the above version of the Laplace method to this integral.

In this case, you're better off using Watson's lemma. Or you could approach it from first principles—the idea in this case is the same as in the usual Laplace method: replace $g$ and $h$ by their approximations near the maximum ($h(\tau) \approx -\tau$ and $g(\tau) \approx 1$) and extend the domain of integration appropriately. With some justification you can prove that

$$ 2\int^1_0 \cos(\tau^2)e^{-x\tau} \, d\tau \sim 2\int_0^\infty e^{-x\tau} \, d\tau = \frac{2}{x}. $$


$$ \int^{1}_{0}t^{-\frac{1}{2}} \cos(t) e^{-xt^{\frac{1}{2}}} \, dt \sim \frac{2}{x}, $$

showing that the original asymptotic you were seeking to verify is incorrect.

Edit: The asymptotic in the question has been edited to agree with my conclusion.

  • $\begingroup$ I see! My lecturer asked for it to be done distinctly via Laplace's method however, would approaching from first principles still be count as "Laplace's Method" ? $\endgroup$ – Evan Feb 12 '18 at 19:16
  • $\begingroup$ Yes, sorry for being unclear, that's what I meant. The "Laplace method" really is a method guided by the simple principle that you try to approximate the integral by approximating the integrand near its largest point. There are several prepackaged results (like the one you gave) which people call "the Laplace method", but it's really much more flexible than that. $\endgroup$ – Antonio Vargas Feb 12 '18 at 19:52
  • $\begingroup$ @N.K The appendix of my thesis may be some help. $\endgroup$ – Antonio Vargas Feb 12 '18 at 21:59

This is a correct result. In fact, since $$ \begin{align} \int_x^\infty\tau^{4k}\,e^{-\tau}\,\mathrm{d}\tau &=\int_x^\infty\tau^{4k}\,e^{-\tau/2}\,e^{-\tau/2}\,\mathrm{d}\tau\\ &\le c_k\int_x^\infty e^{-\tau/2}\,\mathrm{d}\tau\\[3pt] &=2c_ke^{-x/2} \end{align}\\ $$ we can extend this asymptotic expansion $$ \begin{align} 2\int_0^1\cos\left(\tau^2\right)\,e^{-x\tau}\,\mathrm{d}\tau &=2\int_0^1\left(\sum_{k=0}^{n-1}\frac{\left(-\tau^4\right)^k}{(2k)!}+O\!\left(\tau^{4n}\right)\right)\,e^{-x\tau}\,\mathrm{d}\tau\\ &=\sum_{k=0}^{n-1}\frac{2(-1)^k}{(2k)!}\int_0^1\tau^{4k}\,e^{-x\tau}\,\mathrm{d}\tau+O\!\left(\frac1{x^{4n+1}}\right)\\ &=\sum_{k=0}^{n-1}\frac{2(-1)^k}{x^{4k+1}(2k)!}\int_0^x\tau^{4k}\,e^{-\tau}\,\mathrm{d}\tau+O\!\left(\frac1{x^{4n+1}}\right)\\ &=\sum_{k=0}^{n-1}\frac{2(-1)^k}{x^{4k+1}(2k)!}\left[(4k)!-\int_x^\infty\tau^{4k}\,e^{-\tau}\,\mathrm{d}\tau\right]+O\!\left(\frac1{x^{4n+1}}\right)\\ &=\sum_{k=0}^{n-1}\frac{2(-1)^k(4k)!}{x^{4k+1}(2k)!}+O\!\left(\frac1{x^{4n+1}}\right) \end{align} $$ For example, $$ 2\int_0^1\cos\left(\tau^2\right)\,e^{-x\tau}\,\mathrm{d}\tau \sim\frac2x-\frac{24}{x^5}+\frac{3360}{x^9}-\frac{1330560}{x^{13}} $$


The function $h(\tau)=-\tau$ has no interior maximum, so you cannot find an $a$ so that $h'(a)=0$. This means that with the functions you have chosen, the Laplace method is not useful.


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