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I'm still having a little trouble applying Laplace's method to find the leading asymptotic behavior of an integral. Could someone help me understand this? How about with an example, like:

$$\int_0^{\infty} t^{3/4}e^{-x(t^2+2t^4)}dt$$ for $x>0$, as $x\rightarrow\infty$.

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The basic idea is that the maximum contribution of the integral comes from a neighborhood of $t=0$, and near there we have $t^2+2t^4 \approx t^2$. This problem is particularly nice because we can do everything explicitly. I'll do the calculation in two steps (two changes of variables) to illustrate what's going on.

Start with the change of variables $t^2+2t^4 = s^2$, where $s \geq 0$. This gives

$$ t=\frac{1}{2}\sqrt{-1+\sqrt{1+8s^2}}, $$

so that the integral becomes

$$ \int_0^{\infty} t^{3/4}e^{-x(t^2+2t^4)}\,dt = \int_0^\infty s^{3/4} f(s) e^{-xs^2}\,ds, $$

where

$$ f(s) = \frac{2^{1/4}s^{1/4}}{\sqrt{1+8s^2}\left(-1+\sqrt{1+8s^2}\right)^{1/8}} = 1 - \frac{15}{4}s^2 + \frac{713}{32}s^4 + \cdots. $$

Now we can make the second change of variables $s^2 = r$ to put the integral into a form where we can directly apply Watson's lemma. Indeed, this gives

$$ \int_0^\infty s^{3/4} f(s) e^{-xs^2}\,ds = \int_0^\infty r^{-1/8} g(r) e^{-xr}\,dr, $$

where

$$ g(r) = \frac{1}{2}f\left(\sqrt{r}\right) = \frac{1}{2} - \frac{15}{8}r + \frac{713}{64}r^2 + \cdots. $$

Finally

$$ \begin{align*} \int_0^\infty r^{-1/8} g(r) e^{-xr}\,dr &\approx \sum_{n=0}^{\infty} \frac{g^{(n)}(0) \Gamma(n+7/8)}{n! x^{n+7/8}} \\ &= \frac{1}{2}\Gamma\left(\frac{7}{8}\right) x^{-7/8} - \frac{15}{8}\Gamma\left(\frac{15}{8}\right)x^{-15/8} + O\left(x^{-23/8}\right) \end{align*} $$

as $x \to \infty$, by Watson's lemma.

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  • $\begingroup$ Can you elaborate on the advantages of splitting the change of variables in two steps? $\endgroup$ Mar 20 '18 at 22:05
  • $\begingroup$ @ThomasAhle there are no mathematical advantages. $\endgroup$ Mar 21 '18 at 0:40
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In summary, just look for the minimum of the function $2t^4+t^2$ which gives the location of the major contribution to the integral. The minimum of $2t^4+t^2$ is attained at the point $t=0$. So, we have

$$ \int_0^{\infty} t^{3/4}e^{-x(t^2+2t^4)}dt \sim \int_0^{\infty} t^{3/4}e^{-xt^2}dt = \frac{\Gamma \left( {\frac {7}{8}} \right)}{2 {x}^{ {\frac {7}{8}} }\, }.$$

To evaluate the last integral use the change of variables $ u=xt^2 $ to transform the integral to the gamma function.

Note: If instead of the function $t^2+2t^4$, you have $g(t)$ and attains its minimum at the point $0$, then just use the Taylor series to get the leading term.

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    $\begingroup$ In the first step, how did you drop the $t^2$ portion? $\endgroup$
    – Alex
    Jan 5 '13 at 16:23
  • $\begingroup$ @Alex: It is corrected. $\endgroup$ Jan 5 '13 at 22:12

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