Applying Laplace Method for asymptotic approximations 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! 
 A: 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
$$
2\int^1_0 \cos(\tau^2)e^{-x\tau} \, d\tau \sim 2\int_0^\infty e^{-x\tau} \, d\tau = \frac{2}{x}.
$$
So
$$
\int^{1}_{0}t^{-\frac{1}{2}} \cos(t) e^{-xt^{\frac{1}{2}}} \, dt \sim \frac{2}{x},
$$
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.
A: 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,
$$
2\int_0^1\cos\left(\tau^2\right)\,e^{-x\tau}\,\mathrm{d}\tau
\sim\frac2x-\frac{24}{x^5}+\frac{3360}{x^9}-\frac{1330560}{x^{13}}
$$

Comments
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.
