Laplace's Method to verify an asymptotic approximation Question:
Using Laplace’s method, verify the following asymptotic approximations as $x \to \infty$
$$\int_0^\infty t^x e^{-t} \ln t \, dt \sim \left(\frac{2\pi}{x}\right)^{1/2} e^{-x}$$
I am struggling to get this equation into the form of
$$\int_\alpha^\beta g(t)e^{xh(t)} \, dt$$
 A: The proposed asymptotic is very wrong. Where did it come from?
As a rule of thumb, if there is another exponential factor in the integral then it will somehow be involved in the desired $e^{xh(t)}$ factor. Your integral is
$$
\int_0^\infty t^x e^{-t}\ln t\,dt = \int_0^\infty e^{x\ln t} e^{-t}\ln t\,dt = \int_0^\infty \exp(x\ln t - t)\ln t\,dt.
$$
Now we need to balance $x\ln t$ with $-t$, and substituting $t=xs$ does so:
$$
\begin{align}
&\int_0^\infty \exp(x\ln t - t)\ln t\,dt \\
&\qquad= x\int_0^\infty \exp\Bigl[ x\ln(xs) - xs \Bigr] \ln (xs)\,ds \\
&\qquad= x^{x+1} \int_0^\infty \exp\Bigl[ x(\ln s - s) \Bigr] \ln (xs)\,ds \\
&\qquad= x^{x+1}\ln x \int_0^\infty \exp\Bigl[ x(\ln s - s) \Bigr]\,ds + x^{x+1} \int_0^\infty \exp\Bigl[ x(\ln s - s) \Bigr] \ln s\,ds \\
&\tag{$*$}
\end{align}
$$
The zero of $\ln s$ at the critical point $s=1$ in the second integral makes it asymptotically smaller than the first integral.  By the Laplace method we therefore have
$$
\begin{align}
&\int_0^\infty \exp\Bigl[ x(\ln s - s) \Bigr]\,ds \sim e^{-x} x^{-1/2}  \sqrt{2\pi}, \\
&\int_0^\infty \exp\Bigl[ x(\ln s - s) \Bigr] \ln s\,ds = o(x^{-1/2}).
\end{align}
$$
Thus, substituting the first into $(*)$ gives

$$
\int_0^\infty t^x e^{-t}\ln t\,dt \sim \left(\frac{x}{e}\right)^x \sqrt{2\pi x} \ln x, \qquad x \to \infty.
$$

Here's a plot of
$$
\color{blue}{\frac{1}{(x/e)^x\sqrt{x}} \int_0^\infty t^x e^{-t}\ln t\,dt}
$$
in blue versus
$$
\color{red}{\sqrt{2\pi} \ln x}
$$
in red:

A: Since $t = e^{\ln t}$ and hence $t^x = e^{x\ln t}$, we have:
$$
\int_0^\infty t^x e^{-t} \ln t \, dt = \int_0^\infty \underbrace{e^{-t} \ln t}_{g(t)} \  e^{x\ln t} \, dt
$$
