Laplace's Method to verify an asymptotic approximation

Question

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

$$\int_\alpha^\beta g(t)e^{xh(t)} \, dt$$

Answer 1

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:

Answer 2

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