Asymptotic expansion of the integral $\int_0^\infty e^{-xt} \ln(1+\sqrt{t}) dt$ for $x \to \infty$ Consider the following integral:
$$
\int_{0}^{\infty} e^{-xt} \ln(1+\sqrt{t})dt
$$
Calculate its asymptotic expansion to ALL orders as $x\rightarrow\infty$.
It seems the natural thing to do is expand the integrand as a Taylor series and integrate term-by-term. I've been given the hint that I can express difficult integrals in terms of the Gamma function. I also am required to discuss the convergence of the resulting series.
$$
e^{-xt}=\sum_{k=0}^{\infty}\frac{(-xt)^{k}}{k!}
$$
$$
\ln(1+\sqrt{t})=\sum_{k=1}^{\infty} (-1)^{k+1} \frac{t^{k/2}}{k}
$$
I understand how to use this to get the leading order behavior, but how to get the behavior at all orders?
 A: In the present case, there is a shortcut to Watson's lemma (mentioned in the comments), which is to scale properly the variable of integration of the integral $I(x)$ to be evaluated.
Let $x=1/z^2$ with $z\gt0$, hence $z\to0^+$. Using the change of variable $t\to z^2t$, one gets
$$
I(x)=z^2\int_0^{+\infty}\mathrm e^{-t}\log(1+z\sqrt{t})\,\mathrm dt.
$$
For every $N\geqslant0$, an expansion of $\log(1+s)$ up to order $s^N$ when $s\to0$ is
$$
\log(1+s)=\sum_{n=1}^N(-1)^{n-1}\frac1ns^n+o(s^{N}).
$$
This yields
$$
I(x)=z^2\sum_{n=1}^N(-1)^{n-1}I_nz^n+o(z^{N+2})=\sum_{n=1}^N(-1)^{n-1}I_nx^{-1-n/2}+o(x^{-1-N/2}),
$$
where, for every $n\geqslant1$,
$$
I_n=\frac1n\int_0^{+\infty}\mathrm e^{-t}t^{n/2}\,\mathrm dt=\frac1n\Gamma\left(\frac{n}2+1\right)=\frac12\Gamma\left(\frac{n}2\right).
$$
Finally, for every $N\geqslant0$,
$$
I(x)=\sum_{n=1}^N(-1)^{n-1}\frac12\Gamma\left(\frac{n}2\right)x^{-1-n/2}+o(x^{-1-N/2}).
$$
Note that the radius of convergence of the series $\sum\limits_n\Gamma\left(\frac{n}2\right)t^n$ being zero, the formulas above are indeed asymptotic expansions.
