# 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?

-
Spoiler: Watson's Lemma –  Antonio Vargas Oct 27 '12 at 6:53

## 1 Answer

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.

-
I have no idea about what the expansion of the exponential presented in the question is supposed to achieve. –  Did Oct 27 '12 at 8:44
The integral is from $0$ to $1$, not $0$ to $\infty$. This means that $I_n\le\frac2{n(n+2)}$ not nearly $\frac12\Gamma(\frac12n)$. –  robjohn Oct 27 '12 at 13:12
@robjohn I fail to understand your comment (the upper bound of the integral being indeed $+\infty$) but it made me reexamine my solution and realize I had missed the asymptotic expansion aspect. Hence, in the end, thank you. –  Did Oct 27 '12 at 13:25
My comment was made 5 minutes before the edit that changed the upper limit of $I_n$ from $1$ to $\infty$. The subsequent changes have fixed the other issues I was about to comment on. –  robjohn Oct 27 '12 at 13:37
@robjohn Funny to call issues an (obvious) misprint. Whatever. –  Did Oct 27 '12 at 13:44