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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?

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    Spoiler: [Watson's Lemma](http://en.wikipedia.org/wiki/Watson's_lemma)2012-10-27

1 Answers 1

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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.

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    I have no idea about what the expansion of the exponential presented in the question is supposed to achieve.2012-10-27
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    The integral is from $0$ to $1$, not $0$ to $\infty$. This means that $I_n\le\frac2{n(n+2)}$ not nearly $\frac12\Gamma(\frac12n)$.2012-10-27
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    @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.2012-10-27
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    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.2012-10-27
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    @robjohn Funny to call *issues* an (obvious) misprint. Whatever.2012-10-27
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    In the first edit, it was not an obvious misprint. The part of the integral above $t=1$ was not handled; $\log(1+x\sqrt{t})$ was being handled as a power series without error term and so the assumption was that the upper limit was not $\infty$. As I said, the subsequent changes fixed things up.2012-10-27
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    Regarding the convergence: I think it's safe to say the series doesn't converge for sufficiently large x, right? Now, is this what one would expect? My thinking: as x→∞ the $e^{(−xt)}$ factor gets very small, very quickly, and so the integral is dominated by the $ln(1+\sqrt{t})$ term. But $\int_{0}^{\infty} ln(1+\sqrt{t})dt$ diveges, so we expect the asymptotic series to diverge. I'd appreciate some feedback on this thought process. @did2012-10-27
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    As I said, the (complete) series $\sum\limits_n\Gamma(n/2)t^n$ never converges (except when $t=0$) hence the (complete) series $\sum\limits_n(-1)^{n-1}\Gamma(n/2)x^{-1-n/2}$ never converges. // Re your last comment, I find difficult to get what you mean: when $x\to+\infty$, $e^{-xt}\to0$ for every fixed $t\gt0$ and one sees that the whole integral goes to zero. The integral $\int\limits_0^{+\infty}\log(1+\sqrt{t})dt=+\infty$ is related to the limit $x\to0$ (not $x\to+\infty$) since $e^{-xt}\to1$ when $x\to0^+$ (and as a matter of fact $I(x)\to+\infty$ when $x\to0^+$). Apart from that...2012-10-27
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    Hmm, thanks for clarification. Let me rephrase my question: why should we not be surprised that the series does not converge for sufficiently large x? That is, is there an intuition behind expecting this? @did2012-10-27
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    If the series was convergent for $x$ large, the integral in the first displayed equation of my answer would be analytic in a neighborhood of $z=0$. But $z\mapsto\log(1+z\sqrt{t})$ has radius of convergence $1/\sqrt{t}$, which can be made as small as one wants, for large values of $t$. Hence one can expect no positive radius of convergence for the integral.2012-10-27