asymptotic behaviour of the integral without Laplace’s method

Question

I don't know asymptotic behaviour of the integral $$\int_{0}^{\infty}\frac{du}{\sqrt{4\pi u^{3}}}\left(1-\frac{e^{-\Omega u}}{\sqrt{\frac{1-\exp\left(-2u\right)}{2u}}}\right),$$ when I read a physics paper. It says that the integral have asymptotic behaviour $\log\left(\pi\Omega B\right)/\sqrt{2\pi}$, when $\Omega\to 0$. But most of paper about asymptotic behaviour are talking about the Laplace’s method, which is not match this integral. So I want to ask to get the asymptotic behaviour of the integral.

Thank you for your reading.

Answer 1

We may replace $\sqrt{\frac{2u}{1-e^{-2u}}}$ with $\sqrt{2u+1}$ and compute the resulting integral in terms of the Tricomi $U$ function $$ \sqrt{2}\cdot U\left(-\frac{1}{2},1,\frac{\Omega}{2}\right)\approx \frac{1}{\sqrt{2\pi}}\,\log\left(\Omega\cdot\frac{1}{2}e^{2\gamma+\psi\left(-\frac{1}{2}\right)}\right)$$ whose (logarithmic) asymptotic behavior as $\Omega\to 0^+$ is known by Kummer's differential equation.

Answer 2

Please view this as a supplement to @Jack's Answer which avoids the use of special functions and adds another constant contribution which might explain the differences between numerics and former asymptotic calculations.

To make the analysis simpler, let us rescale $\Omega u=x$ to obtain (we drop the $4\pi$ troughout and set $\Omega>0$)

$$ I(\Omega)=\sqrt{\Omega}\int_0^{\infty}\frac{1}{x^{3/2}}\left[1-\sqrt{\frac{2 x}{\Omega}}\frac{e^{-x}}{\sqrt{1-e^{-2x/\Omega}}}\right] $$

the only lengthscale in this problem is determined by $x_1=\Omega/2$ so it is reasonable to splint the range of integration at $x_1$. We call the resulting Integrals $I_<$ and $I_>$.

Starting with the first, we may replace the exponents by their Taylor approximants. We get

$$ I_<(\Omega)\sim \sqrt{2} \int_0^{\Omega/2}\sum_{n\geq 0}c_n\frac{x^{n-1/2}}{\Omega^{n+1/2}}+\mathcal{O}\left(\sqrt{\Omega}\right) $$

with fastly decaying coefficents $c_n$ ($c_0=1/2,c_1=1/24,c_2=1/48,c_3=1/640$...). Here and throughout we mean by $\sim$ asymptotically equal in the limit $\Omega \rightarrow 0_+$.

Obviously upon integration we get a constant contribution (C is defined through the series above)

$$ \color{\green}{I_<=C+\mathcal{O}\left(\sqrt{\Omega}\right)} $$

Now turning to $I_>$ we may expand ind powers of $e^{-2x/\Omega}$, yielding

$$ I_>(\Omega)\sim \sqrt{\Omega}\int_{\Omega/2}^{\infty}\left[\frac{1}{x^{3/2}}-\sqrt{\frac{2 }{\Omega}}\frac{e^{-x}}{x}\left(1+\mathcal{O}\left(e^{-2 x/\Omega}\right)\right)\right] $$

the first part of this integral gives a constant contribution namely $2\sqrt{2}$. Furthermore it holds that

$$ -\int_{\Omega/2}^{\infty}\frac{e^{-x}}{x}\sim\log(\Omega/2)+\gamma+\mathcal{O}(\Omega) $$

which can be shown by integrating the Taylor series of $e^{-x}/x$ and using the representation $\gamma = \int_0^{\infty}dx\left[\frac{1}{e^x x}-\frac{1}{e^x -1}\right]$ of the Euler constant (Note that this series not converge in a classical sense, it is a truely asymptotic power series). Therefore

$$ \color{\red}{I_>\sim \sqrt{2}\log(\Omega)+\sqrt{2}( \gamma-\log(2)+2)+\mathcal{O}(\Omega)} $$

and last but not least

$$ I(\Omega)=\color{\red}{I_>}+\color{\green}{I_<}\sim \sqrt{2} \log( \Omega)+C'+\mathcal{O}(\sqrt{\Omega}) $$

with $C'=C+\sqrt{2}( \gamma-\log(2)+2)$. Note that this seems to match the other answer expect that we included another constant contribution from the region of integration around zero

from which we can obtain an value for your constant $B$. Since $C\approx 1/2$, $e^C\approx 1.6$ which seems to close the gap a bit between Jack's answer and numerical predictions

PS: Because Mathematica is not avaiable at the moment, i can't crosscheck my results properly so please forgive me some minor errors!