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In doing a computation regarding the electroweak phase transition in the early universe, I came across the following integral: $$I=\int_0 ^ {\infty} x^2 \text{ln}(1-\text{exp}(-\sqrt{x^2+u^2}))\text{d}x. $$

I am interested in the behaviour of this parametric integral near $u=0$. Specifically, my reference asks me to show that

$$I = -\frac{\pi^4}{45} + \frac{\pi^2}{12}u^2 - \frac{\pi}{6}u^3 - \frac{1}{32}\bigg(\text{ln}(u^2)-(\frac{3}{2}+2 \text{ln}(4\pi)-2\gamma)\bigg) u^4+\mathcal{O}(u^6) $$

This expression is rather mysterious to me. I'm pretty sure I can obtain the first two terms by putting $u=0$ and differentiating with respect $u^2$, respectively, and using some values for $\zeta(n)$. The cubic term and the logarithm seem to be beyond my abilities.

Any help is appreciated.

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  • $\begingroup$ Please reread my question. I am not interested in the value of this integral at $u=1$. $\endgroup$ – Thomas Bakx May 11 at 9:49
  • $\begingroup$ @ThomasBakx: Are you sure that the coefficient of $u^4$ contains a $\ln u^2$? I would have expected it to be a constant. I am also surprised to see a $u^3$ term, given that $I$ depends on $u^2$, therefore is an even function. $\endgroup$ – Alex M. May 13 at 10:12
  • $\begingroup$ @AlexM. I correctly copied the result from my reference. There is however no background information on how to obtain this series. My lecturer (who supposedly made the exercise himself) stated that the integral is not an analytic function of $u$, and there is some 'branch cut' for the logarithm, whatever that means. For this reason, the series should not be called a Taylor series but a so called 'trans series', hence the title of the question. As I already mentioned in my post, I have no idea how there could ever be a $u^3$ in this expansion. $\endgroup$ – Thomas Bakx May 13 at 10:29
  • $\begingroup$ I am far from this topic, but a few years ago I also had a problem related with integral evaluation. From an answer by Antonio Vargas I have learned about Laplace method. This may be (or may be not) useful to you. $\endgroup$ – Alex Ravsky May 14 at 19:45
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$\color{brown}{\textbf{Solution.}}$

Subsitution $$x=u\sinh y\tag1$$ changes the issue integral to the form of $$I= u^3\int\limits_0^\infty\,\,\ln\left(1-e^{\Large^{\LARGE-u\cosh y}}\right)\sinh^2y\cosh y\,dy.\tag2$$

Integration by parts gives $$I= \dfrac13 u^3\int\limits_0^\infty\,\ln\left(1-e^{\Large^{\LARGE-u\cosh y}}\right)\,d\sinh^3y$$ $$ = \dfrac13 u^3 \ln\left(1-e^{\Large^{\LARGE-u\cosh y}}\right)\,\sinh^3y\bigg|_0^\infty\, -\dfrac13 u^4\int\limits_0^\infty\,\dfrac{\sinh^4y\,e^{\Large^{\LARGE-u\cosh y}}} {1-e^{\Large^{\LARGE-u\cosh y}}}\,dy$$ $$I=-\dfrac13 u^4\int\limits_0^\infty\,\dfrac{\sinh^4y}{e^{\Large^{\LARGE u\cosh y}}-1}\,dy.\tag3$$

On the other hand,

$$I=-\dfrac13 u^4\sum\limits_{k=1}^\infty I_k,\tag4$$ where $$I_k = \dfrac13 u^4\int\limits_0^\infty\,\sinh^4y\,e^{\Large^{\LARGE-ku\cosh y}}\,dy.\tag{5a}$$ If $u\to \infty,$ then the most part of the integral is situated in the area $y >> 1,$ where $$\sinh y \approx \cosh y.$$

So $$I_k \approx \dfrac13 u^4\int\limits_0^\infty\,\sinh^3y\cos y\,e^{\Large^{\LARGE-ku\cosh y}}\,dy\tag{5b},$$ $$I_k = -\dfrac{e^{\Large^{\LARGE-ku\cosh y}}}{12k^4} (24 + 2 k^2u^2 - ku (-24 + k^2 u^2) \cosh(y) + 6 u^2 \cosh(2 y) + k^3u^3 \cosh(3 y))\bigg|_0^\infty$$ $$ = \dfrac2{3k^4} e^{-ku} (k^2u^2 + 3 k u + 3),$$ and summation via polylogarithm changes $(3)$ to the expression $$I\approx -\dfrac23 u^2\operatorname{Li}_2\left(e^{-u}\right) - 2u\operatorname{Li}_3\left(e^{-u}\right) -2 \operatorname{Li}_4\left(e^{-u}\right)\tag6$$ (see also Wolfram Alpha summation), where $\operatorname{Li}$ is polylogarithm.

Similarly from $(3)$

$$I=-\dfrac13 u^4\sum\limits_{k=1}^\infty \dfrac{J_k}k,\tag7$$ where $$I_k = u^3\int\limits_0^\infty\,\sinh^2y\cosh y\,e^{\Large^{\LARGE-ku\cosh y}}\,dy.\tag{8a}$$

If $u\to \infty,$ then more accurate approximation can be used in the form of $$\sinh y\cosh y = \cosh^2y\sqrt{1-\cosh^{-2}y}\approx \cosh^2y\left(1-\dfrac12\cosh^{-2}y\right) = \cosh^2y-\dfrac12.$$

So $$J_k \approx \dfrac12u^3\int\limits_0^\infty\,(2\cosh^2 y -1)\sinh y \,e^{\Large^{\LARGE-ku\cosh y}}\,dy,\tag{8b}$$ $$J_k = \dfrac{k^2u^2 cos(2y)+4ku\cosh(y)+4}{2k^3} e^{\Large^{\LARGE-ku\cosh y}}\bigg|_0^\infty = \dfrac{(ku+2)^2}{2k^3} e^{-ku},$$ and summation via polylogarithm changes $(3)$ to the expression $$I\approx -\dfrac12 u^2 \operatorname{Li}_2\left(e^{-u}\right) - 2 u \operatorname{Li}_3\left(e^{-u}\right) - 2 \operatorname{Li}_4\left(e^{-u}\right).\tag9$$ (see also Wolfram Alpha summation), where $\operatorname{Li}$ is polylogarithm.

$\color{brown}{\textbf{Testing.}}$

\begin{vmatrix} \text{Formulas} & \text{OP (issue)} & (4)+(5b) & (7)+(8b) & (6) & (9)\\ \text{Range} & (0,150) & (0,12) &(0,12) & - & - \\ u=0.1 & -2.15691\,41055 & -2.15691\,41055 & -2.15691\,41055& -2.15938\,14465 & -2.15719\,44641\\ u=0.01 & -2.16456\,47397 & -2.16456\,47397 & -2.17457\,47397 & -2.16459\,18581 & -2.16456\,53771\\ \end{vmatrix}

  1. Formula from the OP reference $\color{brown}{\textbf{has the error about 10x}}$ and looks wrong.
  2. Approximation of $\sinh y$ in the integrals $(5b),(8b)$ works very accurate.
  3. Both of the closed expressions $(6),(9)$ give a good accuracy.
  4. The accuracy of the closed expressions improves when $u\to 0.$
  5. Formula $(9)$ $$\color{green}{\boxed{\phantom{\Big|}\mathbf{I\approx -\dfrac12 u^2 \operatorname{Li}_2\left(e^{-u}\right) - 2 u \operatorname{Li}_3\left(e^{-u}\right) - 2 \operatorname{Li}_4\left(e^{-u}\right)}}\tag9}$$ gives the better approximation.
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Partial answer. Regarding $I$ as a function of $u$, we note that

\begin{align*} I(0) &= \int_{0}^{\infty} x^2 \log(1-e^{-x})\, \mathrm{d}x = -\frac{\pi^4}{45}, \\ I'(0) &= 0, \\ I''(0) &= \int_{0}^{\infty} \frac{x}{e^x - 1} \, \mathrm{d}x = \frac{\pi^2}{6}. \end{align*}

To get the higher terms, write

\begin{align*} I(u) &= I(0^+) + \frac{I''(0)}{2}u^2 + \int_{0}^{\infty} \bigg( \underbrace{ x^2 \log\left(1 - e^{-\sqrt{x^2+u^2}}\right) - x^2 \log\left(1 - e^{-x}\right) - \frac{u^2 x}{2(e^x - 1)} }_{=\text{(*)}} \bigg) \, \mathrm{d}x. \end{align*}

Here, the integrand $\text{(*)}$ can be recast as

\begin{align*} \text{(*)} &= \int_{0}^{u} \left[ \frac{\partial}{\partial s} x^2 \log\left(1 - e^{-\sqrt{x^2+u^2}}\right) - \frac{sx}{(e^x - 1)} \right]\, \mathrm{d}s \\ &= \int_{0}^{u} \bigg[ \frac{sx^2}{\sqrt{x^2+s^2}(e^{\sqrt{x^2+s^2}}-1)} - \frac{sx}{(e^x - 1)} \bigg]\, \mathrm{d}s \\ &= \int_{0}^{u} \int_{0}^{s} s x^2 \frac{\partial}{\partial t} \bigg( \frac{1}{\sqrt{x^2+t^2}(e^{\sqrt{x^2+t^2}}-1)} \bigg) \, \mathrm{d}t \mathrm{d}s \\ &= -\int_{0}^{u} \int_{0}^{s} \frac{s x^2 t}{(t^2 + x^2)^2} f\left(\sqrt{x^2+t^2}\right) \, \mathrm{d}t\mathrm{d}s, \end{align*}

where $f(a) = \frac{a(e^a(1+a) - 1)}{(e^a - 1)^2}$. For the future use, we remark that $a = 0$ is a removable singularity of $f$ with $f(0) = 2$ and $2 \geq f(a) \geq f(b) \geq 0$ for all $0 \leq a \leq b$. Interchanging the order of integration,

\begin{align*} \text{(*)} &= -\int_{0}^{u} \frac{x^2 t (u^2-t^2)}{2(t^2 + x^2)^2} f\left(\sqrt{x^2+t^2}\right) \, \mathrm{d}t. \end{align*}

Plugging this back and applying the substitution $(x, t) \mapsto (utx, ut)$,

\begin{align*} I(u) &= I(0^+) + \frac{I''(0)}{2}u^2 - u^3 \int_{0}^{\infty} \int_{0}^{1} \frac{x^2 (1-t^2)}{2(1 + x^2)^2} f\left(u t \sqrt{x^2+1}\right) \, \mathrm{d}t\mathrm{d}x. \end{align*}

As $u \to 0^+$, monotone convergence tells that

\begin{align*} &\int_{0}^{\infty} \int_{0}^{1} \frac{x^2 (1-t^2)}{2(1 + x^2)^2} f\left(u t \sqrt{x^2+1}\right) \, \mathrm{d}t\mathrm{d}x \\ &\xrightarrow[u \to 0^+]{} \int_{0}^{\infty} \int_{0}^{1} \frac{x^2 (1-t^2)}{(1 + x^2)^2} \, \mathrm{d}t\mathrm{d}x = \frac{\pi}{6}. \end{align*}

So it follows that

$$ I(u) = -\frac{\pi^4}{45} + \frac{\pi^2}{12}u^2 - \frac{\pi}{6}u^3 + o(u^3). $$


Numerical analysis. If $0 < x < \frac{1}{2}$, then $\left| \log (1 - x) \right| = -\log(1-x) \leq \frac{x}{1-x} \leq 2x $. Using this, we note that, for $R > 1$,

$$ \left|\int_{R}^{\infty} x^2 \log\left( 1 - e^{-\sqrt{x^2 + u^2}} \right) \, \mathrm{d}x\right| \leq \int_{R}^{\infty} 2x^2 e^{-x} \, \mathrm{d}x = 2(R^2 + 2R + 2) e^{-R}. $$

For instance, this bound is $< 10^{-30}$ for $R \geq 80$, and so, we may compute

$$ \int_{0}^{\infty} x^2 \log\left( 1 - e^{-\sqrt{x^2 + u^2}} \right) \, \mathrm{d}x = \int_{0}^{80} x^2 \log\left( 1 - e^{-\sqrt{x^2 + u^2}} \right) \, \mathrm{d}x \pm 10^{-30}$$

uniformly in $u > 0$. Since the integrand is continuous on $[0, 80]$, we may use any CAS to easily estimate the truncated integral with a precision of $10^{-30}$. Then for the quantities

\begin{align*} \text{(A)} &= \text{numerical integration of } \int_{0}^{80} x^2 \log\left( 1 - e^{-\sqrt{x^2 + u^2}} \right) \, \mathrm{d}x, \\ \text{(B)} &= \text{OP's asymtotic form} \\ &= -\frac{\pi^4}{45} + \frac{\pi^2}{12}u^2 - \frac{\pi}{6}u^3 + \frac{\frac{3}{2} + 2\log(4\pi) - 2\gamma - 2\log u}{32} u^4, \\ \text{(C)} &= \text{@Yuri Negometyanov's asymptotic form} \\ &= -\frac{u^2}{2}\operatorname{Li}_2(e^{-u}) -2u \operatorname{Li}_3(e^{-u}) - 2\operatorname{Li}_4(e^{-u}), \end{align*}

we obtain the following comparisons of numerical values:

comparison of numerical values

Coinciding leading decimals are highlighted as red.

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I think that you can do what you need, if you carefully use the inequality (due to Yves Coudene (2018) A Strange Inequality Concerning Alternating Series, The American Mathematical Monthly, 125:6, 554-557.): $$ \ln(1+x)\le x - {x^2\over 2}+{x^3\over 3}+x^4 \left ( \ln(2)-1+{1\over2}-{1\over3} \right). $$

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Thanks everyone for all the response. For those interested, I found a derivation in the following paper: http://www.physics.princeton.edu/~steinh/ph564/Dolanjackiw.pdf. Contrary to what Yuri claims, the expression appears to be correct (see eq. 3.13c, 3.16 for the form of the potential and appendix B for the derivation).

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  • 1
    $\begingroup$ Ah, the combination of regularization and partial fraction decomposition. That's really a cool trick to read. Thank you for sharing this (and for the +50 for my incomplete answer)! $\endgroup$ – Sangchul Lee May 20 at 19:44

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