$\DeclareMathOperator{\erfc}{erfc} \DeclareMathOperator{\Ei}{Ei} $ What is the series expansion of $f$ for small $q$? \begin{align} U(q) &= q e^{q^2}\erfc q\\ I(q,q') &= \int_0^{2\pi} \frac{d\phi}{2\pi} U(\sqrt{q^2 + q'^2 -2qq'\cos\phi})-U(q')\\ f(q) &= \int_{0}^{\infty} dq'\,I(q,q') \end{align} Even better, perhaps you can integrate this analytically to find $f$ [or $I(q,q')]$? When I numerically integrate, the results are not grid dependent for small $q$: i.e. I get consistent results with double the gridpoints and/or double the grid maximum, so it seems to be well-behaved (and the result looks approximately quadratic in $q$). Here is my attempt to find the series expansion of $f$ by differentiating under both integrals, then integrating over $\phi$ analytically: \begin{align} D_n(q') &\equiv \left[\frac{\partial ^{n} I(q,q')}{\partial q^n}\right]_{q=0}\\ D_0(q') &= 0\\ D_1(q') &= \int_0^{2\pi} \frac{d\phi}{2\pi} \left[ \frac{U'(\sqrt{q^2 + q'^2 -2qq'\cos\phi})}{2\sqrt{q^2 + q'^2 -2qq'\cos\phi}} (2q - 2q'\cos\phi) \right]_{q=0}\\ &= -U'(q') \int_0^{2\pi} \frac{d\phi}{2\pi} \cos\phi = 0\\ D_2(q') &= \int_0^{2\pi} \frac{d\phi}{2\pi} \left[\sin^2\phi \frac{U'(q')}{q'} + \cos^2\phi \, U''(q')\right] \\ &= \frac{U'(q')}{2q'} + \frac{U''(q')}2 \\&= \left( 2q'^3+4q'+\frac1{2q'} \right)e^{q'^2}\erfc q' -\frac{3+2q'^2}{\sqrt\pi} \\ D_3(q') &= 0\\ f(q) &= \frac{q^2}2\int_{0}^{\infty} dq'\,D_2(q') + O(q^4) \end{align} $D_2(q') = \frac12 q'^{-1} + O(1)$ so the last integral diverges logarithmically for small $q'$. Also it seems that $D_{2n}(q') = O(q'^{1-2n})$ so the higher terms get worse.

I don't know why the Taylor series looks like a sum of divergent terms: perhaps the series expansion of $f$ is in non-integer powers of $q$ rather than a Taylor series?

In case it is helpful here is some Mathematica:

U[q_] := q Exp[q^2] Erfc[q]
Dint[qd_, n_] :=  Simplify[Integrate[(D[U[Sqrt[q^2 + qd^2 - 2 q qd Cos[phi]]] -U[qd], {q, n}]) /. q -> 0, {phi, 0, 2 Pi}]/(2 Pi), qd >= 0]

so that Dint[qd, 2] gives the above expression for $D_2(q')$. Integrate[Dint[qd,2], qd] gives an expression in terms of $_{2}F_2$ and that expression is divergent for $q'\to0$.

EDIT: If I try to evaluate $I_>$ from joriki's answer, I note that $q$ is small and $x$ is large, so $qx$ can be anything, but $\epsilon=q\left(\sqrt{1+x^2-2x\cos\phi}-x\right) = q[-\cos\phi + \sin^2\phi/2x + O(x^{-2})]$ is small, so \begin{align} I_> &= q\int_1^\infty dx \int_0^{2\pi} \frac{d\phi}{2\pi} U(qx +\epsilon) - U(qx) \\ &= \sum_{n=1}^\infty \frac{q^{n+1}}{n!}\int_1^\infty dx \, U^{(n)}(qx) \int_0^{2\pi} \frac{d\phi}{2\pi} \left(\sqrt{1+x^2-2x\cos\phi}-x\right)^n \\ &= \frac{q^2}{2}\int_1^\infty dx \, U'(qx) \int_0^{2\pi} \frac{d\phi}{2\pi} \left(\sqrt{1+x^2-2x\cos\phi}-x\right)\\ &+ \frac{q^3}{6}\int_1^\infty dx \, U''(qx) \int_0^{2\pi} \frac{d\phi}{2\pi} \left(\sqrt{x^2+1-2x\cos\phi}-x\right)^2 + O(q^4)\\ &= \frac{q^2}{2}\int_1^\infty dx \, U'(qx) f(x) + \frac{q^3}{6}\int_1^\infty dx \, U''(qx) [1-2x f(x)] + O(q^4)\\ f(x) &= \frac{(x+1)E[4x/(x+1)^2] + (x-1)E[-4x/(x-1)^2]}{\pi} - x = \frac{1}{4x} + O(x^{-3})\\ \end{align} so, since $U'(q) = -\frac{2 q}{\sqrt\pi} + (1+2 q^2)e^{q^2} \erfc(q)$, the first $x$ integral diverges logarithmically and the next one is worse. Expanding in $\epsilon$ gives an expansion in $q$ which is divergent. Perhaps someone has an idea of how to proceed to get an expansion in $1/q$ for small $q$.

  • $\begingroup$ Why is the $\phi$ dependence in $D_1(q')$ proportional to $\cos\phi$? Shouldn't it include the reciprocal of the square root? $\endgroup$ – joriki Apr 21 '16 at 23:19
  • $\begingroup$ Thanks for looking at my question. $D_1(q')$ is evaluated at $q=0$. When $q=0$ the square root evaluates to $q'$. I'll edit the question to make this more clear. $\endgroup$ – Ramashalanka Apr 21 '16 at 23:30
  • $\begingroup$ Sorry, it was clear enough already before the edit :-) $\endgroup$ – joriki Apr 22 '16 at 4:08
  • $\begingroup$ I added more detail for $D_2(q')$. Any guesses at what is going on would be appreciated. Perhaps the above proves it is not well behaved? $\endgroup$ – Ramashalanka Apr 27 '16 at 5:44
  • 1
    $\begingroup$ I've been thinking about this on and off over the last week. It's a tough problem; I think I have a solution, but I'm pretty busy these days and it'll probably take a few days before I can finish and post it. $\endgroup$ – joriki Apr 27 '16 at 6:07

A simple start (which you may have already done):

Since $\erfc(q) \sim \dfrac{e^{-q^2}}{q\sqrt{\pi}} $ for large $q$, $U(q) = q e^{q^2}\erfc q \sim q e^{q^2}\dfrac{e^{-q^2}}{q\sqrt{\pi}} = \dfrac{1}{\sqrt{\pi}} $ so that $U(q)$ is constant for large $q$.

For small $q$, $\erfc(q) = 1-\dfrac{2q}{\sqrt{\pi}}+O(q^3) $ so that for small $q$,

$\begin{array}\\ U(q) &= q e^{q^2}\erfc q\\ &\sim q e^{q^2}(1-\dfrac{2q}{\sqrt{\pi}}+O(q^3))\\ &\sim q (1+q^2+O(q^4))(1-\dfrac{2q}{\sqrt{\pi}}+O(q^3))\\ &\sim q (1-\dfrac{2q}{\sqrt{\pi}}+q^2+O(q^3))\\ \end{array} $

By seeing the range of arguments to $U$ in the integral, and possibly using more terms of the asymptotic expansion of $\erfc$, you might get some useful approximations.


Here's my answer so far; I'll try to complete it when I have the time, hopefully in a couple of days.

The problem is basically that you have difficult behaviour both for $q'\ll q$ and for $q'\gg q$. For $q'\ll q$, the integrand near $\phi=0$ behaves like $|q'-q|$, which isn't differentiable at $q'=q$. This problem could be avoided by expanding $U$ into a power series at $0$ instead of expanding in a power series in $q$, but this runs into the problem for $q'\gg q$: Here $U$ tends to a constant, but this isn't visible in a series expansion at $0$.

The solution, I believe, is to split the integral into two parts with $q'\lessgtr q$, much as one does in the Laplace expansion, in which the same argument occurs. So

\begin{align} f(q) &= \int_0^\infty\mathrm dq'I(q,q') \\ &= \int_0^\infty\mathrm dq'\int_0^{2\pi} \frac{\mathrm d\phi}{2\pi}\left(U\left(\sqrt{q^2 + q'^2 -2qq'\cos\phi}\right)-U\left(q'\right)\right) \\ &= q\int_0^\infty\mathrm dx\int_0^{2\pi} \frac{\mathrm d\phi}{2\pi}\left(U\left(q\sqrt{1 + x^2 -2x\cos\phi}\right)-U\left(qx\right)\right) \\ &= \overbrace{q\int_0^1\mathrm dx\int_0^{2\pi} \frac{\mathrm d\phi}{2\pi}\left(U\left(q\sqrt{1 + x^2 -2x\cos\phi}\right)-U\left(qx\right)\right)}^{I_\lt} \\ &+ \underbrace{q\int_1^\infty\mathrm dx\int_0^{2\pi} \frac{\mathrm d\phi}{2\pi}\left(U\left(q\sqrt{1 + x^2 -2x\cos\phi}\right)-U\left(qx\right)\right)}_{I_\gt}\;. \end{align}

For $I_\lt$, we can expand $U(t)=t\mathrm e^{t^2}\erfc(t)=t-\frac2{\sqrt\pi}t^2+t^3+O\left(t^4\right)$ and integrate term by term:

\begin{align} q^2\int_0^1\mathrm dx\int_0^{2\pi} \frac{\mathrm d\phi}{2\pi}\left(\sqrt{1 + x^2 -2x\cos\phi}\,-x\right)&\approx0.587336q^2 \end{align}

(Wolfram|Alpha quadrature),

\begin{align} q^3\int_0^1\mathrm dx\int_0^{2\pi} \frac{\mathrm d\phi}{2\pi}\left(\left(1 + x^2 -2x\cos\phi\right)-x^2\right)&=q^3\;, \end{align}

and so on, with the terms corresponding directly to powers of $q$.

I haven't solved the $I_\gt$ part yet, but I suspect that it has a non-analytic behaviour like $\exp\left(-\frac1{q^2}\right)$ at $q=0$, so it or a part of it needs to be expanded in $\frac1q$, so the overall result might be a sum of a series in $q$ and a series in $\frac1q$.

  • $\begingroup$ Thanks joriki. I've had a stab at $I_>$ (edit appended to my question), but I can't see how to proceed. Any more clues? $\endgroup$ – Ramashalanka Jul 12 '16 at 0:46
  • $\begingroup$ Also, how does an $\exp(-1/q^2)$ term fit with $I(0,q')=0 \implies f(0)=0$? $\endgroup$ – Ramashalanka Aug 20 '16 at 21:01

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