Mathematica Asked by yarchik on February 23, 2021
In trying to reproduce results from one paper I stumbled upon a problem with definition of some elliptic integrals (this is my guess of what could be the problem).
I will first present in a simplified form what I am trying to calculate, details are in the original paper (PRL 99, 226801, see Google Scholar for PDF)
The goal is to compute the following 2D integral
$$I(k)=-iint_{Omega_c}frac{mathrm{d}^2vec q}{4pi q}left(1-costheta(vec k,vec {k}-vec {q})right),$$
where $theta(vec a,vec b)$ is the angle between $vec a$ and $vec b$, $q=|vec q|$. For those who would like to compare with the paper, this is essentially Eq.(2a), where for simplicity I set $e=kappa=1$, selected the case $s=1$ and substituted all the definitions into one equation.
The integration domain is $Omega_c: |k|le k_c$, where $k_c$ is a positive number.
The analytic result is known to be [cf. Eq. (3a)]:
$$I(k)=tfrac{1}{pi}k_cleft[h!left(k/k_cright)-f!left(k/k_cright)right],quad I(0)=-tfrac{1}{2}k_c.$$
Assuming, we want to know the result for $k<k_c$, there are following definitions
$$f(x)=E(x),quad h(x)=xleft[tfrac{pi}{4}log(4/x)-tfrac{pi}{8}right]
-xint_{0}^x!mathrm{d}y, y^{-3}left[K(y)-E(y)-tfrac{pi}{4}y^2right].$$
Here $K(x)$ and $E(x)$ are the complete elliptic integral
of the first and second kinds, respectively. I do not know, how this integral can be computed, neither by hand or with mathematica…
What is disturbing is that I was not able verify the integral numerically.
In the following, I will first rewrite all the equations in MA language.
i[1]=Integrate[EllipticK[y^2]-EllipticE[y^2],{y,0,1/x},
Assumptions->x>1]
i[2]=Integrate[(EllipticK[y^2]-EllipticE[y^2]-π/4 y^2)/y^3,{y,0,x},
Assumptions->0<=x<=1]
f[x_]=EllipticE[x^2]
h[x_]=x(π/4Log[4/x]-π/8)-x i[2]
Notice, it takes a while to compute i[2]
. Now, we will be interested in the $Delta I(k)=I(k)-I(0)$ function
xi[k_,kc_]:=kc/π(h[k/kc]-f[k/kc])
Δxi[k_,kc_]:=xi[k,kc]+kc/2
Now we define the numerical integral (adding a small cut-off a
) transforming it to polar coordinates and assuming $vec kparallel vec e_x$
Δni[k_?NumericQ,kc_?NumericQ,a_?NumericQ]:=1/(4 π) NIntegrate[((k- q Cos[θ])/Sqrt[k^2+q^2-2 k q Cos[θ]]),{q,a,kc},{θ,0,2π},PrecisionGoal->4]
and compare
dataI=Table[{k,Δni[k,30,10^-5]},{k,0.1,2,0.1}]
Plot[Δxi[k,30],{k,0,2},Epilog->{PointSize[Medium],Point[dataI]},PlotRange->{0,2.3}]
The points should exactly fall onto the analytic curve, but they are not… I would be happy with any of the answers:
Notice, I can easily verify Fig.1 of that paper with MA. But the integral considered here is not plotted there.
The article "Density Dependent Exchange Contribution to $partial mu/partial n$ and Compressibility in Graphene" by E.H. Hwang,Ben Yu-Kuang Hu, and S. Das Sarma has a typo in the definition of $h$ (there must be a plus before $frac {pi}{8}$). After correction, the results match (I wrote down the finished results for the integrals, so as not to waste time every time to calculate them)
(*i[1]=Integrate[EllipticK[y^2]-EllipticE[y^2],{y,0,1/x},Assumptions
[Rule]x>1]
i[2]=Integrate[(EllipticK[y^2]-EllipticE[y^2]-[Pi]/4
y^2)/y^3,{y,0,x},Assumptions[Rule]0[LessEqual]x[LessEqual]1]*)
i1[x_] := ([Pi] (-HypergeometricPFQ[{-(1/2), 1/2, 1/2}, {1, 3/2}, 1/
x^2] + HypergeometricPFQ[{1/2, 1/2, 1/2}, {1, 3/2}, 1/x^2]))/(
2 x)
i2[x_] :=
3/256 [Pi] x^2 (HypergeometricPFQ[{1, 1, 3/2, 5/2}, {2, 3, 3},
x^2] + 3 HypergeometricPFQ[{1, 1, 5/2, 5/2}, {2, 3, 3}, x^2])
f[x_] := If[x <= 1, EllipticE[x^2],
x EllipticE[1/x^2] - (x - 1/x) EllipticK[1/x^2]]
h[x_] := If[x <= 1, x ([Pi]/4 Log[4/x] + [Pi]/8) - x i2[x], x i1[x]]
xi[k_, kc_] := kc/[Pi] (h[k/kc] - f[k/kc])
[CapitalDelta]xi[k_, kc_] := xi[k, kc] + kc/2
[CapitalDelta]ni[k_?NumericQ, kc_?NumericQ, a_?NumericQ] :=
1/(4 [Pi]) NIntegrate[((k - q Cos[[Theta]])/
Sqrt[k^2 + q^2 - 2 k q Cos[[Theta]]]), {q, a, kc}, {[Theta], 0,
2 [Pi]}]
dataI = Table[{k, [CapitalDelta]ni[k, 30, 10^-10]}, {k, 0.1, 2, 0.1}]
Plot[[CapitalDelta]xi[k, 30], {k, 0, 2},
Epilog -> {PointSize[Medium], Point[dataI]}, PlotRange -> {0, 2.3}]
Correct answer by Alex Trounev on February 23, 2021
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