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How to integrate an unbounded function?

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XIONG Qingxiang
XIONG Qingxiang on 12 Jul 2019
Commented: XIONG Qingxiang on 16 Jul 2019
Hello every one,
since the normal numerical integration cannnot work to an unbounded function, is there any other methods for the integration of unbounded function?.
Thank you.


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Accepted Answer

Jon on 12 Jul 2019
According to the MATLAB documentation MATLAB's integral function can handle singlularities on the boundary (upper or lower limit) of the domain to be integrated. If the singularity is not on the boundary, you could always break up the integral into the sum of two integrals, one with the singularity at the upper bound, the other with the singularity at the lower bound. If you have a finite number of singularities in your domain, you can extend this idea, and break up the overall integration into pieces each with singularities at their endpoints.
I am assuming the function you are integrating is one dimensional. MATLAB also has routines, integral2 and integral3 for two and three dimensional cases. I think these can also handle singularities at the boundaries, but you can check the documentation for details.


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XIONG Qingxiang
XIONG Qingxiang on 15 Jul 2019
%Here is the code,and I made some comments for you understanding. Thank you very much.
syms r
rmin = 1.4824*10^-9;
rmax = 1.6720*10^-4;
sigma = 0.31;
mu = -17.1043;
z = 10^6;
r0 = linspace(rmin,rmax,z);
P0 = lognpdf(r0,mu,sigma);
rc =3e-8;
beta = 3.52e-9;
epsmin = -2*beta/3;
nloop =3;
mystruct(nloop) = struct('Matphid',[],'MatPHID',[],'Mateps',[]);
k =1;
N = zeros(1,nloop);
H = (3/4)*((2*r+beta-2*rc)/beta - (1/3)*((2*r+beta-2*rc)/beta)^3) + 1/2;
dH = (3/2)*(1/beta - ((2*r+beta-2*rc)^2)/(beta^3));
M = (-3/4)*((2*r-beta-2*rc)/beta - (1/3)*((2*r-beta-2*rc)/beta)^3) + 1/2;
dM = (-3/2)*(1/beta - ((2*r-beta-2*rc)^2)/(beta^3));
P = (1/(sigma*sqrt(2*pi)))*(1/r)*exp(-((log(r) - mu)^2)/(2*sigma^2));
while (k <= nloop)
n = k;
N(k) = n;
eps = zeros(1,n+1);
F = zeros(1,n+1);
G = zeros(1,n+1);
phid = zeros(1,n+1);
S3 = zeros(1,n+1);
S4 = zeros(1,n+1);
phid(1) = phi0;
PHID = zeros(1,n);
rn = zeros(n+1,z);
rn(1,:) = r0;
Pn = zeros(n+1,z);
Pn(1,:) = P0;
PHid = linspace(0.219,217,n);
for i = 1:n
if i == 1
F =1/(1+eps(i)*dH);
G =1/(1+eps(i)*dM);
F = F/(1+eps(i)*dH);% (1+eps(i)*dH) is the derivative, it could be zero at two staionary points
G = G/(1+eps(i)*dM);% (1+eps(i)*dH) is the derivative.
f1 = P;
f2 = r*F*P;
f3 = r*G*P;
f4 = r*P;
f5 = H*F*P;
f6 = M*G*P;
Pd1 = P;
Pd2 = F*P;
Pd3 = M*P;
% J1 and J2 are the integrations
J1 = eval(feval(symengine,'numeric::int',f4,'r=lower3..upper3')) + ...
eval(feval(symengine,'numeric::int',f2,'r=lower1..upper1')) + ...
eval(feval(symengine,'numeric::int',f3,'r=lower5..upper5')) + ...
J2 = eval(feval(symengine,'numeric::int',f5,'r=lower1..upper1')) + ...
PHID(i) = phi0*((epsmin*J2 + J1)/r0ave)^2;
phid(i+1) = PHid(i);
eps(i+1) =(sqrt(phid(i+1)/phi0)-(1/r0ave)*J1)/((1/r0ave)*J2);
% S3 and S4 are the function value of two stationary points of function (r+eps(i)*H)
mystruct(k).Matphid = phid;
mystruct(k).Mateps = eps;
k = k + 1;
Bruno Luong
Bruno Luong on 15 Jul 2019
Not sure with those complicated formula (and a lot of them is not important for the discussion) but if I understand
P(r) = (1/(sigma*sqrt(2*pi)))*(1/r)*exp(-((log(r) - mu)^2)/(2*sigma^2))
which is equivalent to O(1/r) * O(1/r^(1/(2*sigma^2))) for r ~0.
The first term already induce non-defined integral. So there is no point trying to compute such thing.
Back to the blackboard.
PS: it looks like you are trying to integrate some EM potential layers.
XIONG Qingxiang
XIONG Qingxiang on 16 Jul 2019
In fact, it's a log-normal distribution and this code is for the propose of solving the parameter epsilon. What I'm working on is porous media. Anyway, thank you for your warming help and I will vote your response as the answer.

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More Answers (1)

Matt J
Matt J on 12 Jul 2019
You cantry use the symbolic toolbox, or you could just take the numerical integral over a sufficiently large finite interval. A numerical integration is an approximation anyway, so who cares about the extra epsilon of error that you get when you truncate the interval to something finite.

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XIONG Qingxiang
XIONG Qingxiang on 15 Jul 2019
Thank you for your idea. But could you please give me more details that how to use symbolic toolbox to solve this problem?

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