Double integration

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Kala
Kala il 20 Mar 2012
Commentato: Walter Roberson il 17 Giu 2023
How to perform double integration of exp(-ax-by)*(x^m)*(y^n)/(cx+dy)where x & y lies between 0 and infinity, a,b,m,n,c,d are positive real numbers.
Thank you.

Risposta accettata

Mike Hosea
Mike Hosea il 20 Mar 2012
If you have the 2012a release, just use INTEGRAL2:
>> a = 2; b = 3; m = 2; n = 3; c = 2; d = 3;
>> f = @(x,y)exp(-a*x-b*y).*(x.^m).*(y.^n)./(c*x+d*y)
f =
@(x,y)exp(-a*x-b*y).*(x.^m).*(y.^n)./(c*x+d*y)
>> integral2(f,0,inf,0,inf)
ans =
0.0030864
Be sure to use .*, .^, and ./ to do elementwise operations instead of matrix ops.
If you don't have 2012a yet, you can use the solution I presented here: http://www.mathworks.com/matlabcentral/answers/14514-double-integral-infinite-limits
  4 Commenti
Kala
Kala il 31 Mar 2012
Modificato: Walter Roberson il 17 Giu 2023
Yes sir.. It has worked. Thanks for your help. Now I have one more doubt.Here is my code..
part1=((xij)^(n1*k+alpha))*((yij)^(n2*r+beta))*k/(gamma(n2*r+beta)*gamma(n1*k+alpha));
I2=0;
for l=0:k-1
for m=0:r
part2=nchoosek(k-1,l)*nchoosek(r,m)*(-1)^(l+m);
I=mydblquad(@(v,u) fun2(v,u,xij, yij, n1, n2, k, r, alpha, beta,l,m),0,inf,0,inf);
I2=I2+part2*I;
end
end
Rpbcap=I2*part1;
%Baye's estimator for Series
part3=((xij)^(n1*k+alpha))*((yij)^(n2*r+beta))*r/(gamma(n2*r+beta)*gamma(n1*k+alpha));
I3=0;
for m=0:r-1
part4=nchoosek(r-1,m)*(-1)^(m);
I1=mydblquad(@(v,u) fun3(v,u,xij, yij, n1, n2, k, r, alpha, beta,m),0,inf,0,inf);
I3=I3+(part4*I1);
end
Rsbcap=I3*part3;
where
function fun2 = fun2(v,u,xij, yij, n1, n2, k, r, alpha, beta,l,m )
fun2=exp(-xij*u-yij*v).*u.^(n1*k+alpha).*v.^(n2*r+beta-1)./((l+1)*u+m*v);
end
and
function fun3 = fun3(v,u,xij, yij, n1, n2, k, r, alpha, beta, m)
fun3 = exp(-xij*u-yij*v).*u.^(n1*k+alpha-1).*v.^(n2*r+beta)./(k*u+(m+1)*v);
end
Here if I give n1 & n2 values >= 15 your mydblquad is showing NaN.. Is it out of range ? Please help..
Mike Hosea
Mike Hosea il 31 Mar 2012
I don't have all your input values to confirm it, but I think your integrand function is returning NaNs for some of the resulting input values. NaNs will be generated when you have overflow and try to calculate inf/inf or inf - inf.

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Più risposte (2)

Thomas
Thomas il 20 Mar 2012
  1 Commento
Mike Hosea
Mike Hosea il 20 Mar 2012
DBLQUAD won't handle infinite limits.

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Hussein Thary
Hussein Thary il 17 Giu 2023
integral
alfa=1;fai=2*pi;theata=2*pi;L=5;k=10;w=2e-2
w = 0.0200
f=@(r, fai)exp(-alfa*L./2).*exp(-j*k*r*theata*cos(fai)).*exp((-r^2./w^2)-(j*fai))
f = function_handle with value:
@(r,fai)exp(-alfa*L./2).*exp(-j*k*r*theata*cos(fai)).*exp((-r^2./w^2)-(j*fai))
s=integral2(f,0,100,0,2*pi)
Error using *
Incorrect dimensions for matrix multiplication. Check that the number of columns in the first matrix matches the number of rows in the second matrix. To operate on each element of the matrix individually, use TIMES (.*) for elementwise multiplication.

Error in solution>@(r,fai)exp(-alfa*L./2).*exp(-j*k*r*theata*cos(fai)).*exp((-r^2./w^2)-(j*fai)) (line 2)
f=@(r, fai)exp(-alfa*L./2).*exp(-j*k*r*theata*cos(fai)).*exp((-r^2./w^2)-(j*fai))

Error in integral2Calc>integral2t/tensor (line 237)
Z1 = FUN(X(VTSTIDX),Y(VTSTIDX)); NFE = NFE + 1;

Error in integral2Calc>integral2t (line 55)
[Qsub,esub] = tensor(thetaL,thetaR,phiB,phiT);

Error in integral2Calc (line 9)
[q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);

Error in integral2 (line 105)
Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);
  1 Commento
Walter Roberson
Walter Roberson il 17 Giu 2023
alfa=1;fai=2*pi;theata=2*pi;L=5;k=10;w=2e-2
w = 0.0200
f=@(r, fai)exp(-alfa.*L./2).*exp(-j.*k.*r.*theata.*cos(fai)).*exp((-r.^2./w.^2)-(j.*fai));
s=integral2(f,0,100,0,2*pi)
s = 0.0000 - 0.0027i

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