Cannot perform integral for a function with two symbols
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Hi everyone, I try to perform integral for a vector (vector t) which each element contains two symbols. The integral operation is respected to z, hence I expect to see only gamma after the integral operation is done. However, the results still have z and gamma. I suspect that I did not define the vector t correctly. Can someone please help out with this. The code is below:
clc, clear all;
lamda1 = 1;
lamda2 = 1;
lamda3 = 1;
lamda4 = 1;
d1 = 1;
d2 = 1;
d3 = 1;
d4 = 1;
m = 2.7;
n = 1;
P = 1;
R = 3;
sigma_sr = 0.01;
sigma_rd = 0.01;
gamma0 = 2^R - 1;
alpha = 0:0.01:1;
a = (4*(n^2)*(P^2))*alpha.^2;
b = 2*n*P*alpha.*(1-alpha)*sigma_sr*(d1^m)*(d3^m);
c = 2*n*P*alpha.*(1-alpha)*sigma_rd*(d2^m)*(d4^m);
d = ((1-alpha).^2)*sigma_sr*sigma_rd*(d1^m)*(d3^m)*(d2^m)*(d4^m);
syms z;
syms gamma;
arg = (gamma*c)./a;
u = sqrt(z/(lamda2*lamda4));
t = ((4*(gamma*b*z + gamma*d)).^(0.5)).* ((a*z - gamma*c)*lamda1*lamda3).^(-0.5);
test = int(t(1,3),z,arg(1,3),Inf)
The result is:
test =
int((z/625 - (112986307451471*gamma)/288230376151711744)^(1/2)*((7086501203356261*gamma)/18446744073709551616 + (112986307451471*gamma*z)/72057594037927936)^(1/2), z, (70616442157169375*gamma)/288230376151711744, Inf)
We can see that z is still in the expression.
Thank you for any help
댓글 수: 2
madhan ravi
2019년 1월 20일
편집: madhan ravi
2019년 1월 20일
Do you have by any chance the range of gamma values ? so that it can converted to numerical integral to be evaluated ? int() is not able to evaluate the integral.
Tommy Huynh
2019년 1월 20일
답변 (1개)
Walter Roberson
2019년 1월 20일
0 개 추천
MATLAB is not able to find a closed form solution for the integral .That is common: it is not difficult to create functions with no known closed form integrand.
댓글 수: 5
Tommy Huynh
2019년 1월 20일
Walter Roberson
2019년 1월 21일
Are there available range restrictions on gamma, or c, or a ? Can we assume that any of them are real-valued? Can we assume that any of them are nonnegative? (This matters because there is a singularity at some negative values.)
Walter Roberson
2019년 1월 21일
My testing using a more powerful package suggests that the integral is either infinite or undefined, depending on the relationships between some of the parameters. In particular plain infinity if I add the assumption that all of the parameters are non-negative.
Tommy Huynh
2019년 1월 21일
Walter Roberson
2019년 1월 22일
Your integral is the product of two square roots.
The second value whose square root is being taken is non-negative under the same condition as the lower bound on z, and the value increases infinitely as z increasing to infinity.
The first value whose square root is being taken is nonnegative for nonnegative z, and increases to infinity as z increases to infinity.
So you are multiplying two terms both of which go to infinity as z goes to infinity. But somehow you are expecting a finite result.
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