I am doing simulation and its code is given. I am getting this kind of error

2 views (last 30 days)
a11=0.665272;
b11=0.395336;
c11=0.002901;
a21=6.546;
b21=-528.965;
c21=542.128;
syms u
eta1=a11.*(t.^b11).*exp(c11.*t);
beta1=a21.*(((1./t) +1).^b21).*exp(c21./t);
L1=3.2.*10.*(u.^(10-1)).*(exp((-3.2).*(u.^10)));
F1=(1-exp((-u./eta1).^beta1));
G=int(F1,L1,0,inf)
TYPE OF ERROR IS
Error using mupadmex
Error in MuPAD command: Can integrate only with respect to identifiers. [int]
Error in sym/int (line 124)
rSym = mupadmex('symobj::intdef',f.s,x.s,a.s,b.s,options);
Error in Untitled (line 13)
G=int(F1,L1,0,inf)

Answers (1)

Walter Roberson
Walter Roberson on 29 Jun 2015
You are asking int() to integrate with respect to a function, and that function does not even occur in the definition of the value being integrated. What is the intended meaning?
One possible intended meaning would be to do a coordinate transformation, to solve for u in terms of L1, substitute that u into your function to be integrated, and then integrate the result with respect to L1 from 0 to inf.
In order for that to work, each positive real L1 value would have to lead to a unique u value. Is that the case? Examine your L1. L1 when u = 0 is 0; L1 when u = 1 is a positive value; the exp(-u^10) is clearly going to approach 0 quickly as u gets larger so L1 approaches 0 as u approaches infinity. No oscillating functions, no singularities: we can see then that L1 must start at 0, climb to some finite peak, and reduce towards 0 again, that L1 values greater than that finite peak do not occur at all for real-valued u, and that all L1 values below that finite peak must occur for two different u. That peak is, by the way, at approximately 4.15. So from L1 = 0 to about 4.15 there are two corresponding u and for L1 from 4.15 to infinity there are no real solutions.
You could switch to complex integration to get L1 values larger than 4.15, but that just makes the problem of multiple solutions even worse, as there are up to 10 complex roots for L1=x with x real: how do you know which root to choose for the integration?
And this is why MATLAB does not support integration with respect to a function: it is almost always too ambiguous.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!