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how to get an analytical solution

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Darcy Chou
Darcy Chou on 26 Jun 2018
Commented: Ameer Hamza on 26 Jun 2018
syms tp t
NN=((exp(-1.45*10^(-4)*tp^1.274))^1.5)/2
z=4/3*pi*NN*(t-tp)^3
Z=int(z,tp,0,t)
output is
Z =
int((2*pi*exp(-(2674777890687885*tp^(637/500))/18446744073709551616)^(3/2)*(t - tp)^3)/3, tp, 0, t)
that is not an analytical solution, then how can I get the analytical solution?
If t=10 ,how to get the numerical solution?
Thanks for reading.
  1 Comment
Walter Roberson
Walter Roberson on 26 Jun 2018
You cannot expect analytic solutions when you use exponents that are floating point numbers, as calculus does not define integrals with floating point exponents.
You should not be asking for an analytic integral for any expression that involves any floating point constants at all. If you want to express uncertainty in value (floating point values need to be considered as uncertain) then replace the floating point numbers with rationals plus a symbolic uncertainty.

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Accepted Answer

Ameer Hamza
Ameer Hamza on 26 Jun 2018
The existence of an analytical solution is not always necessary. I cannot prove mathematically but it might be the case that this function does not have an analytical solution at all. It might also be the limitation of MATLAB symbolic engine. In the first case, there is not much you can do other than resorting to the numerical solution. In the second case, you may try some other symbolic math engine. As far as the numerical result is concerned, you can get it as with integral().
syms tp t
NN=((exp(-1.45*10^(-4)*tp^1.274))^1.5)/2;
z=4/3*pi*NN*(t-tp)^3;
Z=int(z,tp,0,t);
Z_function = matlabFunction(z);
Z_numerical = integral(@(tp) Z_function(10, tp), 0, 10);
  2 Comments

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More Answers (2)

KSSV
KSSV on 26 Jun 2018
syms tp t
NN=((exp(-1.45*10^(-4)*tp^1.274))^1.5)/2 ;
z=4/3*pi*NN*(t-tp)^3 ;
Z=int(z,0,t)
subs(Z,'t',10)
  4 Comments
Darcy Chou
Darcy Chou on 26 Jun 2018
Z=int(z,tp,0,t)
in this syntax, z is the integrand and tp is the variable of integration,
Z is a definite integral whose upper limit is the variable t.

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Walter Roberson
Walter Roberson on 26 Jun 2018
You have
NN=((exp(-1.45*10^(-4)*tp^1.274))^1.5)/2
Remember that exp(A*x)^B is the same as exp(A*B*x), so you can bring the 1.5 inside, forming
NN=exp(-1.5*1.45*10^(-4)*tp^1.274)/2
Now introduce a symbolic variable A and rewrite this as
syms tp nonnegative
syms A positive
NN = exp(-A*tp^sym(1.274)) / 2
for A = 1.5*1.45*10^(-4) but leave A as symbolic at the moment.
Then look at
z=4/3*pi*NN*(t-tp)^3
and see that this can be rewritten as
z = B * exp(-A*tp^sym(1.274)) * (t-tp)^3
for B = 4/3*pi/2
Now you want int(z, tp, 0, t)
which is int(B * exp(-A*tp^sym(1.274)) * (t-tp)^3, tp, 0, t)
which is B * int(exp(-A*tp^sym(1.274)) * (t-tp)^3, tp, 0, t)
And it turns out that MATLAB can calculate
temp = int(exp(-A*tp^sym(1.274)) * (t-tp)^3, tp, 0, t)
fairly quickly.
So the overall result is 4/3*pi/2 * subs(temp, A, 1.5*1.45*10^(-4))
  3 Comments
Ameer Hamza
Ameer Hamza on 26 Jun 2018
Wolfram Alpha does produce a solution for indefinite integral of the form
int(exp(A*tp^1.274)*(t-tp)^3, tp)
in term of gamma function. Although gamma function is itself defined in term of an integral for most cases.
I cannot find the required definite integral because its calculation requires a pro license. But I guess maple or Mathematica might be able to give a solution for the definite integral. Although I am not sure whether such a solution will be useful for any practical purpose.
Also from @Darcy comment, it appears that the result will later be used in a numerical optimization problem. Since you are eventually using a numerical optimization algorithm, it might be better to also use numerical integration techniques instead of searching for a closed form solution.

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