Complex Integration in MATLAB with symbolic integration limits

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Elliott Kitson
Elliott Kitson 2024 年 6 月 25 日
コメント済み: Elliott Kitson 2024 年 6 月 25 日
I'm having trouble with integrating this using MATLAB. When i try and run it as shown below, the answer that comes out i'm pretty sure is wrong. This is the first time i'm using matlab to help with integration so apologies in advance if i'm doing something very silly. Can anyone give me some pointers as to what i'm doing wrong?
syms y b n_1 n_2 R
b_y = (b-((b-(b/4))/(n_1+n_2))*y);
x = (y^2*b_y)/(1+(y/R));
I = int(x,-n_1,n_2)
I = 
  5 件のコメント
Elliott Kitson
Elliott Kitson 2024 年 6 月 25 日
What do you mean by "is -R in the interval of integration"? - Really sorry, i'm not familiar with that term. But eta1 and eta2 are both positive values and all parameters involved are real numbers. Appreciate your time helping! :)
Torsten
Torsten 2024 年 6 月 25 日
編集済み: Torsten 2024 年 6 月 25 日
What do you mean by "is -R in the interval of integration"?
If -R is in the interval [-eta1 eta2], the denominator of your integrand becomes 0 and your integrand +/- Infinity or NaN.
Similar to
syms x
f = 1/(1-x);
int(f,x,0.5,1.5)
ans = 
NaN

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Torsten
Torsten 2024 年 6 月 25 日
編集済み: Torsten 2024 年 6 月 25 日
Does this help ?
Note that no distiction is made about the R-value. But you should keep in mind that the answer is incorrect if -R is in the interval [-eta1 eta2].
syms y b n_1 n_2 R
b_y = (b-((b-(b/4))/(n_1+n_2))*y);
x = (y^2*b_y)/(1+(y/R));
I = int(x,y);
simplify(subs(I,y,n_2)-subs(I,y,-n_1))
ans = 
  3 件のコメント
Torsten
Torsten 2024 年 6 月 25 日
Does this then mean that the solution you provided below would be valid?
Yes.
Elliott Kitson
Elliott Kitson 2024 年 6 月 25 日
Thanks for the help! Massively appreciate it :)

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