If the integral decomposition solution is not obtained, what integral method can be used to get the closest value?
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codes are as this,
if all the value except the r are 1, what integral method can be used to get the closest value?
syms d z r;
syms a;
syms r a z d;
syms a0 a1 a2 a3 a4 alpha0 alpha1 alpha2 alpha3 alpha4;
syms rho hgang;
syms phi;
sum1 =...
...
int(r*(((3*r^2)/4 - z^2)/(2*d) + (pi*alpha1*cos((pi*z)/d)*besseli(1, (pi*r)/d))/(d*besseli(0, (pi*a)/d)) + (2*pi*alpha2*cos((2*pi*z)/d)*besseli(1, (2*pi*r)/d))/(d*besseli(0, (2*pi*a)/d)) + (3*pi*alpha3*cos((3*pi*z)/d)*besseli(1, (3*pi*r)/d))/(d*besseli(0, (3*pi*a)/d)) + (4*pi*alpha4*cos((4*pi*z)/d)*besseli(1, (4*pi*r)/d))/(d*besseli(0, (4*pi*a)/d)))*(alpha0/2 - (- r^3/4 + r*z^2)/(2*d) + (alpha1*cos((pi*z)/d)*besseli(0, (pi*r)/d))/besseli(0, (pi*a)/d) + (alpha2*cos((2*pi*z)/d)*besseli(0, (2*pi*r)/d))/besseli(0, (2*pi*a)/d) + (alpha3*cos((3*pi*z)/d)*besseli(0, (3*pi*r)/d))/besseli(0, (3*pi*a)/d) + (alpha4*cos((4*pi*z)/d)*besseli(0, (4*pi*r)/d))/besseli(0, (4*pi*a)/d)), r, 0, a)
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采纳的回答
madhan ravi
2019-5-15
[d a z d a0 a1 a2 a3 a4 alpha0 alpha1 alpha2 alpha3 alpha4 rho hgang phi]=deal(1); % if all the value except the r are 1
syms r
integral(matlabFunction(r*(((3*r^2)/4 - z^2)/(2*d) +...
(pi*alpha1*cos((pi*z)/d)*besseli(1, (pi*r)/d))...
/(d*besseli(0, (pi*a)/d)) + ...
(2*pi*alpha2*cos((2*pi*z)/d)*besseli(1, (2*pi*r)/d))...
/(d*besseli(0, (2*pi*a)/d)) + (3*pi*alpha3*cos((3*pi*z)/d)...
*besseli(1, (3*pi*r)/d))/(d*besseli(0, (3*pi*a)/d)) +...
(4*pi*alpha4*cos((4*pi*z)/d)*besseli(1, (4*pi*r)/d))...
/(d*besseli(0, (4*pi*a)/d)))*(alpha0/2 - ...
(- r^3/4 + r*z^2)/(2*d) + (alpha1*cos((pi*z)/d)...
*besseli(0, (pi*r)/d))/besseli(0, (pi*a)/d)...
+ (alpha2*cos((2*pi*z)/d)*besseli(0, (2*pi*r)/d))...
/besseli(0, (2*pi*a)/d) + (alpha3*cos((3*pi*z)/d)...
*besseli(0, (3*pi*r)/d))/besseli(0, (3*pi*a)/d) +...
(alpha4*cos((4*pi*z)/d)*besseli(0, (4*pi*r)/d))...
/besseli(0, (4*pi*a)/d))),0, a,'ArrayValued',1)
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