How can I plot integral function for fun3 ?

Question

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fun3 = fun1 - fun 2 , where;

fun1= @(x1) (i*w*x1./U)*(-1./r.^3)+(3*(x1*x1)./(r.^5));

fun2=fun2= @(x1) exp(i*w*x1./U).*(((k2.^2)*x1*x1./r.^3)*exp(-k2*r+k*x1)-(k2*k*x1*1./(r.^2))*exp(-k2*r+k*x1)-(k2*1./(r.^2))*exp(-k2*r+k*x1)...

+(2*k2*x1*x1./(r.^4))*exp(-k2*r+k*x1)-(k2*k*x1*1./(r.^2))*exp(-k2*r+k*x1)+((k.^2)*1./r)*exp(-k2*r+k*x1)...

-(k*1*x1./(r.^3))*exp(-k2*r+k*x1)-(k2*x1*x1./(r.^4))*exp(-k2*r+k*x1)+(k*x1*1./(r.^3))*exp(-k2*r+k*x1)...

+(1./(r.^3))*exp(-k2*r+k*x1)-(3*x1*x1./(r.^5))*exp(-k2*r+k*x1));

where: x2=0; ru=1; mu=1; w=0.01; U=1;

r = sqrt((x1).^2 + (x2).^2 + (x3).^2); k2 =sqrt((k.^2)+(h.^2)); h =sqrt(-i*ru*w./mu);

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Answer 1

Nadir Altinbas 2019-11-4

0 个投票

at the first stage you should define a boundry

laila elatrash 2019-11-4

you means interval values integral?

[a,b] = [-inf , x1]

is it what you means?