Warning: Explicit integral could not be found.

1 view (last 30 days)
Hi Could someone please help with this integration:
syms teta h t z n;
f=exp((-n^2*pi^2*t)/h^2)*cos(0.8*n*pi)*cos(n*pi)*(1/2-z/h);
f1=symsum(f,-n,n);
q=exp(-(tan(teta)^2-(z+03*h)^2)/4*t)*(1+2*f1);
q1=int(q,z,-h/2,h/2);
q3=int((exp(-1/4*t)/t)*q1,t,1,10)
q2=(1/(2*h))*q3;
Warning: Explicit integral could not be found.
q3 =
int((pi^(1/2)*(erf((5*h*(-t)^(1/2))/4) - erf(7/4*h*(-t)^(1/2)))*(exp((2778046668940015*n^2*t)/(281474976710656*h^2)) + cos(pi*n)*cos((4*pi*n)/5) + 2*n*cos(pi*n)*cos((4*pi*n)/5)))/((-t)^(3/2)*exp(t/4)*exp((t*tan(teta)^2)/4)*exp((2778046668940015*n^2*t)/(281474976710656*h^2))), t = 1..10)

Accepted Answer

Walter Roberson
Walter Roberson on 20 Nov 2011
In your symsum, because you did not specify which variable to sum with respect to, symsum will use symvar to determine the summation variable. The rules for symvar will end up choosing z as the summation variable. As you later integrate with respect to z, it seems unlikely that you intended to get rid of the z through the symsum step.
I recommend that you specify the summation variable specifically; it would go as the second parameter, before the summation range.
  2 Comments
Rayanne
Rayanne on 20 Nov 2011
Thanks, I haven't notice this. But now I'm getting the warning since the first integral in function of z.
#syms teta h t z n;
#f=exp((-n^2*pi^2*t)/h^2)*cos(0.8*n*pi)*cos(n*pi)*(1/2-z/h);
#f1=symsum(f,n,-8,8);
#q=exp(-(tan(teta)^2-(z+03*h)^2)/4*t)*(1+2*f1);
#q1=int(q,z,-h/2,h/2);
#q3=int((exp(-1/4*t)/t)*q1,t,1,10)
#q2=(1/(2*h))*q3;
Walter Roberson
Walter Roberson on 21 Nov 2011
Please do not put those '#' characters in: I have to edit them out when I go to work with the code.
Please recheck the 'q' line. Why is there a 03*h ? Should that be 0.3 ?
Even with the above change, the integrals are proving difficult to work with. q1 is coming out with sqrt(-t) in multiple places, and as t is positive, that requires working over the complex plane.

Sign in to comment.

More Answers (0)

Community Treasure Hunt

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

Start Hunting!