How to solve NaN values of integrand defined over a finite interval?

1 view (last 30 days)
Hi, could someone know why quadv () returns NaN (see %r1), although my defined function is well behaved on [-theta; pi/2]?
see below the code:
f=@(k) k .* exp(-k);
x=1
al=1;
bet=0.5;
if al==1
c=exp(-(pi * x) /2 * bet);
c2=1 /(2*abs(bet));
theta0=pi/2
zet=0
v1=@(theta) 2/pi .* ((pi/2 + bet.*theta) ./cos(theta)) .* exp(1/bet .* (pi/2 + bet.*theta)...
.* tan(theta));
g=@(theta) c .* v1(theta);
%p=c2 * quadv(@(theta) f(g(theta)),-theta0,pi/2);
y= @(theta) c2 .* f(g(theta)) ; % values of integrand
%r1= quadv(@(theta) y(theta),-theta0,pi/2)
% check behavior of integrand
n= @(theta) f(g(theta));
a=[-theta0:0.01:pi/2]';
N=n(a);
P=[a;N];
fprintf('%6s %12s\n','x','f(x)');
fprintf('%6.2f %12.8f\n',P);
plot(a,P)
else
alm=al-1;
d=1/alm;
da=al .*d;
theta0=(1/al) .*atan(bet .*tan(pi .*al/2));
u=al .*theta0;
zet=-bet .*tan(pi*al/2)
v=@(theta) (cos(u) .^d) .*((cos(theta) ./sin(al .*(theta0+theta))) .^da) .*...
(cos(u+alm .*theta) ./cos(theta));
c=(x-zet) .^da;
c2=al ./(pi .*abs(al-1) .* (x-zet));
g=@(theta) c .* v(theta);
y= @(theta) f(g(theta)) ; % values of integrand
%y= @(theta) c2 .* f(g(theta)) ;
r2= c2 .* quadv(@(theta) y(theta),-theta0,pi/2)
end
% plot values
%a=[-theta0:0.01:pi/2]';
%Y=y(a);
%U=[a;Y];
%fprintf('%6s %12s\n','x','f(x)');
%fprintf('%6.2f %12.8f\n',U);
%plot(a,Y)
Also for curiosity I am wondering why from the screen x goes beyond the maximum limit pi/2. actually matlab outputs lots of zeros there.
Thanks

Answers (0)

Categories

Find more on MATLAB in Help Center and File Exchange

Community Treasure Hunt

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

Start Hunting!