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

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Isma
Isma 2015-5-26
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

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