Problem of calculating the value of integrals
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Qian Feng
le 27 Août 2017
Modifié(e) : Qian Feng
le 22 Sep 2017
Hi there, I encounter the problem of calculating the value of integrals, the code is given as follows:
r1 = 2; r2 = 4; r3 = r2 - r1;
d = 5;
syms y x real
tri = [];
for i = 1 : d
tri = [tri ; sin(20*i*x)];
end
for i = 1 : d
tri = [tri ; cos(20*i*x)];
end
tri = [1; tri];
trig = [];
for i = 1 : d
trig = [trig ; sin(20*i*x)];
end
for i = 1 : d
trig = [trig ; cos(20*i*x)];
end
trig = [1; trig];
cl1 = -log(abs(2*sin(10*x))); cl2 = int(-log(abs(2*sin(0.5*y))), 0, 20*x);
fi1 = [cl1; cl2]; fi2 = fi1;
ny1 = size(fi1,1); ny2 = size(fi2,1); ny = ny1 + ny2;
Ga1 = sym([ 0 zeros(1,d) 1./sym(1:d); 0 1./(sym(1:d)).^2 zeros(1,d)]); Ga2 = Ga1;
ep1 = fi1 - Ga1*tri; ep2 = fi2 - Ga2*trig;
epp1 = matlabFunction(ep1*ep1'); epp2 = matlabFunction(ep2*ep2');
E1 = integral(epp1,-r1,0,'ArrayValued',true); E2 = integral(epp2,-r2,-r1,'ArrayValued',true);
In my code cl1 and cl2 are the Clausen functions which can be decomposed as series of trigonometric functions. The problem is that there are Inf values in the calculation results of E1 and E2, which should not be there at all.
E1 =
0.1135 -Inf
-Inf Inf
>> E2
E2 =
0.0501 -Inf
-Inf Inf
The function cl2 itself is expressed in terms of a integration, so I am not sure if any complication will be generated by the structure of cl2 here.
Could someone help me to figure out which step I got wrong here ? Thank you !
3 commentaires
Star Strider
le 27 Août 2017
One problem is that this term:
cl2 = int(-log(abs(2*sin(0.5*y))), 0, 20*x);
will evaluate to +Inf at 0.
Réponse acceptée
Star Strider
le 27 Août 2017
One problem is that this term:
cl2 = int(-log(abs(2*sin(0.5*y))), 0, 20*x);
will evaluate to +Inf at 0.
I usually use ‘sqrt(eps)’ or ‘1E-8’ for zero when zero is not an option. The squared value will usually retain some small value at higher powers, eliminating the problem of a NaN or ±Inf result.
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