# Problem of calculating the value of integrals

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Qian Feng on 27 Aug 2017
Edited: Qian Feng on 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 !
Qian Feng on 27 Aug 2017
This is something I did not notice before. Although the integrand has infinite value at zero, the function cl2 in fact is of finite value. I probably should change the lower limits of the integration interval into a small positive number.

Star Strider on 27 Aug 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.