Problem of calculating the value of integrals

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 !

[Jan]

I do not have the Symbolic Toolbox, so I cannot participate in the solution of the problem. If you see any Inf values, where you do not expect them, it would be useful if you explain both: Where do these values appear and why do you assume anything else. Note that all the readers see is the current code, but this does not allow to guess, what your intention is.

[Star Strider]

One problem is that this term:

cl2 = int(-log(abs(2*sin(0.5*y))), 0, 20*x);

will evaluate to +Inf at 0.

[Qian Feng]

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.

Answer 1

[Star Strider]

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.