Gauss-Kronrod quadrature

Please specify the integrand: for the predefined cases the interval of integration is [0,1].
f(x) = x^(1/4) (continuous, not differentiable in x=0) I=0.8
f(x) = 1/((x-1.1)*(x+0.1)) (infinitely smooth on [0,1]) I = -ln(11)/0.6 = -3.99649212
f(x) = 4/(1+x^2) (infinitely smooth on the real axis) I = pi =3.141592653589
f(x) = exp(-pi*x)cos(pi*x) (analytic) I = (1+e^(-pi))/(2pi) = 0.1660326514
f(x) = 2/(2+sin(10*pi*x) periodic and infinitely smooth on the real axis, but not analytic I=1.1547005384
f(x) = -dlog(x)*x**.1d0 f only C0, I = 1/(1.1)2
f(x) = -dlog(x)*x**0.5 f only C0, I=1/(1.5)2
f(x) = -dlog(x)*x**.9d0 f only C0, I=1/(1.9)2
f(x) = -dlog(x)*x**2d0 f only C2, I=1/9
f(x) = -dlog(x)*x**4d0 f only C4, I=1/25
f(x) = -dlog(x)*x**6d0 f only C6, I=1/49
f(x) = -dlog(x)*x**8d0 f only C8, I=1/81
f(x) = -dlog(x)*x**10d0 f only C10, I=1/121
A function of your own Please type the evaluation program of your function here using FORTRAN rules. Your final statement must be 
      fu= some expression you computed before or just here depending on x
You may use the constants pi, e(=exp(1)), sqrt2(=1.414...), the integer variables i,j,k, the logicals bool1,bool2,bool3 and the double precision variables sum,h1,h2,h3,h4,y(100),z(100),a(100,100) which are all intialized with zero resp. .false. . The routine has the parameters x (double, input) and fu (double out). never change x!. first is a local integer and set 0 before calling the function the first time. You may use this in order to initialize some local data and set it 1 afterwards to avoid multiple such initialization. Your settings of the local variables are preserved during program execution.


And specify the interval here:
a = b =

Please specify the desired relative and absolute precision.
epsrel = Important : 1e-12 <= epsrel <= 1e-2 !
epsabs = Important : 1e-16 <= epsabs <= 1e-6 !

Please specify n:
n=7 n=10
n=15 n=20
n=25 n=30

Click on "evaluate", in order to submit your input.

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19.02.2015