Tanh-sinh-transformation method

Please choose a function: for the predefined cases there always holds [a,b]=[0,1]

	f(x) = x^(1/4)	not differentiable at x=0, I=0.8
	f(x) = 1/((x-1.1)*(x+0.1))	infinitely differentiable on [0,1] , I = -ln(11)/0.6 = -3.99649212
	f(x) = 4/(1+x^2)	infinitely differentiable on R, I = pi =3.141592653589
	f(x) = exp(-pi*x)cos(pi*x)	entier, I = (1+e^(-pi))/(2pi) = 0.1660326514
	f(x) = 2/(2+sin(10*pi*x))	periodic, but not entier, 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= 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. C23456789 C don't forget the role of the first 6 columns fu=dlog(x)**2 C this is for the interval [0,1] and gives 2 a = b = I don't know the exact integral value The true integral value is

Specify the summation index k₀ here.
2 k₀+1 function values will be computed for the first estimate
k0 =

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