Re: An exact 1-D integration challenge

From: C W (sylvester7_at_ns.sympatico.ca)
Date: 01/30/05 

Date: Sun, 30 Jan 2005 19:05:11 GMT

> ..................................................................
>
> A correct answer to my challenge
>
> int((Pi-2*arctan(exp(z+ln(tan(1/2)))))^2, z= 0..infinity);
>
> is
>
> -4*I*(-1+Pi)*polylog(2,exp(I))+I*(-1+Pi)*polylog(2,exp(2*I))-4\
> *polylog(3,exp(I))+1/2*polylog(3,exp(2*I))+1/2*I*(-1+Pi)^2*(Pi\
> +2*I*ln(tan(1/2)))-7/2*Zeta(3);
>
> evalf(%, 65);
> evalf(Int((Pi-2*arctan(exp(z+ln(tan(1/2)))))^2, z=0..infinity,\
> 65, _Gquad));
>
> 3.6660906226888345192183155643604104157276559770322760321501468034
> 3.6660906226888345192183155643604104157276559770322760321501468031
>
> Best wishes,
>
> Vladimir Bondarenko

F(x) = 4*arctan(x)^2/arccot(x)^2*int(arccot(x)^2/x,x)
int((Pi-2*arctan(exp(z+ln(tan(1/2)))))^2,z = 0 .. infinity) = F(cot(1/2))

int((Pi-2*arctan(exp(z+ln(tan(1/2)))))^2,z = 0 .. infinity) =
Int(4*arctan(v)^2/v,v = 0 .. cot(1/2))

int((Pi-2*arctan(exp(z+ln(tan(1/2)))))^2,z = 0 .. infinity) =
3.666090622688834519218315564360410415727655977032276032150146803104437301145763430338119718910081736559778743708081886527966092771

Chris