On Wed, 26 Nov 2008 , James Record wrote:
>(%i1) integrate(x^3/(%e^x-1), x, 0, inf);
>Is zeta(3) positive, negative, or zero? positive;
>
>(%o1)
>limit(6*li[4](%e^x)-6*x*li[3](%e^x)+3*x^2*li[2](%e^x)+x^3*log(1-%e^x)-x^
>4/4,x,inf)-%pi^4/15
>
>Any idea on how to simplify this further?
>
>Jim
I have not been able to use Maxima's limit function to
make progress with this approach to a well known
integral (so well known that integrate should have
it in its "lookup table".
As a workaround, you can multiply the numerator
and denominator by exp(-x), which is less than one
over the domain of interest, expand in a power
series in y = exp(-x), interchange the order of
integration and summation to get the correct
answer:
(%i1) display2d:false$
(%i2) ix : x^3/( exp(x) - 1 )$
(%i3) ( exp( -x )*num( ix ) )/expand(exp( -x)*denom( ix ) );
(%o3) x^3*%e^-x/(1 -%e^-x)
(%i4) ratsubst( y, exp( -x ), % );
(%o4) -x^3*y/( y - 1 )
(%i5) powerseries( %, y, 0 );
(%o5) x^3*y*'sum( y^i1, i1, 0, inf )
(%i6) ratsubst( exp( -x ), y, % );
(%o6) x^3*%e^-x*'sum( %e^-(i1*x), i1, 0, inf )
(%i7) intosum(%);
(%o7) 'sum( x^3*%e^(-i1*x - x ), i1, 0, inf )
(%i8) ixx:part( %, 1 );
(%o8) x^3*%e^( -i1*x - x )
(%i9) declare( i1, integer )$
(%i10) assume( x > 0, i1 >= 0 )$
(%i11) integrate( ixx, x, 0, inf );
(%o11) 6/(i1 + 1)^4
(%i12) ratsubst( n, i1+1, % );
(%o12) 6/n^4
(%i13) sum(%, n, 1, inf ), simpsum;
(%o13) %pi^4/15
Ted Woollett