> int(((1-x^2)*ln((1+x)/(1-x))+2*x)^2,x=0..infinity);
But I have the impression it can be done via a number of intermediate
steps (?) Any help will be appreciated.
Han de Bruijn
Your problem is not well-posed. The integrand, f(x) = ((1-
x^2)*ln((1+x)/(1-x))+2*x)^2 is not defined for x > 1 (because ln((1+x)/
(1-x)) is the log of a negative number).
R.G. Vickson
Did you try the idefinite integral on the integrator (very ugly result
- but shows the issue Ray V. mentions).
The integral diverges to -infinity, and Maple 14 knows that: it gives
the result as -infinity. Note that as x -> infinity,
ln((1+x)/(1-x)) -> ln(-1) = I*Pi, so the integrand goes like -Pi^2 x^4.
Or were you thinking of some other integral?
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Yes, sorry. It must be this (still no outcome with Maple 8):
> int(((1-x^2)*ln(abs((1+x)/(1-x)))+2*x)^2,x=0..infinity);
Han de Bruijn
If you don't succeed the first time, try and try again:
> f(x) := ((1-x^2)*ln(abs((1+x)/(1-x)))+2*x)^2;
> int(expand(f(x)),x=0..infinity);
And the outcome is .. : 8/15*Pi^2 . YES !!
Han de Bruijn
I haven't tried Maple 14. I do get this answer in Maple 11, but only
after some manipulation: (1) split the integral into integrations over
[0,1] and [1,infinity), writing the integrand without absolute value
signs in the two cases; (2) changing variables to u = log((1+x)/(1-x))
in the first integral and to u = log((1=x)/(x-1)) in the second
integral. Summing the two results gives the above.
R.G. Vickson