Another integral

[Han de Bruijn]

MAPLE does not directly evaluate the following integral:

> 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

[Ray Vickson]

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

[amzoti]

On Jan 4, 12:24 pm, Han de Bruijn <umum...@gmail.com> wrote:

Did you try the idefinite integral on the integrator (very ugly result
- but shows the issue Ray V. mentions).

http://integrals.wolfram.com/index.jsp

[Robert Israel]

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

[Han de Bruijn]

On Jan 5, 4:28 am, Robert Israel
<isr...@math.MyUniversitysInitials.ca> wrote:

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

[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

[Ray Vickson]

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