Re: [Maxima-discuss] Integration asksign bug

[Raymond T. <toy...@gm...>]

On 7/10/23 09:38, Michel Talon wrote:
>
> Le 10/07/2023 à 16:27, Raymond Toy a écrit :
>> I didn't look at the actual solutions, but this is the approach given 
>> in Stackoverflow: 
>> https://math.stackexchange.com/questions/4646362/about-an-integral-of-the-mit-integration-bee-finals-2023
>
> OK, this is the same method, the answer is given by Anne Bauval who is 
> a French mathematician
>
> very active on Wikipedia math stuff. At the end someone noted one can 
> make an integration by parts
>
> which simplifies integrate(3*z^3/(z^3+1)^2) to 
> -z/(z^3+1)+integrate(1/(z^3+1),z) noting that
>
> 3*z^2 is the derivative of z^3+1.  The first term vanishes at 0 and 
> infinity, and
>
> z^3+1=(z+1)*(z^2-z+1) so that the integral reduces to well known 
> ones.  This way one gets a logarithmic
>
> term  log((z+1)^2/(z^2-z+1)) which vanishes at 0 and infinity and an 
> arc tangent term
>
> atan((2*z-1)/sqrt(3))/sqrt(3) which gives %pi/2/sqrt(3) at infinity 
> and -%pi/6/sqrt(3) at 0 yielding the good result.
>
>
> So finally a quite interesting problem, requiring a lot of work in 
> short time if computations have to be done by hand.
I find the integration bee quite interesting.  Not sure if I would be 
able to solve this myself, but I find it fun to ask maxima to do the 
integral.  If it can't I try to help it along with changevar or 
something and see how far I can get.  I'm too old and slow to want to do 
the basic stuff by hand anymore. :-)