On 3/10/10 4:11 PM, Janek Kozicki wrote:
> Hi,
>
> integrate(sum(1/(n^4+x^2), n, 1, inf),x,0,inf);
>
> by hand + using maxima to calculate some small parts of it,
> I eventually reached the result, which is pi^3/12.
>
> I wonder if it is possible to calculate in maxima straight off. Is it?
>
> What I did, was replacing by hand arctan(0) with 0, separately
> calculating limits and so on. Took a bit of time, and mostly I did it
> myself on paper, just checking some parts with maxima. But still I'm
> not even sure if my final result is correct ;)
>
> question is - how to do it in a "clean" way in maxima?
>
>
Not really clean, but this is how I would do it. The sum in uniformly
convergent, so interchange the order of summation and integration. Then
we have
integrate(1/(n^4+x^2),x,0,inf) -> %pi/2/n^2
after telling maxima that n is positive. Then the sum is
load(simplify_sum);
simplify_sum(sum(%pi/2/n^2,n,1,inf)) -> %pi^3/12.
An aside: I thought that nusum(1/n^2,n,1,inf) knew that the answer is
%pi^2/6, but it doesn't now. Perhaps my memory is wrong.
I guess if there is a way to interchange summation and limits, then you
could have derived the right answer pretty easily too. I don't know how
to tell maxima to interchange the order.
Ray