Daniel Lemire <lemire@ondelette.com> writes:
>
>
> (C24) f(x):= integrate(1/sqrt(2-2*cos(x)),x);
>
> 1
> (D24) f(x) := INTEGRATE(------------------, x)
> SQRT(2 - 2 COS(x))
> (C25) f(x);
>
> ABS(COS(x) + 1)
> (D25) - ASINH(---------------)
> ABS(SIN(x))
>
> Which is quite satisfying this early in the morning. My brain isn't
> working but, using argument likes ....
>
> integrate(1/sqrt(2-2*cos(x)),x,0+0.0001,%PI/2-0.0001);
>
> Seem to lead to the conclusion that it blows up (BTW, I'm sure it is a
> very easy problem). It would seem like at zero, I get f(0) =
> -asinh(Infinity) = -Infinity. I have no problem at PI/2. It would seem
> like the integral is actual infinity.
It is. Using Taylor expansion one sees that the intergrand behaves
like 1/sqrt(2-2(1-x*x/2+...)) = 1/x near 0.
> > correctness of the highest
> > order is important.
> >
> > RJF
> >
May I ask you to think about your citation mode? Probably most
readers (but at least myself) would prefer if you cited the necessary
parts of messages first (including the name of their author).
Thank you,
Nicolas.
--
Dr. Nicolas Neuss
Email: Nicolas.Neuss@IWR.Uni-Heidelberg.De
WWW: <http://www.iwr.uni-heidelberg.de/~Nicolas.Neuss>;