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Re: Series expansion for an integral - involving incomplete elliptic integral
Posted:
Dec 12, 2011 2:14 PM
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On Fri, 09 Dec 2011 10:23:14 -0700, Dan Luecking <LookInSig@uark.edu>
wrote:
>
> On Fri, 09 Dec 2011 06:02:01 -0700, Dan Luecking <LookInSig@uark.edu>
> wrote:
>
> > On Wed, 7 Dec 2011 09:14:30 +0000, victorphy <vbapst@gmail.com> wrote:
> >
> > > Hi,
> > >
> > >
> > > I have a little question about a series expansion, which may be very
> > > basic.
> > >
> > > I would like to expand the integral of ArgCosh[1 + x^2 + a] between 0
> > > and y >0, at first order in the small parameter a.
...
> I think I can show that this function
>
> integral (0 to y) ArgCosh(1 + x^2 + a) dx
>
> has an expansion like
>
> G(y) + H(y)*f(a) + o(a).
>
> where f(a) tends to 0 as a goes to 0, and satisfies
>
> f(a)*(1 - log f(a)) = O(a)
>
> for small a.
May the third time be a charm: That should have been:
a*(1 - log a) = O(f(a))
so that f(a) is no smaller than a*(1 + log(1/a))
In particular, f(a)/a tends to infinity.
And now this is a valid conclusion.
Dan
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