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Re: Series expansion for an integral  involving incomplete elliptic integral
Posted:
Dec 12, 2011 2:14 PM


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 To reply by email, change LookInSig to luecking



