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Re: Series expansion for an integral  involving incomplete elliptic integral
Posted:
Dec 9, 2011 12:23 PM


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 don't see how this is possible. For such a first order > expansion to exist, one would need at least that > > integral(0 to y) ArgCosh[1 + x^2 + a]  ArgCosh[1 + x^2] dx > > be less than a constant times a. By the mean value theorem > (differentiating with respect to a), that would require that > > integral(0 to y) 1/sqrt((1 + x^2 + a)^2  1) dx > > remain bounded as a tends to 0. But now the monotone > convergence theorem says that this integral tends to: > > integral(0 to y) 1/sqrt((1 + x^2)^2  1) dx > > And now a little algebra shows that this is equal to > > integral(0 to y) 1/( x * sqrt(x^2 + 2)) dx > = (1/sqrt(y^2+2)) * integral(0 to y) 1/x dx
Somehow a greaterthan sign disappeared. That equal sign should have been a ">=" (greater than or equal to).
> > = infinity. >
Of course the "smaller" one is already infinite, so they are in fact equal, but one wouldn't necessarily know that until after it is evaluated.
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. In particular, f(a)/a tends to infinity.
Dan To reply by email, change LookInSig to luecking
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