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