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Re: Series expansion for an integral  involving incomplete elliptic integral
Posted:
Dec 9, 2011 8:02 AM


ResentFrom: <bergv@illinois.edu> From: Dan Luecking <LookInSig@uark.edu> Date: December 8, 2011 4:00:31 PM MST To: "scimathresearch@moderators.isc.org" <scimathresearch@moderators.isc.org> Subject: Re: Series expansion for an integral  involving incomplete elliptic integral
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 = infinity.
Dan To reply by email, change LookInSig to luecking



