Series expansion for an integral - involving incomplete elliptic integral

[victorphy]

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.

Obviously a naive approach using first order taylor expansion in the
integral doesn't work as it leads to a logarithmic divergency near 0.

I tried using Mathematica but it gives me a form involving incomplete
ellipitic integral that I do not understand how to deal with :
I*(EllipticE[I ArcSinh[Sqrt[1/a] x], a/(2 + a)] -
EllipticF[I ArcSinh[Sqrt[1/a] x], a/(2 + a)])
(there is an i in the first argument of the function). It looks like
this as the form cst + cst * a near 0 but I don't know how to compute
the cst exactly.

Does anyone know if this is a good point to start (as these forms come
from the exact primitivation of ArgCosh[1+x^2+a]), or if there is a
simpler way to find the answer ?


Thank you very much in advance.

Victor

[Dan Luecking]

Resent-From: <be...@illinois.edu>
From: Dan Luecking <Look...@uark.edu>
Date: December 8, 2011 4:00:31 PM MST
To: "sci-math...@moderators.isc.org"
<sci-math...@moderators.isc.org>
Subject: Re: Series expansion for an integral - involving incomplete
elliptic integral

On Wed, 7 Dec 2011 09:14:30 +0000, victorphy <vba...@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

[Dan Luecking]

On Fri, 09 Dec 2011 06:02:01 -0700, Dan Luecking <Look...@uark.edu>
wrote:

> > On Wed, 7 Dec 2011 09:14:30 +0000, victorphy <vba...@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
To reply by email, change LookInSig to luecking

.

[Dan Luecking]

On Fri, 09 Dec 2011 10:23:14 -0700, Dan Luecking <Look...@uark.edu>

wrote:

>
> On Fri, 09 Dec 2011 06:02:01 -0700, Dan Luecking <Look...@uark.edu>
> wrote:
>
> > On Wed, 7 Dec 2011 09:14:30 +0000, victorphy <vba...@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.