Re: pitfall in analytic function
- To: mathgroup at smc.vnet.net
- Subject: [mg41574] Re: pitfall in analytic function
- From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
- Date: Tue, 27 May 2003 01:47:36 -0400 (EDT)
- References: <31378N138@web2news.com> <20030518.031910@whim.org>
- Sender: owner-wri-mathgroup at wolfram.com
[I've added other groups to which this same question was originally posed.]
rob at whim.org (Rob Johnson) wrote:
> In article <31378N138 at web2news.com>,
> "Narasimham G.L." <google.news.invalid at web2news.net> wrote:
> >I tried to integrate a transcendental function having factor (x-2),
> >analytic on the whole line, even at x=2, but unable to get through with
> >the smooth function at x=2.The following Mathematica prgram was run, and
> >although at x=2 it is a smooth plot ( minimum value is around -1.204 at
> >x=2.1), it proves to be a stumbling block in performing integration.
>
> First let's look at this problem mathematically rather than immediately
> running to Mathematica:
>
> |\ 2^x - x^2
> | --------- dx
> \| x - 2
>
> |\ 2^x - 4
> = | -(x+2) + ------- dx
> \| x - 2
>
> |\ 2^x - 4
> = -(x+2)^2/2 + | ------- dx [1]
> \| x - 2
>
> Let u = (x-2)ln(2), then 4 e^u = 2^x. Therefore, [1]
>
> |\ e^u - 1
> = -(x+2)^2/2 + 4 | ------- du [2]
> \| u
>
> You can easily get the series for the integral in [2] from the series
> for e^u:
>
> |\ e^u - 1 u^2 u^k
> | ------- du = C + u + --- + ... + ---- + ... [3]
> \| u 4 k k!
>
> This series converges for all u.
>
> The integral in [3] can also be given by the the Exponential Integral,
> EI(u):
>
> |\ e^u - 1
> | ------- du = EI(u) - ln(u) + C [4]
> \| u
I suspect that you actually intended to write EI(u) - ln(|u|) + C above.
> Therefore, the indefinite integral could be written as
>
> |\ 2^x - x^2
> | --------- dx
> \| x - 2
>
> = -(x-2)^2/2 + 4 ( EI((x-2)ln(2)) - ln(|x-2|) ) + C [5]
Typo: The first term above should be -(x+2)^2/2.
> >Plot [ (2^x-x^2)/(x-2), { x,0,4 }]
> >y1=Integrate [(2^x-x^2)/(x-2),{x,0,x}]
> >Plot[y1 ,{x,0, 4} ];
> >y2= -2 x - x^2/2 + 4 ExpIntegralEi[-2 Log[2] + x Log[2]] -
> > 4 (ExpIntegralEi[-2 Log[2]] - Log[2]) - 4 Log[2 - x];
> >Plot[y2 ,{x,2.1, 4} ];
> >Plot[y2 ,{x,0, 4} ];
>
> I see that Mathematica did not get the correct integral since it
> gives Log[2-x] instead of Log[Abs[2-x]].
While I certainly like your [5] better than Mathematica's proposed
antiderivative, I think it is debatable whether or not Mathematica got a
correct antiderivative. After all, the integrand has a singularity, albeit
a removable one, at x = 2. That being the case, it seems to me that it is
reasonable to allow an antiderivative to have a singularity at x = 2 also.
Indeed, your [5] and Mathematica's antiderivative both have singularities
at x = 2. The big difference between your [5] and Mathematica's
antiderivative is that the singularity in yours is removable, that in
Mathematica's is not. I certainly think that your [5] is much nicer, but
I'd be reluctant to say that Mathematica's is actually wrong.
> If you replace Log[2-x]
> above with Log[Abs[2-x]], the plot comes out fine.
Agreed (unless we plot so exceedingly close to x = 2 that numerical
troubles near the singularity become apparent).
BTW, an antiderivative which might fully satisfy the OP is
{ -(x+2)^2/2 + 4 ( EI((x-2)ln(2)) - ln(|x-2|) ) if x <> 2
{
{ 4( gamma + ln(ln(2)) - 2 ) if x = 2
where gamma is the Euler-Mascheroni constant.
Regards,
David Cantrell