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Re: A tough Integral
*To*: mathgroup at smc.vnet.net
*Subject*: [mg28555] Re: A tough Integral
*From*: Richard Easther <easther at physics.columbia.edu>
*Date*: Fri, 27 Apr 2001 03:56:09 -0400 (EDT)
*Organization*: Columbia University
*References*: <200104250530.BAA19306@smc.vnet.net> <9c7n4d$sda@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
On 25 Apr 2001, Daniel Lichtblau wrote:
>bobbym1953 wrote:
>>
>> Does anyone know how I can get the following Integral to at least 60 places,
>> using Mathematica?
>>
>> Int(1/(cos(x)+x^2)) between x=0 and x=infinity. Both Integrate and NIntegrate
>> seemed helpless.
>
> 60 places?! I wish you luck.
Actually, it can be done. Break the integral into two pieces, I_1 where
the integral is integrated from O<x<N, and I_2 where we integrate between
N<=x<Infinity.
The answer is then simply I_1 + I_2
If N is not too large, the first integral can be done to the rquired
precision using NIntegrate.
The second integral can be evaluated by noting that the integrand
1/(x^2(1 + Cos[x]/x^2))
can be written as a sum with the general term
(-1)^n (Cos[x]/x^2)^n /x^2
For integer n, Mathematica can evaluate the indefinite integral over the
general term (although it does not seem to be able to give a general
formula) so the series can be integrated term by term. Taking enough
terms will give I_2 to arbitrary accuracy.
The trick is to choose N large enough so that you do not need a huge
number of terms in the series that approximates I_2, but not so large that
NIntegrate cannot do the first sub-integral to the desired accuracy.
If you worked really hard, and extracted the analytic form of the integral
over the general term in I_2, it is just possible you may be able to find
an analytic expression for I_2. If you were really lucky you might then be
able to analytically continue N back to zero, and thus evalute the
integral algebraically.
Richard,
who after doing this started to wonder why this integral needs to be
worked out to 60 significant figures.
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