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Impossible Integral


Date: 11/12/2001 at 19:44:04
From: Tom Parker
Subject: Multivarible Calculus/ Integrals

As our homework for the week we were assigned an integral that no one 
has been able to answer.  I keep getting stuck doing it by parts.  

Here it is: there are no limits.
 
(sin y)/y dy


Date: 11/13/2001 at 03:50:02
From: Doctor Pete
Subject: Re: Multivarible Calculus/ Integrals

Hi,

This is an interesting assignment, because it is well-known that this 
integrand has no elementary antiderivative. Simply put, it's an 
"impossible" integral, and ordinary methods of integration (e.g., by 
parts, trigonometric substitution, partial fraction decomposition), 
will fail. A proof of this fact is beyond the scope of undergraduate 
mathematics courses.

Because of this, the integral from 0 to z of Sin[y]/y dy  is 
considered a special function, and is simply given the name Si[z], or 
the sine-integral of z.

However, this doesn't mean that the function is not integrable; nor 
does this mean there is no series expansion. If we take the Taylor 
series expansion about y = 0 of Sin[y], we obtain

     Sin[y] = y - y^3/3! + y^5/5! - y^7/7! + ... .

Thus,

     Sin[y]/y = 1 - y^2/3! + y^4/5! - y^6/7! + ... ,

and integrating term-by-term with respect to y gives

     Si[y] = y - y^3/(3*3!) + y^5/(5*5!) - y^7/(7*7!) + ... .

The n(th) term of this series is simply

     (-1)^(n-1)*y^(2n-1)/((2n-1)(2n-1)!).

The series expression for Si[y] converges for all complex values y.

- Doctor Pete, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus

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