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/
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