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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 11, Problem 11.4.78b

Sine integral function The function Si(x) = ∫₀ˣ f(t) dt, where f(t) = {(sin t)/t if t ≠ 0, 1 if t = 0, is called the sine integral function.
b. Integrate the series to find a Taylor series for Si.

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1
Recall that the sine integral function is defined as \(\mathrm{Si}(x) = \int_0^x f(t) \, dt\), where \(f(t) = \frac{\sin t}{t}\) for \(t \neq 0\) and \(f(0) = 1\).
Start by expressing \(\sin t\) as its Taylor series expansion around \(t=0\): \(\sin t = \sum_{n=0}^\infty (-1)^n \frac{t^{2n+1}}{(2n+1)!}\).
Divide the series for \(\sin t\) by \(t\) to get the series for \(f(t) = \frac{\sin t}{t}\): \(f(t) = \sum_{n=0}^\infty (-1)^n \frac{t^{2n}}{(2n+1)!}\).
Integrate the series term-by-term from 0 to \(x\) to find the Taylor series for \(\mathrm{Si}(x)\): \(\mathrm{Si}(x) = \int_0^x f(t) \, dt = \int_0^x \sum_{n=0}^\infty (-1)^n \frac{t^{2n}}{(2n+1)!} \, dt\).
Interchange the integral and the summation (justified by uniform convergence on compact intervals) and integrate each term: \(\mathrm{Si}(x) = \sum_{n=0}^\infty (-1)^n \frac{1}{(2n+1)!} \int_0^x t^{2n} \, dt = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)(2n+1)!}\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definition and Properties of the Sine Integral Function

The sine integral function Si(x) is defined as the integral from 0 to x of (sin t)/t dt, with a special value at t=0 to ensure continuity. Understanding this function involves recognizing it as an integral of a function with a removable discontinuity and its role in analysis and applications.
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Taylor Series Expansion of Functions

A Taylor series represents a function as an infinite sum of terms calculated from the derivatives at a single point, usually zero. To find the Taylor series of Si(x), one must express the integrand as a power series and then integrate term-by-term, ensuring convergence within the radius of expansion.
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Term-by-Term Integration of Power Series

Integrating a power series term-by-term is valid within the interval of convergence and allows finding the integral of complex functions by integrating each term individually. This technique simplifies finding the Taylor series of integral-defined functions like Si(x) by integrating the series expansion of (sin t)/t.
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