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- Thread starter Frillth
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[tex] \sin x = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1!}x^{2n+1} [/tex]

[tex] \frac{\sin x}{x} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1!}x^{2n} [/tex]

[tex] \int_{0}^{x} \frac{\sin x}{x}\; dx = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1!(2n+1)}x^{2n+1} [/tex]

[tex] \frac{\sin x}{x} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1!}x^{2n} [/tex]

[tex] \int_{0}^{x} \frac{\sin x}{x}\; dx = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1!(2n+1)}x^{2n+1} [/tex]

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- #3

Hurkyl

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Why do you think that? You're integrating a continuous, bounded function over a finite interval -- that's like the poster child for being integrable!I'm fairly sure that sint/t dt has no integral

The integral most surely exists. The function

[tex]g(x) := \int_0^x \frac{\sin t}{t} \, dt[/tex]

is perfectly well-defined. It just might be that there isn't a "simpler" way to represent

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Hurkyl

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