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If $\frac{\sin(x)}{x}$ is not defined at $x=0$, why is $\operatorname{Si}(x)=\int_0^x \frac{\sin(t)}{t} \, dt$ defined at any value of $x$?

Michael Hardy
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Numox
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  • $$ \begin{align} \sin x = {} & x - \frac{x^3} 6 + \frac{x^5}{120} - \frac{x^7}{5040} + \cdots \ & \text{(The denominators are factorials.)} \ {} \ \text{So } \frac{\sin x} x = {} & 1 - \frac{x^2} 6 + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots. \ {} \ \end{align} $$ That last power series is perfectly well defined when $x=0,$ and in many contexts it makes sense to define $\frac{\sin x}x$ to be equal to that. However, as I mention in a comment under the answer by Parcly Taxel, we don't really need that. $\qquad$ – Michael Hardy Jan 31 '20 at 15:54

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The singularity at $x=0$ is removable; the function can be defined as $1$ at that point, which makes it entire over the whole complex plane. Hence the sine integral can be defined.

Parcly Taxel
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    Even if the singularity were not removable, one would say $ \displaystyle \operatorname{Si}(0) = \int_0^0 \cdots = 0. \qquad$ – Michael Hardy Jan 31 '20 at 15:50