I have seen answers to similar questions that reference a Si(x) function as a solution to the integral of sinc(x). Most of these questions have infinite upper or lower limits, though. I am interested in the definite integral of the sinc-like function like below.

$$ \int_{0}^{R}\frac{\sin(\frac{\pi r}{R})}{r}dr $$

Explanations for the use of Si(x) lead me in circles. Could anyone explain if Si(x) is necessary here and how it is used.

  • 4
    $\begingroup$ This one is just a sinus, not a cardinal sinus, there is no $r$ at denominator... $\endgroup$ – zwim Oct 7 '17 at 2:33
  • $\begingroup$ Yes, it is necessary to have $\operatorname{Si}(x)$ here. That is why this special function is defined :) $\endgroup$ – Sangchul Lee Oct 8 '17 at 19:24
  • $\begingroup$ Sorry I am unfamiliar with Si(x). Would this integral just become Si(pi)? $\endgroup$ – peasqueeze Oct 8 '17 at 19:43
  • 1
    $\begingroup$ Yes, it's value is simply $\operatorname{Si}(\pi)$, independent of $R$. $\endgroup$ – Sangchul Lee Oct 8 '17 at 20:32

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