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How to compute precisely the integral $$\int_0^\pi \frac{\sin x}{x}dx$$ analytically? It is well-known that $$\int_0^{+\infty}\frac{\sin x}{x}dx=\frac{\pi}{2}.$$ One way to compute the above integral using Taylor expansion for $\sin x$, we then obtain a power series value approximately 1.84. Thanks for any useful hints.

linrr
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3 Answers3

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$$\color{#00f}{\large{\rm Si}\left(\pi\right)}$$

$\displaystyle{\rm Si}\left(x\right)$ is the Sine Integral Function.

Felix Marin
  • 94,079
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Beside the answers given in other post, you could develop $\frac{\sin(x)}{x}$ as a Taylor series built around $x=0$ and integrate. For example, if you use $20$ terms, you would arrive to $1.851937051982916663426668$; $30$ terms would give $1.851937051982466170360965$; $40$ terms would give $1.851937051982466170361053$.

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Your number is sometimes called the Gibbs constant and evaluates to

1.85193705198246617036105337015799136334580972898115
  49098047837818769818901663483585327103365029547577

with 100 digit accuracy. It is most likely not expressible in a simple way in terms of other "well-known" constants.

mrf
  • 44,697