What is the integral from 0 to 2pi of (sin(x))/x ?

The integral from 0 to 2pi of (sin(x))/x is \Si(2pi)

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integral from 0 to 2pi of (sin(x))/x

\int_{0}^{2 π} \frac{sin ( x )}{x} d x

Si (2 π)

Solution steps

\int_{0}^{2 π} \frac{sin ( x )}{x} d x

Use the nonelementary integral:

\int \frac{sin ( x )}{x} d x = Si (x)

= [Si (x)]_{0}^{2 π}

Compute the boundaries:

Si (2 π)

= Si (2 π)

\int_{1}^{2} t d t

\int_{0}^{t} (t - w) e^{2 w} d w

\int_{0}^{1} t^{3} 4 + 9 t^{4} d t

\int_{2}^{x} \frac{1}{ln ( x )} d x

\int_{- 1}^{1} 2 x^{4} d x

What is the integral from 0 to 2pi of (sin(x))/x ?
The integral from 0 to 2pi of (sin(x))/x is \Si(2pi)