How can we calculate the integration $\int_0^{2\pi} \sin^x(x) \, dx$ numerically?

I tried calculating it using Simpson's rule and trapezoidal rule using python program; but it gave 'nan'.

Mathematica's NInregrate[(Sin[x])^x,{x,0,2 Pi}] gives 1.77788+ i 0.364092.

What would be the approach to solve this?

(I am quite puzzled, why does this have an imaginary solution? This part is solved)


closed as off-topic by Jack D'Aurizio Oct 20 '17 at 13:27

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    $\begingroup$ Between $\pi$ and $2 \pi$, the equation $\sin^x(x)$ is not defined, as $\sin(x) \lt 0$, and $a^x, a \lt 0$ is not defined for all numbers, only integers:desmos.com/calculator/gpbagrwj7d $\endgroup$ – John Lou Oct 20 '17 at 3:28

The integral cannot be computed over reals as there are many points of discontinuity for $\pi<x< 2\pi$.

Eg. For $\pi<x<2\pi$, $\sin(x) $ is negative. Say $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$. Then your function is of the form $$\sin\left(\frac{7\pi}{6}\right) ^{\frac{7\pi}{6}} = \left(\frac{-1}{2}\right)^{\frac{7\pi}{6}}$$

Which evidently enough is not real.

The problem lies with the fact that $(f(x))^{x}$ is not well defined when $f(x) < 0$ because of many points of discontinuity.

Alternatively, you can easily calculate the integral for $x\in (0,\pi)$ and we would not have this problem.

  • $\begingroup$ Thank you. You explained it making it so easy :) . By the way, what would be the way to compute the integral? $\endgroup$ – Pradip Kattel Oct 20 '17 at 3:34
  • $\begingroup$ I mean $f(z) = \Re(f(z)) + i \Im(f(z))$ $\endgroup$ – samjoe Oct 20 '17 at 6:17

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