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How does one solve the integral $\int_{0}^{2\pi}(\sin(x))^{4/3}\ \mathrm{d}x$, or more generally $\int_{0}^{2\pi}(\sin(x))^{(m+n)/m}\ \mathrm{d}x$ ? Standard CAD math as Maple did not provide answers.

offline
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pivu0
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    Let $I$ be the integral on $[0,,\pi/2]$, which is half a Beta function. Use symmetries to express your integral as a multiple of $I$. – J.G. Feb 10 '21 at 13:46
  • Those parentheses of yours do nothing to disambiguate the expression! Do you mean $(\sin x)^{4/3}$, or $\sin(x^{4/3})$? – TonyK Feb 10 '21 at 13:50
  • @TonyK I put extra parentheses now – pivu0 Feb 10 '21 at 13:52
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    $$\frac{3 \sqrt{\pi } \Gamma \left(\frac{7}{6}\right)}{\Gamma \left(\frac{2}{3}\right)}$$ – Infiniticism Feb 10 '21 at 13:58
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    If you use Mathematica be careful to write $$\int_0^{2 \pi } \sqrt[3]{\sin ^4(x)} , dx$$ and not $(\sin x)^{4/3}$ otherwise you get this https://www.wolframalpha.com/input/?i=Integrate%5B%28sin+x%29%5E%284%2F3%29%5D%2C+%7Bx%2C+0%2C+2+Pi%7D%5D – Raffaele Feb 10 '21 at 14:09

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For $0 \leq t \leq \pi$ and $a>0$ $$\int_0^t \big[\sin(x)\big]^a\,dx=\frac{\sqrt{\pi }\,\, \Gamma \left(\frac{a+1}{2}\right)}{a\,\, \Gamma \left(\frac{a}{2}\right)}- \, _2F_1\left(\frac{1}{2},\frac{1-a}{2};\frac{3}{2};\cos ^2(t)\right)\cos (t)$$ where appears the gaussian hypergeometric function.

For other conditions, use the symmetry properties .

Edit

For $t=\pi$, we have $$\int_0^\pi \big[\sin(x)\big]^a\,dx=\frac{ \pi\,\, \Gamma (a+1)}{2^a\,\,\Gamma \left(\frac{a}{2}+1\right)^2}$$ which most of the times will not make any problem.

For large $a$, a good approximation is

$$\sqrt {\frac {2\pi}a}\,\exp\Bigg[-\frac{1}{4 a}+\frac{1}{24 a^3}-\frac{1}{20 a^5}+O\left(\frac{1}{a^7}\right) \Bigg]$$ which is in a relative error of less than $0.1$% for $a>2$ and less than $0.001$% for $a>4$.

Claude Leibovici
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