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I am wondering if the following integral has a solution in closed form:

$ \int_0^1 \prod_{j=1}^d \sin \left( \left( \frac{\pi}{2}+(m_j-1)\pi \right) x^{t_j}\right)dx $

where $d > 2$, $m_1,...,m_d$ are positive integers and $t_1,...,t_d$ satisfy $t_1+...+t_d=1$, $t_i \geq 0$, $i=1,...,d$. The points $\left( \frac{\pi}{2}+(m_j-1)\pi \right)$ are zeros of the cosine, so I guess that this should help. Yet, having a product with more than two sines sharing tha same $x$ makes things more involved. Am I missing anything stupid?

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    what is the purpose of the $\pi/2$? – 2017-01-27
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    Letting $0$\sin(z_{k_j} x^{t_j})$, with $k_j \geq 1$ – 2017-01-27
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    Because of the fractional powers of $x$ under the sines, a tractable solution is unlikely. Even simplified cases are awful https://www.wolframalpha.com/input/?i=integral+sin(pi+t%5E(3%2F4)%2F2)+for+t+from+0+to+1 – 2017-01-27

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