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I'm trying to obtain a closed formula for the following integral.

\begin{align} I_n = {} & \int_0^h \Bigr[\sum_{r_1=1}^\infty a_{1,r} \cos\left(\frac{2\pi}{h} r_1t_1\right) \\[6pt] & {}+ b_{1,r}\sin\left(\frac{2\pi}{h}r_1t_1\right) \Bigr]\int_0^{t_1} \Bigr[\sum_{r_2=1}^\infty a_{2,r} \cos\left(\frac{2\pi}{h} r_2 t_2 \right) \\[6pt] & {} + b_{2,r} \sin\left(\frac{2\pi}{h}r_2t_2\right) \Bigr]\cdots \int_0^{t_{n-1}} \sum_{r_n=1}^\infty a_{n,r_n} \cos\left(\frac{2\pi}{h}r_nt_n\right) \\[6pt] & {}+ b_{n,r_n} \sin\left(\frac{2\pi}{h}r_nt_n \right) \, dt_n\cdots dt_1 \end{align}

where $a_{i,r},b_{i,r}$ are constant defined in a way that the sums converge.

The case in which $n=2$ should gives

$$I_2=\frac{h}{4 \pi}\sum_{r=1}^\infty \frac{1}{r} (a_{1,r}b_{2,r}-a_{2,r}b_{1,r})$$

but $I_3$ already is incredibly more complex and i can't find any general formula. There is some way to obtain a closed formula or at least understand which of the terms of this integral is actually differnt from zero?

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    $\begingroup$ Is each integral only multiplied by the preceding sine function, or actually by more terms? I.e. is there any bracket missing or not? $\endgroup$ Commented Jan 23, 2023 at 19:38
  • $\begingroup$ @LoïcTeyssier The integral is an $n$-dimensional integral. It can be defined recursively as: $I_1(t)=\int_0^t \sum_{r=0}^\infty a_{1,r} \cos(\frac{2\pi r}{h}s) + b_{1,j}sin(\frac{2 \pi r}{h}s) ds$ $I_n(t)=\int_0^t[\sum_{r=0}^\infty a_{n,r} \cos(\frac{2\pi r}{h}s) +b_{n,r}sin(\frac{2 \pi r}{h}s) ]I_{n-1}(s)ds$ Sorry, i forgot the bracket $\endgroup$
    – Marco
    Commented Jan 24, 2023 at 13:22
  • $\begingroup$ OK, that's better now ;) $\endgroup$ Commented Jan 24, 2023 at 13:38

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