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?