I'm looking for the limit when $n\rightarrow\infty$ of the sum $$S=\sum_{j=1}^{\frac{n}{2}}e^{-2h\sin\left(\frac{\pi j}{n}\right)}$$ The angle in the sine is $\theta=\frac{\pi j}{n}$ and as $j$ varies from one unit in two successive terms, ($\delta j=1$), I set $d\theta=\frac{\pi}{n}\delta j=\frac{\pi}{n}$ It seems that I can now replace the exponential in the sum by : $\frac{1}{d\theta}e^{-2h\sin\left(\frac{\pi j}{n}\right)}d\theta=\frac{n}{\pi}e^{-2h\sin\theta}d\theta$. The sum may thus converges to the integral:
$$\frac{n}{\pi}\int_{0}^{\frac{\pi}{2}}e^{-2h\sin\left(\theta\right)}d\theta$$
(After that the integral is accessible by a series development).
The boundaries come from the fact that for $j=1,\theta\rightarrow 0$ and for $j=\frac{n}{2}, \theta\rightarrow \frac{\pi}{2}$
Is that correct?