$$ I_n = \int \csc^nx dx $$

I find the reduction formula like this: $$ I_n = \frac{n-2}{n-1}I_{n-2} - \frac{\csc^{n-2}x\cot x}{n-1}$$

Now , evaluate $I_{2m}$ and $I_{2m+1}$ in summation notation, I find them seriously ugly

$$I_{2m} = I_2\prod_{k-1}^{m-1}\frac{2k}{2k+1}-\frac{\csc^{2m-2}x\cot x}{2m-1}-\sum_{k=1}^{m-2}\left(\frac{\csc^{2k}x\cot x}{2k+1}\prod_{i=k+1}^{m-1}\frac{2i}{2i+1}\right)$$ and $$I_{2m+1} = I_1\prod_{k=1}^{m}\frac{2k-1}{2k}-\frac{\csc^{2m-1}x\cot x}{2m}-\sum_{k=1}^{m-1}\left(\frac{\csc^{2k-1}x\cot x}{2k}\prod_{i=k+1}^{m}\frac{2i-1}{2i}\right)$$

I think there exists a simpler and clear expression, I have tried my best to solve still have no other good idea. Is there any clever and simpler way to present reduction formula with summation notation?

  • $\begingroup$ Note that you can telescope the products of the form $\Pi \frac{2i-1}{2i}$ to simplify them. This will considerably simplify the expression. $\endgroup$
    – Zoms96
    Nov 16, 2022 at 3:07


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