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I want to evaluate $\,\displaystyle I_{n}=\int_{0}^{\frac{\pi}{4}} \tan^{2n}(x)\,\mathrm dx$.
I proved that $\,I_{n}+I_{n-1}=\dfrac{1}{2n-1}\,,\,$ where $I_{0}=\dfrac{\pi}{4}$.
From that I found that (it's easy to prove by induction) $$I_{n}=(-1)^{n} \frac{\pi}{4}+\sum_{k=1}^{n} (-1)^{k+1} \frac{1}{2(n-k)+1}$$ My question is, is there any way to telescope that sum ? Any help will be greatly appreciated.

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    Those are the remainders of this series I might be remembering wrong, but I think the partial sums of that series cannot be written as a fixed sum of hypergeometric terms. You could express it in terms of other special functions Whether that is any more useful than the original expression or not depends on the final goal. – plop Jan 04 '23 at 21:38
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    Your formula for $I_n - I_{n-1}$ is wrong. – Robert Israel Jan 04 '23 at 21:45
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    $I_{n}-I_{n-1}=\dfrac{1}{2n-1}$ should be $I_{n}+I_{n-1}=\dfrac{1}{2n-1}$. – xpaul Jan 04 '23 at 21:49
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    Bakugany, you should write $;I_n+I_{n-1}=\dfrac1{2n-1};$ instead of $;I_n-I_{n-1}=\dfrac1{2n-1};$ which is wrong. – Angelo Jan 04 '23 at 21:51
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    I think that it would be better to write your (correct) formula for $I_n$ in the following way :$$I_n=(-1)^n\left[\dfrac{\pi}4-\sum_{k=1}^n(-1)^{k+1}\dfrac1{2k-1}\right]$$ So we can get that $$I_n=(-1)^n\sum_{k=n+1}^\infty(-1)^{k+1}\dfrac1{2k-1}$$ that is $$I_n=\sum_{k=n+1}^\infty(-1)^{k-n+1}\dfrac1{2k-1}$$ or also $$I_n=\sum_{h=1}^\infty(-1)^{h+1}\dfrac1{2(n+h)-1}$$ – Angelo Jan 05 '23 at 01:01
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    Does this help? – Accelerator Jan 06 '23 at 05:55

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