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$$\int_{0}^{1}{x^2-1\over x^2+1}{x^{2n}\over \ln(x)}{\mathrm dx\over x}=F(n)\tag1$$

$n\ge 1.$

How do we show that $$F(n)=(-1)^{n-1}\ln\left({\pi\over 2}\prod_{k=1}^{n-1}\left({k+1\over k}\right)^{(-1)^k}\right)?$$

$x=\tan(u)$ then $\mathrm dx=\sec^2(u)\mathrm du$

$\cos(2x)={1-\tan^2(x)\over 1+\tan^2(x)}$

$$-\int_{0}^{\pi\over 4}\cos(2u){\tan^{2n}(u)\over \ln(\tan(u))}{\mathrm du\over \cos^2(u)\tan(u)}\tag2$$

$\tan(u)\cos^2(u)={2\over \sin(2u)}$

$$-{1\over 2}\int_{0}^{\pi\over 4}\cot(2u){\tan^{2n}(u)}{\mathrm du\over \ln\tan(u)}\tag3$$

$${\cos(2x)\over \sin(2x)}{\sin^{2n}(x)\over cos^{2n}(x)}=\cos(2x){\sin^{2n-1}(x)\over \cos^{2n+1}(x)}=\tan^{2n-1}(x)-\tan^{2n+1}(x)$$

$$-{1\over 4}\int_{0}^{\pi\over 4}{\tan^{2n-1}(x)\over \ln\tan(x)}\mathrm du+{1\over 4}\int_{0}^{\pi\over 4}{\tan^{2n+1}(x)\over \ln\tan(x)}\mathrm du\tag4$$

gymbvghjkgkjkhgfkl
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    Just as a side note, you have already asked many similar questions: https://math.stackexchange.com/questions/2240831/how-may-we-show-that-int-011-x-over-1x-cdot2k3x2-over-1x2-cdot , https://math.stackexchange.com/questions/2283669/how-to-evaluate-the-closed-form-for-int-0-pi-2-mathrm-dx-over-sin2x , https://math.stackexchange.com/questions/2281935/equivalent-int-01x2-x-over-x21-cdot-lnx-over-x31dx-int-0 . At this point you shoud be confident in using the "standard techniques". – Jack D'Aurizio Oct 31 '17 at 15:32

1 Answers1

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For any $s > 0$, we have

\begin{align*} F(s) &\stackrel{u=x^2}{=} \int_{0}^{1} \frac{u-1}{u+1} \frac{u^{s-1}}{\log u} \, du = \int_{0}^{1} \frac{u^{s-1}}{u+1} \left( \int_{0}^{1} u^x \, dx \right) du \\ &= \int_{0}^{1} \int_{0}^{1} \frac{u^{x+s-1}}{u+1} \, dudx = \int_{0}^{1} \left( \sum_{k=0}^{\infty} (-1)^k \int_{0}^{1} u^{k+x+s-1} \, du \right) \, dx \\ &= \int_{0}^{1} \left( \sum_{k=0}^{\infty} \frac{(-1)^k}{k+x+s} \right) \, dx = \sum_{k=0}^{\infty} (-1)^k \log\left( \frac{k+s+1}{k+s} \right). \end{align*}

Now for positive integers $n$, we get

\begin{align*} F(n) &= (-1)^n \sum_{k=n}^{\infty} (-1)^k \log\left( \frac{k+1}{k} \right) \\ &= (-1)^{n-1} \left( \sum_{k=1}^{n-1} (-1)^k \log\left( \frac{k+1}{k} \right) - \sum_{k=1}^{\infty} (-1)^k \log\left( \frac{k+1}{k} \right) \right). \end{align*}

Finally the last infinite sum can be computed by Wallis product, hence the identity follows.

Sangchul Lee
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