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Consider integral $(1)$

$$\int_{0}^{\pi/2}\sqrt{1-\cos x\over 1+\cos x}\ln^{n+1}{\left({1-\cos x\over 1+\cos x}\right)}{\mathrm dx\over \sin x}={2(2n)!\over (2^{2n}-1)B_{n}^{*}}(-1)^{n}\tag1$$ $n\ge1$

Where $B_{2n}^{*}$ is the Bernoulli number

An attempt:

$u={1-\cos x\over 1+\cos x}$ $\implies$ ${(1+\cos x)^2\over 2\sin x}\mathrm du=\mathrm dx$

$(1)$ becomes

$${1\over 2}\int_{0}^{1}\left({1+\cos x\over \sin x}\right)^2\sqrt{u}\ln^{n+1}u\mathrm du\tag2$$

I can't eliminate the variable $x$ in $(2)$

Alternatively another attempt:

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

$(1)$ becomes

$$2^{n+1}\int_{0}^{\pi/2}\tan\left({x\over 2}\right)\ln^{n+1}\left[\tan\left({x\over 2}\right)\right]{\mathrm dx\over \sin x}\tag3$$

$v=\tan\left({x\over 2}\right)$ $\implies$ $2\cos^2\left({x\over 2}\right)dv=dx$

$$2^{n+2}\int_{0}^{1}{v\over 1+v^2}\ln^{n+1}(v){\mathrm dv\over \sin x}\tag4$$

$\sin(x/2)={v\over \sqrt{1+v^2}}$, then $(4)$ becomes

$$2^{n+1}\int_{0}^{1}\sqrt{{1+v^2}}\ln^{n+1}(v)\mathrm dv \tag5$$

I am not sure how to proceed the next step.

How else can we confirm $(1)$?

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  • $\begingroup$ Are you sure the Bernoulli numbers appear in the denominator of the RHS? It looks really odd to me... By $(5)$ and differentiation under the integral sign, we just have to compute the derivatives at the origin of $$ I(\alpha) = \int_{0}^{1}\frac{v^{\alpha+1}}{(1+v^2)^{3/2}}\,dv $$ $\endgroup$ – Jack D'Aurizio Mar 9 '17 at 20:36
  • $\begingroup$ Use $v=e^{-x}$ to change variable and then try to get $$\int_0^\infty \sqrt{1+e^{-2x}}x^{n+1}e^{-x}dx.$$ Also use the Taylor series of $\sqrt{1+e^{-2x}}$. $\endgroup$ – xpaul Mar 9 '17 at 20:42
  • $\begingroup$ You right there @Jack D'. $\endgroup$ – gymbvghjkgkjkhgfkl Mar 10 '17 at 6:20
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Hint: For each natural number $n\in\mathbb{N}$, we have

$$\begin{align} I{\left(n\right)} &=\int_{0}^{\frac{\pi}{2}}\sqrt{\frac{1-\cos{\left(x\right)}}{1+\cos{\left(x\right)}}}\ln^{n+1}{\left(\frac{1-\cos{\left(x\right)}}{1+\cos{\left(x\right)}}\right)}\,\frac{\mathrm{d}x}{\sin{\left(x\right)}}\\ &=\int_{0}^{\frac{\pi}{2}}\frac{1}{\sin^{2}{\left(x\right)}}\sqrt{\frac{1-\cos{\left(x\right)}}{1+\cos{\left(x\right)}}}\ln^{n+1}{\left(\frac{1-\cos{\left(x\right)}}{1+\cos{\left(x\right)}}\right)}\,\sin{\left(x\right)}\,\mathrm{d}x\\ &=\int_{0}^{1}\frac{1}{1-t^{2}}\sqrt{\frac{1-t}{1+t}}\ln^{n+1}{\left(\frac{1-t}{1+t}\right)}\,\mathrm{d}t;~~~\small{\left[\cos{\left(x\right)}=t\right]}\\ &=\int_{0}^{1}\ln^{n+1}{\left(u\right)}\,\frac{\mathrm{d}u}{2\sqrt{u}};~~~\small{\left[\frac{1-t}{1+t}=u\right]}\\ &=\int_{0}^{1}\ln^{n+1}{\left(v^{2}\right)}\,\mathrm{d}v;~~~\small{\left[\sqrt{u}=v\right]}\\ &=2^{n+1}\int_{0}^{1}\ln^{n+1}{\left(v\right)}\,\mathrm{d}v.\\ \end{align}$$

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  • $\begingroup$ What part of this requires $n \in \mathbb{N}$? $\endgroup$ – Brevan Ellefsen Mar 9 '17 at 22:28
  • $\begingroup$ @BrevanEllefsen Strictly speaking, we don't really need to require $n$ be an integer. But because the logarithmic term is negative, taking non-integer powers of this term will require complex numbers in general. $\endgroup$ – David H Mar 9 '17 at 22:43

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