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Let

$$x_1={1 \over \tan(x)}$$

$$x_2={1+\tan^2(x)\over 2\tan(x)}$$

$$x_3={1+3\tan^2(x)\over 3\tan(x)+\tan^3(x)}$$

$$x_4={1+6\tan^2(x)+\tan^4(x)\over 4\tan(x)+4\tan^3(x)})$$

$$x_n={{n\choose 0}+{n\choose 2}t^2+{n\choose 4}t^4+\cdots\over {n\choose 1}t+{n\choose 3}t^3+{n\choose 5}t^5}, t=\tan(x)$$

$$\int_{0}^{\pi\over 4}{\ln(x_n)\over \cos^2(x)}\mathrm dx=F(n)\tag1$$

$$F(1)=1$$ $$F(2)={\pi\over 2}-1$$ $$F(3)=1-{\pi\over 3\sqrt{3}}$$ $$F(4)=(\sqrt{2}-1)\pi-1$$

I am unable to work out the closed form for $(1)$.

How do we evaluate the closed form for $(1)?$

gymbvghjkgkjkhgfkl
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  • Can you be more clear about what the pattern for $x_{n}$ is? – David H Oct 29 '17 at 20:05
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    Is it $$x_n={{n\choose 0}+{n\choose 2}t^2+{n\choose 4}t^4+\cdots\over {n\choose 1}t+{n\choose 3}t^3+{n\choose 5 }t^5+\color{red}{...}}, t=\tan(x)$$ –  Oct 30 '17 at 00:38

1 Answers1

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Removing the syntactic sugar, we want to compute

$$ \int_{0}^{\pi/4}\log\left(\frac{(1+\tan x)^n+(1-\tan x)^n}{(1+\tan x)^n-(1-\tan x)^n}\right)\frac{dx}{\cos^2 x}=\int_{0}^{1}2\,\text{arctanh}\left(\frac{1-t}{1+t}\right)^n\,dt $$ or $$ 4\int_{0}^{1}\frac{\text{arctanh } u^n}{(1+u)^2}\,du \stackrel{\text{IBP}}{=}2n\int_{0}^{1}u^{n-1}\cdot\frac{1-u}{1+u}\cdot\frac{du}{1-u^{2n}}$$ and the last integral can be computed by performing a partial fraction decomposition.
The douple pole at $u=-1$ is responsible for the rational part of the outcome (which is simpler to compute), then a combination of the arguments of the remaining poles completes the job.

Jack D'Aurizio
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