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I was experimenting with series and numerically found this gem:

$$S=\sum_{n=0}^\infty \frac{2^{2n+1}}{(2n+1)^2 \binom{4n+2}{2n+1}}= \frac14 \left(\frac{\pi^2}{4}+\log^2 (2+\sqrt{3} ) \right)$$

Or, rewriting in hypergeometric form:

$${_4 F_3} \left(\frac12, \frac12, 1, 1; \frac34, \frac54, \frac32; \frac14 \right)= \frac14 \left(\frac{\pi^2}{4}+\log^2 (2+\sqrt{3} ) \right)$$

How can we prove this result?

I know that:

$$\int_0^\infty \frac{t^{2n}}{(1+t)^{4n+2}}dt=\frac{2}{(2n+1) \binom{4n+2}{2n+1}}$$

This gives us:

$$S=\int_0^\infty \frac{dt}{(1+t)^2} \sum_{n=0}^\infty \frac{(2t)^{2n}}{(2n+1) (1+t)^{4n}}$$

$$S=\frac12 \int_0^\infty \frac{dt}{t} \tanh^{-1} \left(\frac{2t}{(1+t)^2} \right)$$

Not sure how to find the closed form from here.

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$$I=\int_0^\infty \operatorname{actanh}\left(\frac{2x}{(1+x)^2}\right)\frac{dx}{x}=\frac12 \int_0^\infty \frac{\ln(1+4x+x^2)-\ln(1+x^2)}{x}dx$$ Now we will consider the following integral: $$I(a)=\frac12 \int_0^\infty \frac{\ln(1+ax+x^2)-\ln(1+x^2)}{x}dx\Rightarrow I'(a)=\frac12 \int_0^\infty \frac{dx}{1+ax+x^2}$$ $$=\frac{1}{\sqrt{4-a^2}}\arctan\left(\frac{a+2x}{\sqrt{4-a^2}}\right)\bigg|_0^\infty =\frac{1}{\sqrt{4-a^2}}\arctan\left(\frac{\sqrt{4-a^2}}{a}\right)$$ We have $I(0)=0$ and we are looking to find $I(4)$, then: $$I=\int_0^4 \frac{1}{\sqrt{4-a^2}}\arctan\left(\frac{\sqrt{4-a^2}}{a}\right)da=-\frac12 \arctan^2\left(\frac{\sqrt{4-a^2}}{a}\right)\bigg|_0^4$$ $$=\lim_{a\to 4}\frac12 \operatorname{arctanh}^2\left(\frac{\sqrt{a^2-4}}{a}\right)+\lim_{a\to 0}\frac12 \arctan^2\left(\frac{\sqrt{4-a^2}}{a}\right)$$ $$\Rightarrow \boxed{\int_0^\infty \operatorname{actanh}\left(\frac{2x}{(1+x)^2}\right)\frac{dx}{x}=\frac12 \ln^2(2+\sqrt 3)+\frac{\pi^2}{8}}$$


Another way to differentiate under the integral sign: $$I= \int_0^\infty \operatorname{arctanh} \left(\frac{2t}{(1+t)^2} \right)\frac{dt}{t}\overset{t=\tan \frac{x}{2}}=\int_0^\pi \operatorname{arctanh} \left(\frac{\sin x}{\sin x+1}\right)\frac{dx}{\sin x}$$ $$\operatorname{arctanh} x=\frac12 \ln\left(\frac{1+x}{1-x}\right)\Rightarrow I=\frac12 \int_0^\pi \frac{\ln(1+2\sin x)}{\sin x}dx$$ $$I(a)=\frac12 \int_{0}^\pi \frac{\ln(1+\sin a\sin x)}{\sin x}dx\Rightarrow I'(a)=\frac12 \int_0^{\pi}\frac{\cos a}{1+\sin a\sin x}dx$$ $$\overset{\tan \frac{x}{2}=t}=\int_0^\infty \frac{\cos a}{(t+\sin a)^2+\cos^2 a}dt=\arctan\left(\frac{t+\sin a}{\cos a}\right)\bigg|_0^\infty =\frac{\pi}{2}-a$$ $$I(0)=0\Rightarrow I(a)=\int_0^a \left(\frac{\pi}{2}-x\right)dx=\frac{a}{2}(\pi-a)$$ $$\Rightarrow \boxed{I=\frac{\arcsin 2}{2}(\pi-\arcsin 2)=\frac{\pi^2}{8}+\frac12 \ln^2(2+\sqrt 3)}$$

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