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Let $H_n$ be the harmonic series. I want to find the value of $A=\displaystyle\sum_{n=0}^\infty \binom{2n}{n}\left(\frac{1}{4}\right)^n\frac{H_n}{n} $.

From this paper : https://cs.uwaterloo.ca/journals/JIS/VOL15/Boyadzhiev/boyadzhiev6.pdf , I found that $$\sum_{n=0}^\infty \binom{2n}{n}H_n x^n=\frac{1}{\sqrt{1-4x}}\ln\left({\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}}\right)=f(x)$$

So I get $A=\displaystyle\int_0^{1/4}\frac{f(x)}{x}dx$.

How to find the exact value of $$\int_0^{1/4}\frac{1}{x\sqrt{1-4x}}\ln\left({\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}}\right)dx$$.

Thank in advances.

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    $\begingroup$ Hint: Let $u=\sqrt{1-4x}$ and the rest should be pretty simple. $\endgroup$
    – M.N.C.E.
    Feb 1, 2015 at 13:01

2 Answers 2

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Hint :

by $$u=\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}$$

$$\Rightarrow \sqrt{1-4x}=\frac{1}{2u-1}$$

$$\Rightarrow x=\frac{1}{4}(1-\frac{1}{(2u-1)^2})$$

$$dx=\frac{1}{(2u-1)^3}du$$

you get $$I=\int_1^{\infty}\frac{\ln u}{u(u-1)}du=-\sum_{n=0}^{\infty}\int_1^{\infty}u^{n-1}\ln u du=-\sum_{n=0}^{\infty}-\frac{1}{n^2}=\frac{\pi^2}{6}$$

for the last integral use integration by part

notice that $$\frac{1}{u-1}=-\sum_{n=0}^{\infty}u^n$$

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  • $\begingroup$ $\frac{1}{u-1}=-\sum_{n=0}^{\infty}u^n, \quad |u|<1$ and the interval of the integral is from $1$ to $\infty$. To avoid this issue, let $u=1/y$ in the integral then expand in series. You will get $I=\int_0^1 \frac{\ln y}{y-1}dy=\sum_{n=1}^\infty \frac{1}{n^2}=\zeta(2)$ $\endgroup$ Jan 22, 2023 at 18:18
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Set $x=1$ in

$$ \sum_{n=1}^\infty \frac{{2n\choose n}}{4^n}\frac{H_n}{n} x^n=2\operatorname{Li}_2\left(\frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}\right)$$

we get $$\sum_{n=1}^\infty \frac{{2n\choose n}}{4^n}\frac{H_n}{n}=2\operatorname{Li}_2(1)=2\zeta(2)$$

For a different result, pick $x=-1$,

$$\sum_{n=1}^\infty (-1)^n\frac{{2n\choose n}}{4^n}\frac{H_n}{n}=2\operatorname{Li}_2\left(\frac{1-\sqrt{2}}{1+\sqrt{2}}\right)=2\operatorname{Li}_2(2\sqrt{2}-3).$$

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