TITLE: How to prove that $\int_0^1\left(\sum\limits_{k=n}^\infty {x^k\over k}\right)^2\,dx = \int_0^1 2x^{n-1}\log\left(1+{1\over\sqrt{x}}\right)\,dx$
QUESTION [30 upvotes]: American Mathematical Monthly problem 11611 essentially asks you to show that 
$$\lim_n\ n  \int_0^1\left(\sum_{k=n}^\infty {x^k\over k}\right)^2\,dx=2\log(2).\tag1$$ 
This would follow easily from (2) below, which is true for small values of $n$
 according to Maple. 
But I couldn't prove equation (2) in general, so I found a direct solution for (1) instead.   
$$\int_0^1\left(\sum_{k=n}^\infty {x^k\over k}\right)^2\,dx = \int_0^1 2x^{n-1}\log\left(1+{1\over\sqrt{x}}\right)\,dx.\tag2$$
But I'm still curious about (2). What am I missing? How can equation (2) be proven?

REPLY [12 votes]: One can avoid $\psi$ functions and even the expansion of the logarithm (right, for the logarithm, I am kind of cheating, try to find where...). Call $L_n$ and $R_n$ the LHS and the RHS of (2).

Expand the square of the series in $L_n$ and integrate term by term the resulting double series to get
$$
L_n=\sum_{i\geqslant n}\sum_{j\geqslant n}\frac1{ij(i+j+1)}.
$$
Since $\color{red}{\dfrac{i+1}{j(j+i+1)}=\dfrac1j-\dfrac1{i+j+1}}$,
$$
L_n=\sum_{i\geqslant n}\frac1{i(i+1)}\sum_{j\geqslant n}\left(\frac1j-\frac1{i+j+1}\right)=\sum_{i\geqslant n}\frac1{i(i+1)}\sum_{j\geqslant n}\frac1j[i\geqslant j-n].
$$
Exchanging the order of summations and using the identity $\color{red}{\dfrac1{i(i+1)}=\dfrac1i-\dfrac1{i+1}}$, one gets
$$
L_n=\sum_{j\geqslant n}\frac1j\frac1{\max(n,j-n)}=\sum_{j=n}^{2n-1}\frac1j\frac1n+\sum_{j\geqslant2n}\frac1{j(j-n)}.
$$
Since $\color{red}{\dfrac{n}{j(j-n)}=\dfrac1{j-n}-\dfrac1j}$,
$$
L_n=\frac1n\sum_{j=n}^{2n-1}\frac1j+\frac1n\sum_{j\geqslant2n}\left(\frac1{j-n}-\frac1j\right)=\frac2n\sum_{j=n}^{2n-1}\frac1j.
$$
Use the change of variables $x=t^2$ and an integration by parts to get
$$
nR_n=2\int_0^1\frac{1+t^{2n-1}}{1+t}\mathrm dt.
$$
Compute the differences
$$
(n+1)L_{n+1}-nL_n=2\left(\frac1{2n+1}-\frac1{2n}\right),
$$
and
$$
(n+1)R_{n+1}-nR_n=2\int_0^1\frac{t^{2n+1}-t^{2n-1}}{1+t}\mathrm dt=2\int_0^1(t^{2n}-t^{2n-1})\mathrm dt=2\left(\frac1{2n+1}-\frac1{2n}\right).
$$
Compute the initial terms $L_1=R_1=2$ and conclude that $nL_n=nR_n$ for every $n\geqslant1$.