How to prove that $\int_0^1\left(\sum\limits_{k=n}^\infty {x^k\over k}\right)^2\,dx = \int_0^1 2x^{n-1}\log\left(1+{1\over\sqrt{x}}\right)\,dx$ 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? 
 A: I did some symbol calculation on MATHEMATICA, and I found equation (2) might be deduced in such a way:
1)Let $F(n)=2\int_{0}^{1}x^{n-1}\log(1+1/\sqrt{x})dx$
then $F(n)=\frac{1+n\log(4)}{n^2}-\frac{2}{n}\Phi(-1,1,2n+1)$
$=\frac{1+n\log(4)}{n^2}+\frac{1}{n}(\psi(1/2+n)-\psi(1+n))$
where $\Phi(z,s,a)$ is the Lerch transcendent,$\psi(z)$ is the polygamma function.
2)LHS is equal to the double series $G(n)=\sum_{i,j\geq n}\frac{1}{ij(i+j+1)}$
then $G(n)-G(n+1)=\frac{1}{n^2(2n+1)}+2\sum_{i=n+1}^{\infty}\frac{1}{ni(i+n+1)}$
$=\frac{1}{n^2(2n+1)}-2\frac{(\psi(1+n)-\psi(2+2n))}{n(n+1)}$
If $G(n)-G(n+1)-(F(n)-F(n+1))=0$,then induction works.
3)The last step is to prove an equality of polygamma function.I use MATHEMATICA's FullSimplify function, and the result is 0.
This is not a rigorous proof. I hope this will help.
A: Let $\mathcal{R}_n$ denote the integral on the right-hand-side of eq. (2):
$$
    \mathcal{R}_n = \int_0^1 2 x^{n-1} \log\left(1+\frac{1}{\sqrt{x}}\right) \mathrm{d}x
$$
Consider
$$
\begin{eqnarray}
 (n+1) \mathcal{R}_{n+1} - n \mathcal{R}_n &=& \int_0^1 2 \left( (n+1) x^{n} - n x^{n-1} \right) \log\left(1+\frac{1}{\sqrt{x}}\right) \mathrm{d}x \\
   &=& \int_0^1 2 \log\left(1+\frac{1}{\sqrt{x}}\right) \mathrm{d} \left( -x^n \left(1-x\right)\right) \\
    &=&  \int_0^1 \left(\sqrt{x}-1\right)x^{n-1} \mathrm{d} x = \frac{2}{2n+1} - \frac{1}{n} 
\end{eqnarray}
$$
Therefore:
$$
   n \mathcal{R}_n = \mathcal{R}_1 + \sum_{m=1}^{n-1} \left( \frac{2}{2m+1} - \frac{1}{m} \right) = \psi\left(n+\frac{1}{2}\right) - \psi(n) + 2 \left(\log(2)-1\right) + \mathcal{R}_1
$$
Integral $\mathcal{R}_1$ can be easily integrated by parts:
$$
  \mathcal{R}_1 = 2 \int_0^1 \log\left(1+\frac{1}{\sqrt{x}}\right)\mathrm{d} x\stackrel{\text{by parts}}{=} \left. 2\left(  x \log\left( 1 + \frac{1}{\sqrt{x}}\right) + \sqrt{x} - \log\left(1+\sqrt{x}\right) \right) \right|_{x \downarrow 0}^{x = 1} = 2
$$
Thus
$$
  n \mathcal{R}_n = 2 \log(2) + \psi\left(n+\frac{1}{2}\right) - \psi(n)
$$
Similarly, denoting $\mathcal{L}_n = \int_0^1 f_n(x)^2 \mathrm{d} x$ the integral on the left-hand-side of eq. (2):
$$ \begin{eqnarray}
  n \left(\mathcal{L}_{n+1} - \mathcal{L}_n\right) &=& \int_0^1 \left( n f_{n+1}(x)^2 - n \left( \frac{x^n}{n} + f_{n+1}(x) \right)^2 \right)\mathrm{d} x \\
   &=& \int_0^1 \left( -2 x^n f_{n+1}(x) - \frac{x^{2n}}{n}\right)\mathrm{d} x \\
   &=& \color\maroon{2 \int_0^1 x^n \log(1-x) \mathrm{d} x} + {\color\blue{2 \int_0^1 x^n \sum_{k=1}^{n} \frac{x^k}{k} \mathrm{d} x}} - \frac{1}{n(2n+1)} \\
   &=& \color\maroon{-\frac{2}{n+1} H_{n+1}} + 2 \sum_{k=1}^n \frac{1}{k(k+n+1)} - \frac{1}{n(2n+1)} \\
   &=& \frac{2}{n+1} \left( \psi(n+1) - \psi(2+2n) \right) - \frac{1}{n(2n+1)} \\
   &=& \frac{1}{n (n+1)} \left( \psi(n+1)-\psi\left(n+\frac{3}{2}\right)-2 \log (2) \right) -\frac{1}{n^2 (2 n+1)}
\end{eqnarray}
$$
where $f_n(x) = \sum_{k=n}^\infty \frac{x^k}{k}$. Now since $\mathcal{L}_1 = \int_0^1 \log^2(1-x)\mathrm{d} x = 2$, and since $\mathcal{R}_{n+1} - \mathcal{R}_n$ equals to 
$\mathcal{L}_{n+1} -\mathcal{L}_n$, we establish that $\mathcal{L}_n = \mathcal{R}_n$:
$$
  \mathcal{R}_{n+1} - \mathcal{R}_n = 2\log(2) \left(  \frac{1}{n+1}- \frac{1}{n} \right) 
   + \frac{1}{n+1} \left( \psi\left(n+\frac{3}{2}\right) - \psi (n+1)\right) - 
     \frac{1}{n} \left( \psi\left(n+\frac{1}{2}\right) - \psi (n)\right) \stackrel{\color\maroon{\text{use recurrence equation for } \psi}}{=} \mathcal{L}_{n+1} - \mathcal{L}_n
$$
A: Here's a way to do it by brute force. First write
$$
\begin{align}
\int_0^1\left(\sum_{k = n}^\infty \frac{x^k}{k}\right)^2\,dx & = \sum_{k,m\geq n}\frac{1}{km}\int_0^1x^{k+m}\,dx \\
& = \sum_{k,m \geq n} \frac{1}{km(k+m+1)}.
\end{align}
$$
Put $r = k+m$, so that $m = r-k$, and transform the sum:
$$
\begin{align}
\sum_{k,m\geq n} \frac{1}{km(k+m+1)} &= \sum_{r = 2n}^\infty \frac{1}{r+1}\sum_{k = n}^{r-n}\frac{1}{k(r-k)} \\
&= \sum_{r = 2n}^\infty \frac{1}{r(r+1)}\sum_{k=n}^{r-n}\left(\frac{1}{k} + \frac{1}{r-k}\right) \\
& = \sum_{r = 2n}^\infty \frac{2}{r(r+1)}\sum_{k = n}^{r-n}\frac{1}{k} \\
& = 2\sum_{r = 2n}^\infty\left(\frac{1}{r}\sum_{k = n}^{r-n}\frac{1}{k} - \frac{1}{r+1}\sum_{k = n}^{r-n}\frac{1}{k}\right) \\
& = 2\sum_{r = 2n}^\infty\left(\frac{1}{r}\sum_{k = n}^{r - n}\frac{1}{k} - \frac{1}{r+1}\sum_{k = n}^{r+1 - n} \frac{1}{k}\right) + 2 \sum_{r = 2n}^\infty \frac{1}{(r+1)(r+1 - n)}. \\
\end{align}
$$
The first sum in the last line telescopes, so it can be evaluated as
$$
2\sum_{r = 2n}^\infty \left(\frac{1}{r}\sum_{k = n}^{r - n}\frac{1}{k} - \frac{1}{r+1}\sum_{k = n}^{r+1 - n}\frac{1}{k}\right) = \lim_{r \to \infty} \left(\frac{1}{n^2} - \frac{2}{r+1}\sum_{k = n}^{r+1 - n}\frac{1}{k}\right) = \frac{1}{n^2}.
$$
Thus we need to prove that
$$
\frac{1}{n^2} + 2\sum_{r = 2n}^\infty \frac{1}{(r+1)(r+1 - n)} = 2\int_0^1 x^{n-1}\log(1+x^{-1/2})\,dx. $$
Since
$$
\begin{align}
2\int_0^1 x^{n-1}\log(1+x^{-1/2})\,dx &= 2\int_0^1 x^{n-1}\log(1 + x^{1/2})\,dx - \int_0^1 x^{n-1}\log{x}\,dx \\
& = 2\int_0^1 x^{n-1}\log(1+x^{1/2})\,dx + \frac{1}{n^2},
\end{align}
$$
we need only prove that
$$
\sum_{r = 2n}^\infty \frac{1}{(r+1)(r+1 - n)} = \int_0^1 x^{n-1}\log{(1+x^{1/2})}\,dx.
$$
This can be done by developing $\log{(1+x^{1/2})}$ in powers of $x^{1/2}$:
$$
\begin{align}
\int_0^1 x^{n-1}\log{(1+x^{1/2})}\,dx & = \sum_{m = 1}^\infty\frac{(-1)^{m+1}}{m} \int_0^1 x^{n- 1 + m/2}\,dx \\ 
& = 2\sum_{m = 1}^\infty \frac{(-1)^{m+1}}{m(2n+m)} \\
& = 2 \sum_{\text{$m$ odd}} - 2\sum_{\text{$m$ even}} \frac{1}{m(2n+m)} \\
& = 2 \sum - 4\sum_{\text{$m$ even}} \frac{1}{m(2n+m)} \\
& = 2 \sum_{m = 1}^\infty \frac{1}{m(2n+m)} - \sum_{m = 1}^\infty \frac{1}{m(n+m)} \\
& = 2 \sum_{m = 1}^\infty \frac{1}{m}\left(\frac{1}{2n+m} - \frac{1}{2n+2m}\right) \\
& =  \sum_{m = 1}^\infty \frac{1}{(2n+m)(n+m)} \\
& = \sum_{r = 2n}^\infty \frac{1}{(r + 1)( r +1 - n)},
\end{align}
$$
which proves the identity. The other methods are obviously more concise, but after working through this I couldn't resist posting it.
A: 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 $n\geqslant1$.

