Is there a closed form for: $\int_0^1 \left(\frac{x^p − x^q}{\ln x}\right)^2 dx $ $$I(p,q)=\int_0^1
\left(\frac{x^p − x^q}{\ln x}\right)^2 dx
$$
I have already prove that $I(p,q)$ converges iff $p,q\in(-\frac{1}{2},\infty)$
Now I need to fine the closed expression of $I(p,q)$ could anyone help?

My first attempt was to write

$$\frac{x^p − x^q}{\ln x}=\int_q^px^tdt$$
What next?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\mrm{I}\pars{p,q} & \equiv
\int_{0}^{1}\bracks{x^{p} − x^{q} \over \ln\pars{x}}^{2}\,\dd x =
\int_{0}^{1}\pars{x^{2p} − 2x^{p + q} + x^{2q}}\
\overbrace{\int_{0}^{\infty}x^{t}\,
t\,\dd t}^{\ds{{1 \over \ln^{2}\pars{x}}}}\ \,\dd x
\\[5mm] & =
\int_{0}^{\infty}t\int_{0}^{1}
\pars{x^{2p + t} - 2x^{p + q + t} + x^{2q + t}}\,\dd x\,\dd t
\\[5mm] & =
\int_{0}^{\infty}\pars{{t \over t + 2p + 1} -
2\,{t \over t + p + q + 1} + {t \over t + 2q + 1}}\,\dd t
\\[5mm] & =
\int_{0}^{\infty}\pars{-\,{2p + 1 \over t + 2p + 1} +
{2\pars{p +q + 1} \over t + p + q + 1} - {2q + 1 \over t + 2q + 1}}\,\dd t
\\[5mm] & =
\left.\ln\pars{\bracks{t + p + q + 1}^{2p + 2q + 2} \over
\bracks{t + 2p + 1}^{2p + 1}\bracks{t + 2q + 1}^{2q + 1}}
\right\vert_{\ t\ =\ 0}^{\ t\ \to\ \infty}
\\[5mm] & =
\bbx{\ln\pars{ \bracks{2p + 1}^{2p + 1}\bracks{2q + 1}^{2q + 1}\over
\bracks{p + q + 1}^{2p + 2q + 2}}}
\end{align}
A: Building off of your idea....
$$
\begin{align}
\int_0^1\left(\frac{x^p − x^q}{\ln x}\right)^2\,dx
&=\int_0^1\left(\int_q^px^tdt\right)^2\,dx\\
&=\int_0^1\left(\int_q^p x^t\,dt\right)\left(\int_q^px^s\,ds\right)\,dx\\
&=\int_0^1\int_q^p\int_q^p x^tx^s\,dt\,ds\,dx
\end{align}
$$
Now change the order of integration. Can you take it from here?
