$$I(p,q)=\int_0^1 \left(\frac{x^p − x^q}{\ln x}\right)^2 dx $$
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?
On
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?
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}