calculate $\int_{0}^{\infty}\log^n(x)\log(1+\frac{b^{2}}{x^{2}})dx=? (n\in N^* , b\geq2)$

Question

calculate $$\int_{0}^{\infty}(\log(x))^{n}\log(1+\frac{b^{2}}{x^{2}})dx=? (n\in N^*, b\geq2)$$

for $n=1$:
\begin{align*} \int_{0}^{\infty}\log(x)\log(1+\frac{b^2}{x^2})dx&=\int_{0}^{\infty}\log(x)\log(\frac{x^2+b^2}{x^2})dx\\ &=\int_{0}^{\infty}\log(x)(\log(x^2+b^2)-\log(x^2))dx\\ &=\int_{0}^{\infty}\log(x)\{\log(x^2+b^2)\}_{0}^{b}dx\\ &=\int_{0}^{\infty}\log(x)\int_{0}^{b}\frac{2t}{x^2+t^2}dtdx\\ &=\int_{0}^{b}\int_{0}^{\infty}\frac{2t\log(x)}{t^2+x^2}dxdt\\ &=\int_{0}^{b}\frac{2t\pi\log(t)}{2t}dt\\ &=\pi\int_{0}^{b}\log(t)dt=\pi\{t\log(t)-t\}_{0}^{b}\\ &=\pi b(\log(b)-1)\\ \end{align*}

Wait for your help $n\geq2$

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}{\ln^{n}\pars{x} \over x^{2} + t^{2}}\,\dd x} = \int_{0}^{1}{\ln^{n}\pars{x} \over x^{2} + t^{2}}\,\dd x + \int_{1}^{0}{\ln^{n}\pars{1/x} \over \pars{1/x}^{2} + t^{2}}\, \pars{-\,{\dd x \over x^{2}}} \\[5mm] = &\ \int_{0}^{1}{\ln^{n}\pars{x} \over x^{2} + t^{2}}\,\dd x + {\pars{-1}^{n} \over t^{2}} \int_{0}^{1}{\ln^{n}\pars{x} \over x^{2} + 1/t^{2}}\,\dd x \\[5mm] = &\ \bbox[10px,#eef]{\int_{0}^{1}{\ln^{n}\pars{x} \over x^{2} + t^{2}}\,\dd x} + {\pars{-1}^{n} \over t^{2}}\pars{t \mapsto {1 \over t}} \end{align}

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\begin{align} &\bbox[10px,#eef]{\int_{0}^{1}{\ln^{n}\pars{x} \over x^{2} + t^{2}}\,\dd x} = -\,{1 \over t}\,\Im\int_{0}^{1}{\ln^{n}\pars{x} \over \ic t - x}\,\dd x = -\,{1 \over t}\,\Im\int_{0}^{\large -\ic/t}{\ln^{n}\pars{\ic tx} \over 1 - x}\,\dd x \\[5mm] = &\ {n \over t}\,\Im\int_{0}^{\large -\ic/t} \mrm{Li}_{2}'\pars{x}\ln^{n - 1}\pars{\ic tx}\,\dd x = -\,{n\pars{n - 1} \over t}\,\Im\int_{0}^{\large -\ic/t} \mrm{Li}_{3}'\pars{x}\ln^{n - 2}\pars{\ic tx}\,\dd x \\[5mm] = &\ {n\pars{n - 1}\pars{n - 2} \over t}\,\Im\int_{0}^{\large -\ic/t} \mrm{Li}_{4}'\pars{x}\ln^{n - 3}\pars{\ic tx}\,\dd x \\[5mm] &\ = \cdots = {n\pars{n - 1}\pars{n - 2}\cdots 1 \over t}\,\pars{-1}^{n + 1} \,\Im\int_{0}^{\large -\ic/t}\mrm{Li}_{n + 1}'\pars{x}\,\dd x \\[5mm] = &\ \pars{-1}^{n + 1}n!\,{\Im\mrm{Li}_{n + 1}\pars{-\ic/t} \over t} \end{align} Then, \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}{\ln^{n}\pars{x} \over x^{2} + t^{2}}\,\dd x} \\[5mm] = &\ {\pars{-1}^{n + 1}n! \over t}\,\Im\mrm{Li}_{n + 1}\pars{-\,{\ic \over t}} + {\pars{-1}^{n} \over t^{2}}{\pars{-1}^{n + 1}n! \over 1/t} \,\Im\mrm{Li}_{n + 1}\pars{-\,{\ic \over 1/t}} \\[5mm] = &\ n!\,{\pars{-1}^{n + 1}\,\Im\mrm{Li}_{n + 1}\pars{-\ic/t} - \Im\mrm{Li}_{n + 1}\pars{-\ic t} \over t} \end{align}

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\begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}\ln^{n}\pars{x} \ln\pars{1 + {b^{2} \over x^{2}}}\,\dd x} \\[5mm] = &\ 2n!\int_{0}^{b}\bracks{\pars{-1}^{n + 1} \,\Im\mrm{Li}_{n + 1}\pars{-\,{\ic \over t}} - \Im\mrm{Li}_{n + 1}\pars{-\ic t}}\,\dd t \\[5mm] = &\ \bbox[10px,#ffd]{2n!\pars{-1}^{n + 1}\,\Re\int_{-\infty\ic}^{-\ic/b} {\mrm{Li}_{n + 1}\pars{x} \over x^{2}}\,\dd x - 2n!\pars{-1}^{n + 1}\,\Re\int_{0}^{-\ic b}\mrm{Li}_{n + 1}\pars{x}\,\dd x} \\[5mm] = &\ \bbox[10px,#ffd]{2n!\pars{-1}^{n + 1}\, \mathcal{I}_{n + 1}\pars{-\,{\ic \over b}} - 2n!\pars{-1}^{n + 1}\,\mathcal{J}_{n + 1}\pars{\ic b}} \\[5mm] &\ \left.\mbox{where}\ \left\{\begin{array}{rclcl} \ds{\mathcal{I}_{m}\pars{z}} & \ds{\equiv} & \ds{\int_{-\infty\ic}^{z} {\mrm{Li}_{m}\pars{x} \over x^{2}}\,\dd x} & \ds{=} & \ds{-\,{\mrm{Li}_{m}\pars{z} \over z} + \mathcal{I}_{m - 1}\pars{z}} \\[2mm] \ds{\mathcal{J}_{m}\pars{z}} & \ds{\equiv} & \ds{\int_{0}^{z}\mrm{Li}_{m}\pars{x}\,\dd x} & \ds{=} & \ds{\phantom{-}z\,\mrm{Li}_{m}\pars{z} - \mathcal{J}_{m - 1}\pars{z}} \end{array}\right.\right\vert_{\ m\ \geq\ 2} \end{align}

Answer 2

$$\int _0^{\infty }\ln ^n\left(x\right)\:\ln \left(1+\frac{b^2}{x^2}\right)\:dx=\int _0^{\infty }\ln ^n\left(x\right)\:\ln \left(b^2+x^2\right)\:dx-2\int _0^{\infty }\ln ^{n+1}\left(x\right)\:dx$$ Now consider the following definition of the beta function $$\int _0^{\infty }\frac{x^{\alpha -1}}{\left(1+x\right)^{\beta }}\:dx=\text{B}\left(\alpha ,\beta -\alpha \right)$$ Now using the sub $x=\frac{x^2}{b^2}$ gets $$\int _0^{\infty }\frac{x^{2\alpha -1}}{\left(b^2+x^2\right)^{\beta }}\:dx=\frac{1}{2}b^{2\alpha -2\beta }\text{B}\left(\alpha ,\beta -\alpha \right)$$ Using these definitions gets us $$\int _0^{\infty }\ln ^n\left(x\right)\:\ln \left(b^2+x^2\right)\:dx=-\frac{1}{2^{n+1}}\lim_{\alpha\rightarrow 1/2 \\\beta\rightarrow 0}\frac{\partial ^{n+1}}{\partial \alpha ^n\partial \beta }b^{2\alpha -2\beta }\text{B}\left(\alpha ,\beta -\alpha \right)$$ $$-2\int _0^{\infty }\ln ^{n+1}\left(x\right)\:dx=-2\lim_{\alpha\rightarrow 1 \\\beta\rightarrow 0}\frac{\partial ^{n+1}}{\partial \alpha ^{n+1}}\text{B}\left(\alpha ,\beta -\alpha \right)$$ Collecting both results gives us the not so nice generalizaton for your integral $$\int _0^{\infty }\ln ^n\left(x\right)\:\ln \left(1+\frac{b^2}{x^2}\right)\:dx$$ $$=-\frac{1}{2^{n+1}}\lim_{\alpha\rightarrow 1/2 \\\beta\rightarrow 0}\frac{\partial ^{n+1}}{\partial \alpha ^n\partial \beta }b^{2\alpha -2\beta }\text{B}\left(\alpha ,\beta -\alpha \right)-2\lim_{\alpha\rightarrow 1 \\\beta\rightarrow 0}\frac{\partial ^{n+1}}{\partial \alpha ^{n+1}}\text{B}\left(\alpha ,\beta -\alpha \right)$$ Putting this on mathematica for $n=1$ $b=1$ gives me $$-\frac{1}{4}\left(2\pi \psi \left(-\frac{1}{2}\right)-2\pi \psi \left(\frac{1}{2}\right)\right)-0$$ $$=-\frac{1}{4}\left(4\pi -2\pi \gamma -4\pi \ln \left(2\right)+2\pi \gamma +4\pi \ln \left(2\right)\right)=-\pi $$ Which agrees with the result mentioned by FDP.