A bridge between unit square and single integrals for real functions of the form \(f(x\cdot y)\) (Q2188796)

This paper deals with a simple formula that allows a direct conversion of unit square integrals of real functions of the form \(f(x,y)=f(x\cdot y )\) into single integrals involving just \(f(p)\cdot \log p\) where \(p=xy\). One has the following result: \begin{itemize} \item Let \(f: [0,1]\to \mathbb{R}\) be a function such that \(f(p) \ln p\) is Riemann-integrable on~\([0,1]\) and \(f(x\cdot y)\) is Riemann-integrable on the unit square \([0,1]\times [0,1]\). Then \[ \int^{1}_{0}\int^{1}_{0}f(x,y)\,dx\,dy=-\int^{1}_{0}f(p)\ln p\,dp. \] \end{itemize} The above formula can be applied to calculate several double integrals, some of which has been obtained by other means. An example is \[ \int^{\alpha}_{0}\int^{\alpha}_{0} \frac{1}{1-xy}\,dx\,dy=\operatorname{Li}_{2}(\alpha^{2}), \] where \(\operatorname{Li}_{2}\) is the dilogarithmic function. A similar integral that can be obtained is \[ \int^{\alpha}_{0}\int^{\beta}_{0} \frac{1}{1-x^{2}y^{2}}\,dx\,dy=\chi_{2}(\alpha\beta),\quad 0<\alpha,\beta\le 1, \] where \(\chi_{2}(\alpha)=\operatorname{Li}_{2}(\alpha)-\frac{1}{4}\operatorname{Li}_{2}(\alpha^{2})\). Another example is \[ \int^{1}_{0}\int^{1}_{0}\frac{9+xy}{9-x^{2}y^{2}}\,dx\,dy =\frac{\pi^{2}}{6}-\frac{\ln^{2}3}{2}. \] If \(G=\sum_{n\ge 0}\frac{(-1)^{n}}{(2n+1)^{2}}\) is Catalan's constant, it is known that \(G=-\int^{1}_{0}\frac{\ln p}{1+p^{2}}\,dp\) and this equality leads to \[ \int^{1}_{0}\int^{1}_{0}\frac{1}{1+x^{2}y^{2}}\,dx\,dy=G. \] The stated result also holds for Riemann improper integrals: \begin{itemize} \item Let \(f: [1,\infty)\to \mathbb{R}\) be a function such that \(f(p) \ln p\) is Riemann-integrable on~\([1,\infty)\) and \(f(x\cdot y)\) is Riemann-integrable on \([1,\infty)\times [1,\infty)\). Then \[ \int^{\infty}_{1}\int^{\infty}_{1}f(x\cdot y)\,dx\,dy=-\int^{\infty}_{1}f(p)\ln p\,dp. \] \end{itemize} As an application one gets \[ \int^{\infty}_{1}\int^{\infty}_{1} \frac{e^{-xy}}{xy\log (xy)}\,dx\,dy =-\operatorname{Ei}(-1), \] where \(\operatorname{Ei}(z)=-\int^{\infty}_{-z}\frac{e^{-t}}{t}\,dt\) is the Euler integral. One gives also a result to reduce multiple integrals over \([0,1]^{N}\) to single integrals over \([0,1]\): \begin{itemize} \item Let \(f: [0,1]\to \mathbb{R}\) be a continuous function such that \(f(p) \ln p\) is Riemann-integrable on~\([0,1]\). Then, for any integer \(N>1\), \[ \int^{1}_{0}\dotsi \int^{1}_{0}f(x_{1}x_{2}\dotsm x_{N})\,dx_{1}\,dx_{2}\dotsm dx_{N}= \frac{(-1)^{N-1}}{(N-1)!} \int^{1}_{0}f(p)(\ln p)^{N-1}\,dp. \] \end{itemize} The above theorem has a consequence for the Mellin transform of continuous functions: \begin{itemize} \item Let \(f: [0,1]\to \mathbb{R}\) be a continuous function such that \(f(p) \ln p\) is Riemann-integrable on~\([0,1]\). Then, \[ \sum^{\infty}_{N=1}C_{N}z^{N-1}=\int^{1}_{0}f(p) p^{-z}\,dp, \] holds for all complex \(z\) with \(|z|<1\), where \(C_{N}:=\int^{1}_{0}\dotsi \int^{1}_{0}f(x_{1}x_{2}\dotsm x_{N})\,dx_{1}dx_{2}\dotsm dx_{N}\). \end{itemize}