\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x dx

In this short post we will estabilish the following result



\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x dx=-\pi^2\left(4\,\zeta'(-1)+\frac23\right)




\begin{aligned} I&=\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x dx\\ &=\int_0^\infty \ln\left(1+\frac{x^2}{4\,\pi^2}\right)\ln(1-e^{-x}) dx \qquad \left(x \to e^{-x} \right)\\ &=2 \pi\int_0^\infty \ln\left(1+x^2\right)\ln(1-e^{-2 \pi x}) dx \qquad \left(x \to 2 \pi x \right)\\ &=-4 \pi^2 \left(\ln(1-e^{-2 \pi x})\left( x\ln(1+x^2)-2x+2\arctan \left(x \right)\right)\Bigg|_0^\infty+\int_0^\infty \frac{x \ln(1+x^2)}{e^{2 \pi x}-1}\,dx-2\int_0^\infty \frac{x }{e^{2 \pi x}-1}\,dx+2\int_0^\infty \frac{\arctan (x) }{e^{2 \pi x}-1}\,dx \right) \qquad (*)\\ &=-4 \pi^2 \left(\int_0^\infty \frac{x \ln(1+x^2)}{e^{2 \pi x}-1}\,dx-2\int_0^\infty \frac{x }{e^{2 \pi x}-1}\,dx+2\int_0^\infty \frac{\arctan (x) }{e^{2 \pi x}-1}\,dx \right) \qquad (**)\\ &=-4 \pi^2\left(\zeta^\prime(-1)-\frac34+\frac{\ln 2 \pi}{2}-2\left(\frac{1}{4 \pi^2}\int_0^\infty \frac{e^{ -x} x }{1-e^{ -x}}\,dx\right)+2\left(\frac12-\frac{\ln 2 \pi}{4}\right)\right)\\ &=-4 \pi^2\left(\zeta^\prime(-1)+\frac14-2\left(\frac{1}{4 \pi^2}\sum_{k=1}^\infty\frac{1}{k^2}\int_0^\infty e^{ -x} x \,dx\right)\right)\\ &=-4 \pi^2\left(\zeta^\prime(-1)+\frac14-2\left(\frac{1}{4 \pi^2}\frac{\pi^2}{6}\right)\right)\\ &=-4 \pi^2\left(\zeta^\prime(-1)+\frac14-\frac{1}{12}\right)\\ &=-\pi^2\left(4\,\zeta'(-1)+\frac23\right) \qquad \blacksquare \end{aligned}


Where in (**) we used the previous established results (I and II):


\int_0^\infty \frac{x \ln(1+x^2)}{e^{2 \pi x}-1}\,dx=\zeta^\prime(-1)-\frac34+\frac{\ln 2 \pi}{2}


\int_0^\infty \frac{\arctan (x) }{e^{2 \pi x}-1}\,dx=\frac12-\frac{\ln 2 \pi}{4}

In (*) above we used the following result


\begin{aligned} \int \ln(z^2+x^2)\,dx&=uv-\int v\,du\\ &=x\ln(z^2+x^2)-2\int \frac{x^2}{z^2+x^2}\,dx\\ &=x\ln(z^2+x^2)-2\int \frac{z^2+x^2}{z^2+x^2}\,dx+2z^2\int\frac{1}{z^2+x^2}\,dx\\ &=x\ln(z^2+x^2)-2x+\frac{2z^2}{z}\arctan \left(\frac{x}{z} \right)\\ &=x\ln(z^2+x^2)-2x+2z\arctan \left(\frac{x}{z} \right) \qquad \blacksquare\\ \end{aligned}

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