Log Integral with zeta function

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Log Integral with zeta function

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#1 h Dec 14, 2024, 3:52 PM

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$\color{blue}{\int_0^1\ln^3x\left(\frac{\ln(1-x)}{1-x}-\frac{2\ln(1+x)}{1+x}\right)dx=\frac 98\zeta(5).}$

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#2 h Jun 7, 2025, 3:40 PM

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Found few more cool ones.
$\color{blue}{\int_0^\infty\frac{dx}{x^2+\sqrt{1+x^4}}=\frac{1}{6\sqrt\pi}\Gamma^2\left(\frac{1}{4}\right).}$ $\color{blue}{\lim_{x\to\infty}xe^{x^2}\int_x^\infty e^{-t^2}dt=\lim_{x\to\infty}xe^{-x^2}\int_0^x e^{t^2}dt=\frac 12.}$ $\color{blue}{\int_0^{\pi/2}\cos\left(2s\tan^{-1}\left(\frac{x}{\ln\cos x}\right)\right)\frac{dx}{(x^2+\ln^2(\cos x))^s}=\frac{\pi}{2\ln^{2s}2}.}$ $\color{blue}{\int_0^{\pi/2}\sqrt[8]{x^2+\ln^2(\cos x)}\sqrt{\frac 12+\frac 12\sqrt{\frac 12+\frac 12\sqrt{\frac{\ln^2(\cos x)}{x^2+\ln^2(\cos x)}}}}dx=\frac{\pi\sqrt[4]{\ln 2}}{2}.}$ $\color{blue}{\int_0^{\pi/2}\frac{1}{\sqrt[8]{x^2+\ln^2(\cos x)}}\sqrt{\frac 12+\frac 12\sqrt{\frac 12+\frac 12\sqrt{\frac{\ln^2(\cos x)}{x^2+\ln^2(\cos x)}}}}dx=\frac{\pi}{2\sqrt[4]{\ln 2}}.}$ $\color{blue}{\int_0^{\pi/2}\frac{\ln\cos x}{\tan x}\ln\left(\frac{\ln^2(\sin x)}{\pi^2+\ln^2(\sin x)}\right)dx=\frac 32\zeta(2).}$ $\color{blue}{\int_0^{\pi/2}\frac{\ln\cos x}{\tan x}\ln\left(\frac{\ln^2(\cos x)}{\pi^2+\ln^2(\sin x)}\right)dx=\zeta(2).}$ $\color{blue}{\int_0^{\pi/2}\frac{\ln\cos x}{\tan^2x}\ln\left(\frac{\ln^2(\cos x)}{\pi^2+\ln^2(\sin x)}\right)dx=3\zeta(3).}$ $\color{blue}{\prod_{n=1}^\infty(1-e^{-\pi n})=\frac{\sqrt[4]{\pi}}{\sqrt[8]{8}\Gamma(3/4)}e^{\pi/24}.}$ $\color{blue}{\prod_{n=1}^\infty(1-e^{-2\pi n})=\frac{\sqrt[4]{\pi}}{\sqrt{2}\Gamma(3/4)}e^{\pi/12}.}$ $\color{blue}{\prod_{n=1}^\infty(1-e^{-4\pi n})=\frac{\sqrt[8]{2}\sqrt[4]{\pi}}{2\Gamma(3/4)}e^{\pi/6}.}$ $\color{blue}{\sum_{n=0}^\infty\frac{(-1)^n(2n+1)!}{4^{2n+3}n!(n+2)!}=\frac{13}{8}-\phi.}$

This post has been edited 4 times. Last edited by Entrepreneur, Jun 8, 2025, 10:03 AM

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