Math formulas: Logarithmic definite integrals


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$$ \int^1_0 x^m(\ln x)^n dx = \frac{(-1)^n n!}{(m+1)^{n+1}} , \quad m>-1,\, n=0,1,2,\dots $$

$$ \int^1_0 \frac{\ln x}{1+x} dx = -\frac{\pi^2}{12} $$

$$ \int^1_0 \frac{\ln x}{1-x}dx = -\frac{\pi^2}{6} $$

$$ \int^1_0 \frac{\ln (1+x)}{x} dx = \frac{\pi^2}{12} $$

$$ \int^1_0 \frac{\ln (1 - x)}{x} dx = - \frac{\pi^2}{6} $$

$$ \int^1_0 \ln x \ln (1+x) \,dx = 2 - 2\ln 2 - \frac{\pi^2}{12} $$

$$ \int^1_0 \ln x \ln (1 - x) \,dx = 2 - \frac{\pi^2}{6} $$

$$ \int^\infty_0 \frac{x^{p-1}\,\ln x}{1+x} dx = -\pi^2\, \csc (p\pi)\,\cot (p\pi), 0 < p < 1 $$

$$ \int^1_0 \frac{x^m - x^n}{\ln x} dx = \ln \frac{m+1}{n+1} $$

$$ \int^\infty_0 e^{-x}\,\ln x\,dx = - \gamma $$

$$ \int^\infty_0 e^{-x^2} \ln x \, dx = -\frac{\sqrt{\pi}}{4} ( \gamma + 2\,\ln 2 ) $$

$$ \int^\infty_0 \ln \left(\frac{e^x+1}{e^x-1} \right) dx = \frac{\pi^2}{4} $$

$$ \int^{\pi/2}_0 \ln (\sin x) dx = \int^{\pi/2}_0 \ln (\cos x) dx = -\frac{\pi}{2} \ln 2 $$

$$ \int^{\pi/2}_0 ( \ln (\sin x))^2 dx = \int^{\pi/2}_0 ( \ln (\cos x))^2 dx = \frac{\pi}{2}(\ln 2)^2 + \frac{\pi^3}{24} $$

$$ \int^\pi_0 x\,\ln(\sin x) dx = -\frac{\pi^2}{2}\,\ln 2 $$

$$ \int^{\pi/2}_0 \sin x \ln (\sin x) dx = \ln 2 - 1 $$

$$ \int^{2\pi}_0 \ln (a + b \sin x) dx = \int^{2\pi}_0 \ln (a + b \cos x) dx = 2\pi \ln\left(a + \sqrt{a^2-b^2} \right)$$

$$ \int^\pi_0 \ln (a + b\cos x) dx = \pi \ln \left( \frac{a + \sqrt{a^2 - b^2}}{2} \right) $$

$$ \int^\pi_0 \ln\left( a^2 - 2ab \cos x + b^2 \right) dx = \begin{cases} 2\pi\ln a & a \geq b > 0 \\ 2\pi \ln b & b \geq a > 0 \end{cases} $$

$$ \int^{\pi/4}_0 \ln(1 + \tan x)dx= \frac{\pi}{8} \ln 2 $$

$$ \int^\frac{\pi}{2}_0 \sec x \,\ln\left(\frac{1+b\cos x}{1+ a \cos x}\right) dx = \frac{1}{2}\left(\arccos^2 a - \arccos^2b\right)$$

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As for everything else, so for a mathematical theory: beauty can be perceived but not explained.

Arthur Cayley

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