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Math formulas: Logarithmic definite integrals

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01xm(lnx)ndx=(1)nn!(m+1)n+1,m>1,n=0,1,2, \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
01lnx1+xdx=π212 \int^1_0 \frac{\ln x}{1+x} dx = -\frac{\pi^2}{12}
01lnx1xdx=π26 \int^1_0 \frac{\ln x}{1-x}dx = -\frac{\pi^2}{6}
01ln(1+x)xdx=π212 \int^1_0 \frac{\ln (1+x)}{x} dx = \frac{\pi^2}{12}
01ln(1x)xdx=π26 \int^1_0 \frac{\ln (1 - x)}{x} dx = - \frac{\pi^2}{6}
01lnxln(1+x)dx=22ln2π212 \int^1_0 \ln x \ln (1+x) \,dx = 2 - 2\ln 2 - \frac{\pi^2}{12}
01lnxln(1x)dx=2π26 \int^1_0 \ln x \ln (1 - x) \,dx = 2 - \frac{\pi^2}{6}
0xp1lnx1+xdx=π2csc(pπ)cot(pπ),0<p<1 \int^\infty_0 \frac{x^{p-1}\,\ln x}{1+x} dx = -\pi^2\, \csc (p\pi)\,\cot (p\pi), 0 < p < 1
01xmxnlnxdx=lnm+1n+1 \int^1_0 \frac{x^m - x^n}{\ln x} dx = \ln \frac{m+1}{n+1}
0exlnxdx=γ \int^\infty_0 e^{-x}\,\ln x\,dx = - \gamma
0ex2lnxdx=π4(γ+2ln2) \int^\infty_0 e^{-x^2} \ln x \, dx = -\frac{\sqrt{\pi}}{4} ( \gamma + 2\,\ln 2 )
0ln(ex+1ex1)dx=π24 \int^\infty_0 \ln \left(\frac{e^x+1}{e^x-1} \right) dx = \frac{\pi^2}{4}
0π/2ln(sinx)dx=0π/2ln(cosx)dx=π2ln2 \int^{\pi/2}_0 \ln (\sin x) dx = \int^{\pi/2}_0 \ln (\cos x) dx = -\frac{\pi}{2} \ln 2
0π/2(ln(sinx))2dx=0π/2(ln(cosx))2dx=π2(ln2)2+π324 \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}
0πxln(sinx)dx=π22ln2 \int^\pi_0 x\,\ln(\sin x) dx = -\frac{\pi^2}{2}\,\ln 2
0π/2sinxln(sinx)dx=ln21 \int^{\pi/2}_0 \sin x \ln (\sin x) dx = \ln 2 - 1
02πln(a+bsinx)dx=02πln(a+bcosx)dx=2πln(a+a2b2) \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)
0πln(a+bcosx)dx=πln(a+a2b22) \int^\pi_0 \ln (a + b\cos x) dx = \pi \ln \left( \frac{a + \sqrt{a^2 - b^2}}{2} \right)
0πln(a22abcosx+b2)dx={2πlnaab>02πlnbba>0 \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}
0π/4ln(1+tanx)dx=π8ln2 \int^{\pi/4}_0 \ln(1 + \tan x)dx= \frac{\pi}{8} \ln 2
0π2secxln(1+bcosx1+acosx)dx=12(arccos2aarccos2b) \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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