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I have recently taken an interest in evaluating logarithmic integrals and would really love practice problems and especially, theorems, series expansions, and identities that have helped any of ya’ll in evaluating integrals like these, especially ones with products of logarithms with different argument combined with ratios of polynomials.

To hopefully give a sense as to the level I am at in the evaluation of these, I can evaluate integrals like:

$$\int_{0}^{\frac{\pi}{4}}\log(\cos(x))\,dx$$

$$\int_{0}^{1}\frac{\log^n(x)}{x^2+1}\,dx$$

$$\int_{0}^{1}\log(x)\log(1\pm x)\,dx$$

$$\int_{0}^{1}\frac{\log(x)}{1-x^2}\,dx$$

$$\int_{0}^{\infty}\frac{\log(x^4+x^2+1)}{x^2+1}\,dx$$

$$\int_{0}^{\infty}\frac{\log(x)\sin(x)}{x}\,dx$$

$$\int_{0}^{\infty}\log(x)e^{-x^2}\,dx$$

$$\int_{0}^{1}\log(x)\arctan(x)\,dx$$

Anything is appreciated!

Tian Vlašić
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Person
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    A lot of logarithmic integrals on $[0,1]$ can be evaluated through the theory of coloured zeta values. Some other fun logarithmic integrals: $$\int_{0}^{\infty}\log\left(1+\frac{2\cos(x)}{x^2}+\frac{1}{x^4}\right), dx=4\pi W\left(\frac{1}{2}\right)$$ with $W$ being the principal branch of the Lambert $W$ function. Another family of neat examples are Herglotz-type integrals such as $$\int_{0}^{1} \frac{\log\left(1+x^{2+\sqrt{3}}\right)}{1+x} , dx=\frac{\pi^2}{12} (1-\sqrt{3})+\log (2) \log(1+\sqrt{3}).$$ – KStarGamer Jun 23 '23 at 23:56
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    Here you can find listed a few of them. But you can start with this one: $$\int_0^\infty \frac{\ln\left(\frac{1-x^2+x^4}{(1-x^2)^2}\right)}{(1+x^2)^2}dx=\frac{\pi}{2}\ln\left(\frac32\right)$$ – Zacky Jun 24 '23 at 00:13
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    Here is a classic family of integrals related to your first example: $$\int_0^{\pi/2}\ln \lvert\sin(mx)\rvert \cdot \ln \lvert\sin(nx)\rvert , dx = \frac{\pi^3}{24} \frac{\gcd^2(m,n)}{mn}+\frac{\pi\ln^2(2)}{2}$$ – Christian E. Ramirez Jun 24 '23 at 00:43
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    A trig-log integral$$\int_{0}^{\infty}\cos x\ln(1+\frac1{x^2}),dx=\pi(1-\frac1e)$$ – Quanto Jun 24 '23 at 00:56
  • One of my favorites is $$\int_0^\infty \frac{dx}{x}\log\left(\frac{x^2+2bx+1}{x^2+2ax+1}\right) = \begin{cases} \left(\cos^{-1}a\right)^2-\left(\cos^{-1}b\right)^2 & a,b\in[-1,1] \ \left(\cosh^{-1}b\right)^2-\left(\cosh^{-1}a\right)^2 & a,b \in [1,\infty)\end{cases}$$ – Ninad Munshi Jun 24 '23 at 01:32
  • $\int_{-\frac{1}{2}}^{0}\frac{1}{1+x^{2}}\ln\left(\frac{\sqrt{5}+1+2x}{\sqrt{5}-1-2x}\right)dx = \cot^{-1}\left(2\right)\operatorname{csch}^{-1}\left(2\right)$ – Accelerator Jun 24 '23 at 06:33

2 Answers2

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Below are a family of $\int_0^1\ln(\cdot)\ln(\cdot) dx$ integrals \begin{align} &\int_0^1 \ln^2 (1-x)\ dx=2 \\ &\int_0^1 \ln^2 (1+x)\ dx=2(1-\ln 2)^2\\ &\int_0^1 \ln^2 (1+x^2)\ dx=-4G+\ln^2 2+2\pi(\ln2-1)+8\\ &\int_0^1 \ln x\ln(1-x)\ dx=2 -\frac{\pi^2}6\\ &\int_0^1 \ln x\ln(1+x)\ dx=2 -\frac{\pi^2}{12}-2\ln2\\ &\int_0^1 \ln x\ln(1+x^2)\ dx=-2G-\frac\pi2-\ln2+4\\ &\int_0^1 \ln(1-x)\ln(1+x)\ dx=-\frac{\pi^2}6+\ln^22-2\ln2+2\\ &\int_0^1 \ln(1-x)\ln(1+x^2)\ dx=4-2G-\frac{5\pi^2}{48}-\frac{\pi}4(2-\ln2)+\frac14\ln^22-\ln2\\ &\int_0^1 \ln(1+x)\ln(1+x^2)\ dx=4-\frac{\pi^2}{48}-\frac{\pi}4(2-\ln2)+\frac74\ln^22-5\ln2\\ \end{align}

Quanto
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You should look into exponential generating functions. The technique is useful when dealing with logarithms raised to some powers. You will definetly like this paper which covers plenty of examples such as, $$\begin{align*}\int_0^\infty e^{-x}\log(x)\ dx&=-\gamma\\ \int_0^{\pi/2}\log^{n}(\tan(x))\ dx&=\left(\frac{\pi}{2}\right)^{n+1}|E_n| \\ \int_0^{\pi/2}\sin^{2n}(x)\log(\sin(x))\ dx&=\frac{\pi}{2^{2n+1}}\binom{2n}{n}(H_{2n}-H_{n}-\log(2))\end{align*}$$ where $E_n$ and $H_n$ are Euler and Harmonic numbers respecitely.

bob
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