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Evaluate this integration : $$\Omega = \int^1_0 \frac{\ln^3(1-x^2)}{1+x}dx$$ My attempt : \begin{align*} \Omega & =\int^1_0 \frac{\ln^3(1-x^2)}{1+x}dx\overset{t=\frac{1-x}{1+x}}{=}\int^1_0\frac{\ln^3\left({\frac{4t}{(t+1)^2}}\right)}{t+1}dt \\ &= \underbrace{\int^1_0\frac{\ln^3(4t)}{t+1}dt}_{I_1} -8\underbrace{\int^1_0\frac{\ln^3(1+t)}{1+t}dt}_{I_2} -6\underbrace{\int^1_0 \frac{\ln^2(4t)\ln(t+1)}{t+1}dt}_{I_3}+12\underbrace{\int^1_0\frac{\ln(4t)\ln(1+t)}{t+1}dt}_{I_4} \end{align*}

$$I_2 = \int^1_0\frac{\ln^3(1+t)}{1+t}dt = \frac{1}{4}\ln^4(2)$$ \begin{align*} I_1 = \int^1_0\frac{\ln^3(4t)}{t+1}dt &= \ln^3(4)\int^1_0\frac{1}{t+1}dt +\int^1_0\frac{\ln^3(t)}{t+1}dt + 3\ln^2(4)\int^1_0\frac{\ln(t)}{t+1}dt+3\ln(4)\int^1_0\frac{\ln^2(t)}{t+1}dt \\ &= \ln^3(4)\ln(2) -\frac{7}{120}\pi^4-\frac{\ln^2(4)}{4}\pi^2+\frac{9}{2}\zeta(3)\ln(4)\\&= 8\ln^4(2)-\frac{7}{120}\pi^4-\pi^2\ln^2(2)+9\zeta(3)\ln(2) \end{align*}

\begin{align*} I_3 = \int^1_0 \frac{\ln^2(4t)\ln(t+1)}{t+1}dt &= \ln^2(4)\int^1_0\frac{\ln(t+1)}{t+1}dt+\int^1_0 \frac{\ln^2(t)\ln(t+1)}{t+1}dt+2\ln(4)\int^1_0 \frac{\ln(t)\ln(t+1)}{t+1}dt\\ &= 2\ln^4(2) + \int^1_0 \frac{\ln^2(t)\ln(t+1)}{t+1}dt +2\ln(4)\int^1_0 \frac{\ln(t)\ln(t+1)}{t+1}dt \end{align*}

\begin{align*} I_4 &= \int^1_0\frac{\ln(4t)\ln(1+t)}{t+1}dt = \ln(4)\int^1_0 \frac{\ln(1+t)}{1+t}dt+\int^1_0\frac{\ln(1+t)\ln(t)}{1+t}dt\\&= \ln^3(2) +\int^1_0\frac{\ln(1+t)\ln(t)}{1+t}dt \end{align*} My question : Is there a way to evaluate these two integrals?

$$\int^1_0 \frac{\ln(t)\ln(1+t)}{1+t}dt $$ $$\int^1_0 \frac{\ln^2(t)\ln(1+t)}{1+t}dt$$

Quanto
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Delta
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2 Answers2

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To solve the original integral, start with substitution $x=\sqrt u$ $$I=\int_0^1\frac{\ln^3(1-x^2)}{1+x}dx=\int_0^1\frac{\ln^3(1-x)}{2\sqrt x(1+\sqrt x)}dx=\int_0^1\frac{\ln^3(1-x)}{1-x}\frac{1-\sqrt x}{2\sqrt x}dx$$ Now use integration by parts $$u=\frac{x^{-1/2}-1}{2}, \space dv=\frac{\ln^3(1-x)}{1-x}dx$$ $$du=-\frac{x^{-3/2}}{4},\space v=-\frac{1}{4}\ln^4(1-x)$$ $$I=-\frac{1}{16}\int_0^1x^{-3/2}\ln^4(1-x)dx=-\frac{1}{16}\lim_{(x,y)→(-1/2,1)}\left(\frac{\partial}{\partial y}\right)^4B(x,y)$$ Calculating the 4th derivative of the beta function by hand is tedious but doable. Use $$(\Gamma(x))'=\Gamma(x)\psi_0(x)$$ $$(\psi_n(x))'=\psi_{n+1}(x)$$ where $\Gamma(x)$ is the gamma function and $\psi_n(x)$ is the polygamma function

polychroma
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    $$-\frac{1}{16}\lim_{(x,y)→(-1/2,1)}\left(\frac{\partial}{\partial y}\right)^4B(x,y)=\frac{\sqrt\pi}8\left(\frac{\partial}{\partial y}\right)^4,\bigg|_{y=0}e^{\ln\Gamma(1+y)-\ln\Gamma \big(\frac12+y\big)}$$ and then we can use (or evaluate) $\psi^{(k)}(x)=\left(\frac d{dx}\right)^{k+1}\ln\Gamma(x)$ representations for $,x=1,,\frac12$ $$\psi^{(k)}(1)=(-1)^{k+1}k!\zeta(k+1);,,\psi^{(k)}(1/2)=(-1)^{k+1}k!(2^{k+1}-1)\zeta(k+1)$$ https://mathworld.wolfram.com/PolygammaFunction.html – Svyatoslav Sep 14 '24 at 04:04
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\begin{align}J&=\frac{\ln(1+x)\ln^2x}{1+x}dx=\frac{1}{3}\int_0^1\frac{\ln^3x}{1+x}dx-\frac{1}{3}\int_0^1\frac{\ln^3\left(1+x\right)}{1+x}dx+\\&\underbrace{\int_0^1\frac{\ln^2\left(1+x\right)\ln x}{1+x}dx}_{\text{IBP}}-\frac{1}{3}\underbrace{\int_0^1\frac{\ln^3\left(\frac{x}{1+x}\right)}{1+x}dx}_{u=\frac{x}{1+x}}\\ &=-\frac{7\zeta(4)}{4}-\frac{\ln^42}{12}-\frac{1}{3}\underbrace{\int_0^1\frac{\ln^3(1+x)}{x}dx}_{u=\frac{1}{1+x}}-\frac13\int_0^{\frac12}\frac{\ln^3u}{1-u}du\\ &=-\frac{7\zeta(4)}{4}-\frac{\ln^42}{12}+\frac{1}{3}\int_{\frac{1}{2}}^1\frac{\ln^3u}{u(1-u)}du-\frac13\int_0^{\frac12}\frac{\ln^3u}{1-u}du\\ &=-\frac{7\zeta(4)}{4}-\frac{\ln^42}{6}+\frac{1}{3}\int_{\frac{1}{2}}^1\frac{\ln^3u}{1-u}du-\frac13\int_0^{\frac12}\frac{\ln^3u}{1-u}du\\ &=-\frac{15\zeta(4)}{4}-\frac{\ln^42}{6}-\frac23\underbrace{\int_0^{\frac12}\frac{\ln^3u}{1-u}du}_{z=\frac z 2}\\ &=-\frac{15\zeta(4)}{4}-\frac{\ln^42}{6}-\frac{1}{3}\int_0^1\frac{\ln^3\left(\frac z2\right)}{1-\frac z2}dz\\ &=-\frac{15\zeta(4)}{4}-\frac{\ln^42}{6}-\frac13\int_0^1\frac{\ln^3z}{1-\frac z2}dz+\frac13\int_0^1\frac{\ln^32}{1-\frac z2}dz-\int_0^1\frac{\ln^22\ln z}{1-\frac z2}dz+\\&\int_0^1\frac{\ln 2\ln^2 z}{1-\frac z2}dz\\ &=-\frac{15\zeta(4)}{4}+\frac{\ln^42}{2}+4\text{Li}_4\left(\frac{1}{2}\right)-\left(\ln^2 2-\frac{\pi^2}{6}\right)\ln^2 2+\\&\left(\frac{7\zeta(3)}{2}+\frac{2}{3}\ln^3 2-\frac{\pi^2\ln 2}{3}\right)\ln 2\\ &=\boxed{4\text{Li}_4\left(\frac{1}{2}\right)-\frac{\pi^4}{24}+\frac{7\zeta(3)\ln 2}{2}-\frac{\pi^2\ln^22}6+\frac{\ln^42}6} \end{align}

NB: Same method is applicable to the integral $\displaystyle \int_0^1\frac{\ln(1+x)\ln x}{1+x}dx$

FDP
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