Integral $\int_0^{1/2}\arcsin x\cdot\ln^3x\,dx$

Question

It's a follow-up to my previous question. Can we find an anti-derivative $$\int\arcsin x\cdot\ln^3x\,dx$$ or, at least, evaluate the definite integral $$\int_0^{1/2}\arcsin x\cdot\ln^3x\,dx$$ in a closed form (ideally, as a combination of elementary functions and polylogarithms)?

Answer 1

As can be checked by differentiation, there is an antiderivative continuous on $(0,1)$:

$$\begin{align}&\int\arcsin x\cdot\ln^3x\,dx=\\ &\hspace{1cm}\frac32\left[\operatorname{Li}_3\left(\frac{\alpha}2\right)-\operatorname{Li}_3\left(\frac\beta2\right)\right]+3\,(2-\ln x)\cdot\operatorname{Li}_2\left(\frac\alpha2\right)-\frac{\ln^3\alpha}2+24\,\beta\\ &\hspace{1cm}+3\,(\ln x-1)\cdot\ln^2\alpha-\left(\pi^2+12\ln^2x-6\ln^22+24\ln2-72\right)\cdot\frac{\ln\alpha}4\\ &\hspace{1cm}-\beta\,\ln^3x+3\,(2\,\beta+\ln2)\cdot\ln^2x+\left(\frac{\pi^2}2-18\,\beta-3\ln^22\right)\cdot\ln x\\ &\hspace{1cm}+x\left(\ln^3x-3\ln^2x+6\ln x-6\right)\cdot\arcsin(x)\color{gray}{+C},\end{align}$$ where $$\alpha=1+\sqrt{1-x^2},\quad\beta=1-\sqrt{1-x^2}.$$

Here is an outline of an approach leading to this result:

• Integrate by parts to get rid of $\arcsin$.
• Change variable $y=\sqrt{1-x^2}$ to get rid of $\sqrt{1-x^2}$ in the denominator.
• Use identity $\ln(1-y^2)=\ln(1+y)+\ln(1-y)$ and expand parentheses, this will result in a sum of integrals with powers and products of $\ln(1+y),\, \ln(1-y)$ terms.
• Evaluate those integrals in terms of polylogarithms using CAS, WolframAlpha or integral tables.
• Simplify dilogarithm terms.

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Bonus:

$$\begin{align}&\int\arcsin x\cdot\ln^4x\,dx=\\ &\hspace{1cm}120\alpha-\ln^4\alpha-6\operatorname{Li}_4\left(\frac\alpha2\right) +6\operatorname{Li}_4\left(\frac\beta2\right)-3\operatorname{Li}_4\left(-\frac{\alpha^2}{x^2}\right)\\ &\hspace{1cm}+6\,(\ln x-2)\cdot\left[\operatorname{Li}_3\left(\frac\alpha2\right)-\operatorname{Li}_3\left(\frac\beta2\right)\right] -6\,\left(\ln^2x-4\ln x+6\right)\cdot\operatorname{Li}_2\left(\frac\alpha2\right)\\ &\hspace{1cm}+(\alpha-4)\cdot\ln^4x+(4\ln x+4-2\ln2)\cdot\ln^3\alpha\\ &\hspace{1cm}-\left[\vphantom{\Large|}6\,(\ln x+8-2\ln2)\cdot\ln x+\pi^2-36\right]\cdot\frac{\ln^2\alpha}2\\ &\hspace{1cm}+\left[\vphantom{\Large|}(\ln x+2-\ln2)\cdot\pi^2+6\,(4-\ln2)\cdot\ln^2x\right]\cdot\ln\alpha\\ &\hspace{1cm}+\left[\vphantom{\Large|}6\zeta(3)-96+2\ln^32-12\ln^22+36\ln2\right]\cdot\ln\alpha\\ &\hspace{1cm}+\left[\vphantom{\Large|}96\beta-6\zeta(3)-2\ln^32+24\ln^22+(\ln2-4)\cdot\pi^2\right]\cdot\ln x\\ &\hspace{1cm}-3\left(12\beta+\ln^22+8\ln2\right)\cdot\ln^2x+8\,(\beta+\ln2)\cdot\ln^3x\\ &\hspace{1cm}+x\left(\ln^4x-4\ln^3x+12\ln^2x-24\ln x+24\right)\cdot\arcsin(x)\color{gray}{+C},\end{align}$$ where $$\alpha=1+\sqrt{1-x^2},\quad\beta=1-\sqrt{1-x^2}.$$

Answer 2

The indefinite integral can be expressed in terms of hypergeometrics.

We start with integration by parts with $dv=\sin^{-1}(x)$ and $u=\log^3x$.

$$\begin{align} I&=\int \arcsin x \cdot \ln^3x\,dx\\ &=\ln^3x\cdot\left(\sqrt{1-x^2}+x\arcsin x\right)-\int \left(\sqrt{1-x^2}+x\arcsin x\right)\cdot \frac{3\ln^2x}{x}\,dx \end{align}$$

Expanding the integrand we get

$$\frac{3\sqrt{1-x^2}\cdot\ln^2x}{x}+3\arcsin(x)\ln^2(x)$$

The integration of the second term is addressed in your previous question so I will focus on just the first term.

$$K=\int\frac{3\sqrt{1-x^2}\ln^2x}{x}\,dx$$

Mathematica gives the result $K=$

$$\frac{3\sqrt{1-x^2}}{x\sqrt{1-\frac{1}{x^2}}} \left[2 x \, _4F_3\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2};\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{1}{x^2}\right)-2 x \log (x) \, _3F_2\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{1}{x^2}\right)+x \sqrt{1-\frac{1}{x^2}} \log ^2(x)+\log ^2(x) \csc ^{-1}(x)\right]$$

Therefore

$$I=\ln^3x\cdot\left(\sqrt{1-x^2}+x\arcsin x\right)-K-3\int\arcsin x\ln^2x\,dx$$

Here are the sum representations of the hypergeometrics

$$\, _3F_2\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{1}{x^2}\right)=-\frac{1}{\sqrt \pi}\sum_{k=0}^\infty\frac{x^{-2k}}{(2k-1)^3}\cdot\frac{\Gamma\left(k+\frac 1 2\right)}{\Gamma(k+1)}$$

$$\, _4F_3\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2};\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{1}{x^2}\right)=\frac{1}{\sqrt \pi}\sum_{k=0}^\infty \frac{x^{-2k}}{(2k-1)^4}\cdot\frac{\Gamma\left(k+\frac 1 2\right)}{\Gamma(k+1)}$$

Answer 3

We will outline of a way forward leaving some of the work to the reader.

Denote the integral of interest by $I$ where

$$I=\int \arcsin(x) \log^3(x)\,dx \tag 1$$

Integrating $(1)$ by parts by letting $u=\arcsin(x)$ and $v=x\left(\log^3(x)-3\log^2(x)+6\log(x)-6\right)$, we find that

$$\begin{align} I&=x\arcsin(x)\left(\log^3(x)-3\log^2(x)+6\log(x)-6\right)\\\\&-\int \left(\frac{\log^3(x)-3\log^2(x)+6\log(x)-6}{\sqrt{1-x^2}}\right)\,x\,dx \tag 2 \end{align}$$

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Next, denote the integral on the right-hand side of $(2)$ by $J$. Enforcing the substitution $x=\sqrt{1-y^2}$ yields

$$\begin{align} J&=-\int \left(\frac{\log^3(x)-3\log^2(x)+6\log(x)-6}{\sqrt{1-x^2}}\right)\,x\,dx\\\\ &=J_3+J_2+J_1+J_0 \end{align}$$

where

$$\begin{align} J_3&=\int \log^3(\sqrt{1-y^2})\,dy \tag 3\\\\ J_2&=-3\int \log^2(\sqrt{1-y^2})\,dy \tag 4\\\\ J_1&=6\int \log(\sqrt{1-y^2})\,dy \tag 5\\\\ J_0&=-6\int 1\,dy \tag 6 \end{align}$$

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The integrals in $(5)$ and $(6)$ can be evaluated in terms of elementary functions with

$$J_0=-6y$$

and

$$J_1=3y\log(1-y^2)-6y-3\log(1-y)+3\log(1+y)$$

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The integrals in $(3)$ and $(4)$ can be expressed in terms of polylogarithm functions. For $J_2$ we can write

$$\begin{align} J_2&=-3\int \log^2(\sqrt{1-y^2})\,dy\\\\ &=-\frac34 \left(K_1+K_2+K_3\right) \end{align}$$

where

$$\begin{align} K_1&=\int \log^2(1-y)\,dy \tag 7\\\\ K_2&=\int \log^2(1+y)\,dy \tag 8\\\\ K_3&=2\int \log(1-y)\log(1+y)\,dy \tag 9 \end{align}$$

The integrals $K_1$ and $K_2$ can be written in closed form with

$$\begin{align} K_1&=(y-1)\left(\log^2(1-y)-2\log(1-y)+2\right)\\\\ \end{align}$$

and

$$\begin{align} K_2&=(y+1)\left(\log^2(1+y)-2\log(1+y)+2\right)\\\\ \end{align}$$

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For $K_3$ we integrate by parts with $u=\log(1-y)$ and $v=(y+1)\log(y+1)-y$ and obtain

$$\begin{align} K_3&=2(y+1)\log(1-y^2)-2y\log(1-y)+2\int \frac{(y+1)\log(y+1)-y}{1-y}\,dy\\\\ &=2(y+1)\log(1-y^2)-2y\log(1-y)+2y+2\log(1-y)+2\int \frac{(y+1)\log(y+1)}{1-y}\,dy\\\\ &=2(y+1)\log(1-y)+2y\left(1-\log(1-y)\right)+4\int \frac{\log(1+y)}{1-y}\,dy \tag{10} \end{align}$$

To evaluate the integral in $(10)$, we make the substitution $y=1-2z$. Then,

$$\begin{align} \int \frac{\log(1+y)}{1-y}\,dy&=-\log(2)\log(w)-\int \frac{\log(1-w)}{w}\,dw\\\\ &=-\log(2)\log\left(\frac{1-y}{2}\right)+\text{Li}_2\left(\frac{1-y}{2}\right) \end{align}$$

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The integral $J_3$ can be evaluated in terms of the dilogarithm function $\text{Li}_2$ and trilogarithm function $\text{Li}_3$ using a similar approach to the one used herein to evaluate $K_2$. We will leave that very tedious analysis to the reader.