Terms of Closed Form for Logarithmic Integrals

Question

Does the closed-form expression for the integral $$I_n = \int_{0}^{1} \log^n(1+x)\log^n(1-x) \text{ $dx$}$$ only involve terms that can be written solely in terms of zeta functions, powers of logarithms of 2, and rational numbers?

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$I_0$ is evidently $1$.

For $n=1$, we have $$I_1 = \int_{0}^{1} \log(1+x)\log(1-x) \text{ $dx$} = \frac{1}{2} \int_{-1}^{1} \log(1+x)\log(1-x) dx \overset{1+x=2t}{=} \int_{0}^{1} \log{2t} \log{(2(1-t))} dt$$ $$= \log^2(2)\int_{0}^{1} dt + 2\log(2)\int_{0}^{1} \log(1-t) dt + \log(2) \int_{0}^{1} \log(t)dt + \int_{0}^{1} \log(t) \log(1-t) dt$$ $$ = \log^2(2) - \log(2) - \log(2) + 2 - \zeta(2) = \boxed{\log^2(2) - 2\log(2) + 2 - \zeta(2)}$$ (approximately $-0.5507$)

Each term in $I_1$ can be written solely in terms of zeta functions, powers of logarithms of 2, and rational numbers.

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For $n=2$, we have

$$I_2 = \int_{0}^{1} \log^2(1+x)\log^2(1-x) \text{ $dx$}$$

Consider the algebraic identity $A^2B^2 = \frac{1}{12}((A+B)^4 + (A-B)^4 + 2A^4 - 2B^4)$.

Let $A = \log(1-x)$, and $B = \log(1+x)$. Using the aforementioned identity and logarithmic properties, we see that

$$I_2 = \dfrac{1}{12}\underbrace{\int_0^1\log^4(1-x^2)dx}_{\Gamma_1}+\dfrac{1}{12}\underbrace{\int_0^1\log^4\left(\dfrac{1-x}{1+x}\right)dx}_{\Gamma_2}-\dfrac{1}{6}\underbrace{\int_0^1\log^4(1-x)dx}_{\Gamma_3}-\dfrac{1}{6}\underbrace{\int_0^1\log^4(1+x)dx}_{\Gamma_4}$$

See this post on the evaluation of $\Gamma_1, \Gamma_2, \Gamma_3, \Gamma_4$. OP required the evaluation of $\Gamma_1, \Gamma_2$. $\Gamma_1$ can be nicely evaluated using the reduction formula method. The other integrals have nicer closed forms, in terms of the zeta function.

$$\Gamma_1 = 384 -384\log(2) + 192\log^2(2) - 16\pi^2 - 96\zeta(3) - 64\log^3(2) + 16\pi^2\log(2) + 96\zeta(3)\log(2) - 8\pi^2\log^2(2) + 16\log^4(2)-\frac{3}{5}\pi^2$$ $$\Gamma_2 = 42\zeta(4)$$ $$\Gamma_3 = 24$$ $$\Gamma_4 = 2\log^42-8\log^32+24\log^22-48\log2+24$$

I will not go into the full details of the simplification. By replacing terms involving $\pi^2$ with those using $\zeta(2)$, and combining like-terms, we see that

$$I_2 = \frac{1}{12}\Gamma_1 + \frac{1}{12}\Gamma_2 - \frac{1}{6}\Gamma_3 - \frac{1}{6}\Gamma_4 = \boxed{24 - 8\zeta(2) - 8\zeta(3) - \zeta(4) + 8\log(2)\zeta(2) - 4\log^2(2)\zeta(2) + 8\log(2)\zeta(3) - 24\log(2) + 12\log^2(2) - 4\log^3(2) + \log^4(2)}$$ (approximately $0.7962$)

Each term in $I_2$ can be written solely in terms of zeta functions, powers of logarithms of 2, and rational numbers.

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For $n$, we have

$$I_n = \int_{0}^{1} \log^n(1+x)\log^n(1-x) \text{ $dx$}$$

Consider the definition of the Beta function $$B(x,y) = \int_{0}^{1} u^{x-1}(1-u)^{y-1} du \overset{1-2u = t}{\Rightarrow} \int_{-1}^{1} (1-t)^{x-1} (1+t)^{y-1} dt = 2^{x+y-1}B(x,y)$$

$$\int_{-1}^{1} \frac{\partial^{2n}}{\partial x^n \partial y^n}((1-t)^{x-1} (1+t)^{y-1}) dt = \frac{\partial^{2n}}{\partial x^n \partial y^n}(2^{x+y-1}B(x,y))$$

$$\int_{-1}^{1} (1-t)^{x-1} (1+t)^{y-1} \log^n(1-t) \log^n(1+t) dt = \frac{\partial^{2n}}{\partial x^n \partial y^n}(2^{x+y-1}B(x,y))$$

$$\int_{-1}^{1} \log^n(1-t) \log^n(1+t) dt = \lim_{\substack{x\to 1 \\ y\to 1}} \frac{\partial^{2n}}{\partial x^n \partial y^n}(2^{x+y-1}B(x,y))$$

$$\int_{0}^{1} \log^n(1-t) \log^n(1+t) dt = \frac{1}{4} \lim_{\substack{x\to 1 \\ y\to 1}} \frac{\partial^{2n}}{\partial x^n \partial y^n}(2^{x+y}B(x,y))$$

Can each term in $\frac{1}{4} \lim_{\substack{x\to 1 \\ y\to 1}} \frac{\partial^{2n}}{\partial x^n \partial y^n}(2^{x+y}B(x,y))$ (for all $n$) only be written solely in terms of zeta functions, powers of logarithms of 2, and rational numbers?

Thanks.

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Edit

This post was closed for being a duplicate. I have since changed the problem (it has not received any answer yet). The problem focuses on just the terms of the closed for expression for $I_n$, instead of the general case of $I_n$.

Now, these are two different problems. For this post, a solution will involve proving $\frac{1}{4} \lim_{\substack{x\to 1 \\ y\to 1}} \frac{\partial^{2n}}{\partial x^n \partial y^n}(2^{x+y}B(x,y))$ will only contain terms soley using zeta functions, powers of logarithms of 2, and rational numbers. The "duplicate post" focuses on a general form to $\frac{1}{4} \lim_{\substack{x\to 1 \\ y\to 1}} \frac{\partial^{2n}}{\partial x^n \partial y^n}(2^{x+y}B(x,y))$

Thank you.

Answer 1

I do not want this question to be browsing the list of 'unanswered' questions: Quý Nhân has given a nice response in the comments that answers the question of the terms of the closed form evaluation of the above integral.

Observe the first some partial derivatives of the beta function.

\begin{align*} &\frac{\partial}{\partial x} B(x,y) = B(x, y)[\psi_0(x) - \psi_0(x+y)]\\\\ &\frac{\partial^2}{\partial y} B(x,y) = B(x, y)[\psi_0(y) - \psi_0(x+y)]\\\\ &\frac{\partial^2}{\partial y^2} B(x,y) = B(x, y)[[\psi_0(y) - \psi_0(x+y)]^2 + \psi_1(y) - \psi_1(x+y)] \end{align*}

Note that $\frac{\partial^n}{\partial x^a \partial y^{n-x}} B(x,y)$ will always be in terms of polygamma functions at integer points.

Recall the recurrence relation for the polygamma function of order $m$.

$${\frac {\psi ^{(m)}(n)}{(-1)^{m+1}\,m!}}=\zeta (1+m)-\sum _{k=1}^{n-1}{\frac {1}{k^{m+1}}}=\sum _{k=n}^{\infty }{\frac {1}{k^{m+1}}}\qquad m\geq 1$$

Thus, derivatives of the Beta functions can be expressed in terms of zeta functions. It should also be mentioned that $n$th derivatives of $2^x$ will be in terms of $\log(2)$ and $2^x$. Combining these results, we see that

$$\int_{0}^{1} \log^n(1-t) \log^n(1+t) dt = \frac{1}{4} \lim_{\substack{x\to 1 \\ y\to 1}} \frac{\partial^{2n}}{\partial x^n \partial y^n}(2^{x+y}B(x,y))$$

will always be in terms of zeta functions, powers of logarithms of 2, and rational numbers.