Generalizing $\int_{0}^1 \ln^n(1-x)\ln^n(1+x)\,dx$

Question

Note : Apologies, this post is a long read

$$I=\int_{0}^1 \ln^2(1-x)\ln^2(1+x)\,dx$$

I have tried differentiating the beta function but for that no substitution seems to work and also tried different identities but none of them seem to break this product into simpler integrals.

This seems the only similar integral I could find, from which I have tried the above methods.

How do I proceed in this integral?

My attempt:

Here is one identity that made split the product to "slightly" doable form,

$$x^2y^2=\frac{(x+y)^4+(x-y)^4-2x^4-2y^4}{12}$$

$$x=\ln(1-t)\,||||\,y=\ln(1+t)$$

$$(\ln(1-t))^2(\ln(1+t))^2=\frac{(\ln(1-t)+\ln(1+t))^4+(\ln(1-t)-\ln(1+t))^4-2(\ln(1-t))^4-2(\ln(1+t))^4}{12}$$

$$\ln^2(1-t)\ln^2(1+t)=\frac{(\ln(1-t)+\ln(1+t))^4+(\ln(1-t)-\ln(1+t))^4-2\ln^4(1-t)-2\ln^4(1+t)}{12}$$

$$\ln^2(1-t)\ln^2(1+t)=\frac{(\ln(1-t)+\ln(1+t))^4}{12}+\frac{(\ln(1-t)-\ln(1+t))^4}{12}-\frac{\ln^4(1-t)}{6}-\frac{\ln^4(1+t)}{6}$$

$$I=\int_{0}^1 \ln^2(1-t)\ln^2(1+t)\,dt$$

The above integral becomes,

$$I=\int_{0}^1 \frac{(\ln(1-t)+\ln(1+t))^4}{12}+\frac{(\ln(1-t)-\ln(1+t))^4}{12}-\frac{\ln^4(1-t)}{6}-\frac{\ln^4(1+t)}{6}\,dt$$

$$I=\frac{1}{12}I_1+\frac{1}{12}I_2-\frac{1}{6}I_3-\frac{1}{6}I_4\tag1$$

Where,

$$I_1=\int_{0}^1 (\ln(1-t)+\ln(1+t))^4\,dt=\int_{0}^1 \ln^4(1-t^2)\,dt\tag2$$ $$I_2=\int_{0}^1 (\ln(1-t)-\ln(1+t))^4\,dt=\int_{0}^1 \ln^4\left(\frac{1-t}{1+t}\right)\,dt\tag3$$ $$I_3=\int_{0}^1 \ln^4(1-t)\,dt\tag4$$ $$I_4=\int_{0}^1 \ln^4(1+t)\,dt\tag5$$

For $(4)$, I have found this below generalization, $$\int_{0}^1\ln^n(1-x)dx=(-1)^n n!$$

Is there any text/post with the generalisation for this?

$$I=\int_{0}^1 \ln^n(1-x)\ln^n(1+x)\,dx$$

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Edit 1 :

$$I_1=\int_{0}^1 \ln^4(1-t^2)\,dt$$

$t^2=x\implies dt=\frac{1}{2}x^{-\frac{1}{2}}\,dx$

$$I_1=\frac{1}{2}\int_{0}^1 x^{-\frac{1}{2}}\ln^4(1-x)\,dx$$

This can be rewritten as $(x\to t)$,

$$I_1=\frac{1}{2}\int_{0}^1 (1-t)^{-\frac{1}{2}}\ln^4(t)\,dt$$

Inspired from the answers to one of my old question,

Considering the fourth derivative,

$$I_1=\frac{1}{2}\int_0^1 (1-t)^{-\frac{1}{2}}t^n\,dt=\frac{1}{2}\beta\left(n+1,\frac{1}{2}\right)$$

$$\frac{\partial^4{I_1}}{\partial n^4}=\frac{1}{2}\int_{0}^1 (1-t)^{-\frac{1}{2}}\ln^4(t)t^n\,dt$$

Put $n=0$,

$$\frac{\partial^4{I_1}}{\partial n^4}=\int_{0}^1 \ln^4(1-t^2)\,dt=\frac{1}{2}\lim_{n\to 0}\frac{\partial^4}{\partial n^4}\beta\left(n+1,\frac{1}{2}\right)$$

Now took some help of WA to compute this fourth derivative,

$$\int_{0}^1 \ln^4(1-t^2)\,dt=\frac{1}{2}\lim_{n\to 0}\frac{\partial^4}{\partial n^4}\beta\left(n+1,\frac{1}{2}\right)$$

$$ \boxed{I_1 = -96 (\zeta(3) - 4) + \ln(4) \Big( 48 (\zeta(3) - 4) + \ln(4) \big( 48 + (\ln(4) - 8)\ln(4) \big) \Big) - \frac{3 \pi^4}{5}-2 \pi^2 \big( 8 + (\ln(4) - 4)\ln(4) \big)} $$

Taken from here

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Edit 2 :

From @Claude Leibovici's comment,

$$\int_{0}^{1} \ln^n(1+t) \, dt = (-1)^n (\Gamma(n+1, -\log(2)) - n \Gamma(n))$$

Put $n=4$,

$$\boxed{I_4= \Gamma(5, -\log(2)) - 4 \Gamma(4)}$$

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Edit 3 :

Put $n=4$ in the above found generalization,

$$\boxed{I_3=24}$$

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Edit 4 : Regarding the below integral,

$$I_2=\int_{0}^1 (\ln(1-t)-\ln(1+t))^4\,dt=\int_{0}^1 \ln^4\left(\frac{1-t}{1+t}\right)\,dt\tag3$$

I put $\frac{1-t}{1+t}=t$

$$I_2=2\int_{0}^1\frac{\ln^4(t)}{(1+t)^2}\,dt$$

Here, I used the series expansion of $\frac{1}{(1+t)^2}$ and solved to get, $$\boxed{I_2=42\zeta(4)=\frac{7}{15}\pi^4}$$

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Edit 5 :

Using the other integral representation of the Beta function,

$$\int_{-1}^1(1-t)^{a-1}(1+t)^{b-1}\,dt=2^{a+b-1}\beta(a,b)$$

Double Derivative with respect to $b$ then Double Derivative with respect to $a$

$$2\int_{-1}^1(1-t)^{a-1}(1+t)^{b-1}\ln^2(1+t)\ln^2(1-t)\,dt=\frac{\partial^2}{\partial a^2}\left(\frac{\partial^2}{\partial b^2}\left(2^{a+b}\beta(a,b)\right)\right)$$

\begin{aligned} &= \psi^{(0)}(a + b)^2 + \psi^{(0)}(b) (\ln(4) - 2 \psi^{(0)}(a + b)) - \ln(4) \psi^{(0)}(a + b) - \psi^{(1)}(a + b) \\ &\quad + \psi^{(0)}(b)^2 + \psi^{(1)}(b) + \ln^2(2) \left( \ln^2 2 \cdot 2^{a+b} \cdot B(a, b) \right) \\ &\quad + 2^{a+b} \left[ \left( \psi^{(0)}(a) - \psi^{(0)}(a + b) \right)^2 B(a, b) + \left( \psi^{(1)}(a) - \psi^{(1)}(a + b) \right) B(a, b) \right] \\ &\quad + \ln(2) \cdot 2^{a+b+1} \left( \psi^{(0)}(a) - \psi^{(0)}(a + b) \right) B(a, b) \\ &\quad + 2 \left[ -2 \psi^{(0)}(b) \psi^{(1)}(a + b) + 2 \psi^{(0)}(a + b) \psi^{(1)}(a + b) - \psi^{(2)}(a + b) \right] \\ &\quad - \ln(4) \psi^{(1)}(a + b) \left( \ln(2) \cdot 2^{a+b} \cdot B(a, b) + 2^{a+b} (\psi^{(0)}(a) - \psi^{(0)}(a + b)) B(a, b) \right) \\ &\quad + \left( 2^{a+b} \cdot B(a, b) \right) \left[ 2 \psi^{(1)}(a + b)^2 - 2 \psi^{(0)}(b) \psi^{(2)}(a + b) + 2 \psi^{(0)}(a + b) \psi^{(2)}(a + b) \right] \\ &\quad - \ln(4) \psi^{(2)}(a + b)- \psi^{(3)}(a + b) \end{aligned}

Now putting $a=b=1$ and with help of this and this,

Using the fact that the integrand is even,

$$\int_{0}^1\ln^2(1-t)\ln^2(1+t)\,dt= 24 - 8\zeta(2) - 8\zeta(3) - \zeta(4) + 8\zeta(2) \ln(2) - 4\ln^2(2) \zeta(2) + 8\ln(2) \zeta(3) - 24\ln(2) + 12\ln^2(2) - 4\ln^3(2) + \ln^4(2)$$

Numerical check - here

Perhaps the generalization is as follows,

$$\int_{0}^1\ln^n(1+t)\ln^n(1-t)\,dt=\frac{1}{4}\lim_{(a,b)\to (1,1)}\frac{\partial^n}{\partial a^n}\left(\frac{\partial^n}{\partial b^n}\left(2^{a+b}\beta(a,b)\right)\right)$$

Now the question, is there any closed form available for that above expression?

Answer 1

You also have $$\int_{0}^1\ln^n(1+x)dx=(-1)^n (\Gamma (n+1,-\log (2))-n \Gamma (n))$$

Keeping your notations and using Mathematica $$I_1=-96 (\zeta (3)-4)+\log (4) (48 (\zeta (3)-4)+\log (4) (48+(\log (4)-8) \log (4)))-$$ $$\frac{3\pi ^4}{5}-2 \pi ^2 (8+(\log (4)-4) \log (4))$$

$$I_2=\frac{7 \pi ^4}{15}\qquad \qquad I_3=24$$

$$I_4=24+(\log (2) (12+(\log (2)-4) \log(2))-24) \log (4)$$

As a total, with $\color{red}{L=\log(2)}$ $$I=\left(24-8 \zeta (3)-\frac{4 \pi^2}{3}-\frac{\pi ^4}{90}\right)+ \left(8 \zeta (3)-24+\frac{4 \pi^2}{3}\right)L+\left(12-\frac{2 \pi^2}{3}\right) L^2-4L^3+L^4$$ which is the same as @AmrutAyan's result

Answer 2

A physicist's approach:

$$I(n,m)=\int_{0}^1\ln^n(1+t)\ln^m(1-t)\,dt$$

We use an approximation

$$\ln(1+t)\approx \frac{\ln 2}{3}t(4-t)$$ $$0\leqslant t \leqslant 1$$

Thus $$I(n,m)\approx \left ( \frac{\ln 2}{3} \right )^n\int_{0}^1t^n(4-t)^n\ln^m(1-t)\,dt$$

The calculation of this integral is elementary and I will not bring it out here, but I will write the result:

$$I(n,m)\approx \left ( \frac{\ln 2}{3} \right )^n(-1)^{n+m}m! \sum_{k=0}^{n}\binom{n}{k}4^kf(n,k)$$ where

$$f(n,k)=\sum_{i=0}^{2n-k}\binom{2n-k}{i}\frac{(-1)^i}{(2n+1-k-i)^{m+1}}$$

Let's look at some numerical examples

$n=2,m=2$

$I_{exact}=0.796205$

$I_{approx}=\frac{404807}{243000}\ln^22=0.800373$

$n=2,m=3$

$I_{exact}=-2.6305$

$I_{approx}=-\frac{26705809}{4860000}\ln^22=-2.6401$

A slightly more extreme case:

$n=7,m=1$

$I_{exact}=-0.030831$

$I_{approx}=-\frac{2077023165589}{5071470604200}\ln^72=-0.031484$

Answer 3

Integrate[Log[1 + t]^2*Log[1 - t]^2, {t, 0, 1}]

Mathematica shows the result is

$$ \int_0^1 \log ^2(t+1) \log ^2(1-t) \, dt = -8 \zeta (3)+8 \zeta (3) \log (2)-\frac{4 \pi ^2}{3}-\frac{\pi ^4}{90}+24+\log ^4(2)-4 \log ^3(2)-\frac{2}{3} \pi ^2 \log ^2(2)+12 \log ^2(2)-24 \log (2)+\frac{2}{3} \pi ^2 \log (4) $$