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I came across this problem in the book (Almost) Impossible Integrals, Sums and Series Problem 1.8

$$\int_0^1\log^2(1-x)\log^2(1+x)dx=24-8\zeta(2)-8\zeta(3)-\zeta(4)+8\log2\zeta(2)-4\log^22\zeta(2)+8\log2\zeta(3)-24\log2+12\log^22-4\log^32+\log^42$$

In the book, this has been solved by using Beta Function. It is said that it can also be solved by the following method, although the solution is not provided.
$A=\log(1-x)$
$B=\log(1+x)$
So, $A^2B^2=\dfrac{1}{12}\left[(A+B)^4+(A-B)^4-2A^4-2B^4\right]$

Hence, we can write the integral as $$\int_0^1\log^2(1-x)\log^2(1+x)dx=\dfrac{1}{12}\underbrace{\int_0^1\log^4(1-x^2)dx}_{I_1}+\dfrac{1}{12}\underbrace{\int_0^1\log^4\left(\dfrac{1-x}{1+x}\right)dx}_{I_2}-\dfrac{1}{6}\underbrace{\int_0^1\log^4(1-x)dx}_{I_3}-\dfrac{1}{6}\underbrace{\int_0^1\log^4(1+x)dx}_{I_4}$$ I am able to solve $I_3$ and $I_4$ but am stuck in the first two integrals. Can these be solved without using the beta function?

Edit-
The solutions found till now are $$I_2=42\zeta(4)$$ $$I_3=24$$ $$I_4=2\log^42-8\log^32+24\log^22-48\log2+24$$ A solution is provided in book, although I couldn't really get it.

Souparna
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  • Interestingly, the second integral seems to evaulaute to a nice number but the first one is a bit uglier. – Accelerator Oct 12 '23 at 15:16
  • @Accelerator Similarly $I_3=24$ and $I_4$ is a long expression. – Souparna Oct 12 '23 at 15:19
  • Considering $$I_1=12\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{2^{2n}\left(n+1\right)^{5}}$$

    I thought of Integrating 5 Times but encountered Di-Logarithms midway, so there is probably a cleaner way.

     – Miracle Invoker Oct 12 '23 at 18:09
  • @BlackEmperor How do we get to the expression you have provided? – Souparna Oct 13 '23 at 02:48
  • @Souparna $$\frac{1}{2}\int_{0}^{1}\frac{\left(\ln u\right)^{4}}{\sqrt{1-u}}du$$ Can be converted to this and then use Binomial Expansion on Denominator. Also check out my second edit once. – Miracle Invoker Oct 13 '23 at 16:49
  • @BlackEmperor Wolfram gives the answer of the sum, but no idea how to solve. https://www.wolframalpha.com/input?i2d=true&i=Sum%5B%5C%2891%29Divide%5B%5C%2840%292n%5C%2841%29%21%2CPower%5B2%2C2n%5DPower%5B%5C%2840%29n%2B1%5C%2841%29%2C5%5DPower%5B%5C%2840%29n%21%5C%2841%29%2C2%5D%5D%5C%2893%29%2C%7Bn%2C0%2C%E2%88%9E%7D%5D – Souparna Oct 14 '23 at 07:38

2 Answers2

6

I only have the solution for $I_2$ :

$$\int_{0}^{1}\ln\left(\frac{1-x}{1+x}\right)^{4}dx$$

$\frac{2x}{1+x}\to u$ : $$2\int_{0}^{1}\frac{\left(\ln u\right)^{4}}{\left(1+u\right)^{2}}du$$

Considering: $$\frac{1}{\left(u+1\right)^{2}}=\sum_{r=1}^{\infty}\left(-1\right)^{r-1}ru^{r-1}$$

$$I_2=2\sum_{r=1}^{\infty}r\left(-1\right)^{r-1}\int_{0}^{1}\left(\ln u\right)^{4}u^{r-1}du$$

What's interesting is the Integral has an Anti Derivative, though I just used Wolfram to evaluate it. (I will update in a bit with its solution).

$$\int_{0}^{1}\left(\ln u\right)^{4}u^{r-1}du=\frac{24}{r^5}$$

Therefore,

$$I_2=-48\sum_{r=1}^{\infty}\frac{\left(-1\right)^{r}}{r^{4}}$$

Considering: $$\left(\frac{1}{2^{a-1}}-1\right)\sum_{r=1}^{\infty}\frac{1}{r^{a}}=\sum_{r=1}^{\infty}\frac{\left(-1\right)^{r}}{r^{a}}$$ and, $\zeta(4)=\frac{\pi^4}{90}$.

We have, $$I_2=\frac{7}{15}\pi^4$$

EDIT 1:

The Integral evaluated by Wolfram is really just a Repeated Application of Integration by Parts then Putting in the Limits, Considering the User's knowledge I will skip on the derivation.


EDIT 2:

I kind of have an Incomplete Solution for $I_1$: (Though I hardly believe that's how the Author intended to do the exercise.)

Consider:

$$I\left(m,n\right)=\int_{0}^{1}\frac{\ln\left(u\right)^{m}\ln\left(1+u\right)^{n}}{\left(1+u\right)^{2}}du$$

Then:

$$I_1=$$ $$16\int_{0}^{1}\left(\ln\left(1-t\right)\right)^{4}dt+\int_{0}^{1}\left(\ln\left(\frac{1+t}{1-t}\right)\right)^{4}dt-64\ln\left(2\right)^{3}I\left(1,0\right)+192\ln^{2}\left(2\right)I\left(1,1\right)-192\ln\left(2\right)I\left(1,2\right)+64I\left(1,3\right)-144\ln^{2}\left(2\right)I\left(2,0\right)+288\ln\left(2\right)I\left(2,1\right)-144I\left(2,2\right)-112\ln\left(2\right)I\left(3,0\right)+112I\left(3,1\right)-32I\left(4,0\right)$$

Miracle Invoker
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1

Too long for a comment

Again, like in this post https://math.stackexchange.com/q/4788831, one might elegantly exploit The Master Theorem of Series given in (Almost) Impossible Integrals, Sums, and Series (2019), which will immediately reveal that

$$\int_0^1\log^4(1-x^2) \textrm{d}x$$ $$\small =\frac{1}{4}\lim_{n\to -1/2 }\frac{H_n^5+10H_n^3 H_n^{(2)}+15 H_n (H_n^{(2)})^2+20 H_n^2 H_n^{(3)}+20 H_n^{(2)} H_n^{(3)}+30 H_n H_n^{(4)}+24 H_n^{(5)}}{5 n}$$ $$\small +\frac{1}{4}\lim_{n\to 1/2 }\frac{H_n^5+10H_n^3 H_n^{(2)}+15 H_n (H_n^{(2)})^2+20 H_n^2 H_n^{(3)}+20 H_n^{(2)} H_n^{(3)}+30 H_n H_n^{(4)}+24 H_n^{(5)}}{5 n},$$

and done. Surely, one would like to use the Polygamma function notation combined with its recurrence relation (and then get to the same point where one may observe the special cases with the argument $1/2$ are well-known).

Clever examples of the use of The Master Theorem of Series may be found both in (Almost) Impossible Integrals, Sums, and Series (2019) and More (Almost) Impossible Integrals, Sums, and Series (2023). Also, other new ones (related to extremely difficult results) will come soon to the surface of the literature.

End of story

user97357329
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