MHB How Does One Solve the Integral of Logarithmic Functions Over a Unit Interval?

[alyafey22]

logarithm integral # (2)

Solve the following

$$\int^1_0 \frac{\log(1+x)\log(1-x)}{x}\,dx$$​

 

[evinda]

Solve the following

$$\int^1_0 \frac{\log(1+x)\log(1-x)}{x}\,dx$$​

Spoiler

$$\log (1-x)=-\sum_{n=1}^{\infty} \frac{x^n}{n} \text{ for } |x|<1$$

$$\log (1+x)=\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} \text{ for } |x|<1$$

So:

$$\frac{\log(1-x)\log(1+x)}{x}=\frac{- \sum_{n=1}^{\infty} \frac{x^n}{n} \cdot \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}}{x}=\frac{\sum_{n=1}^{\infty} (-1)^{n+2} \frac{x^{2n}}{n^2}}{x}=\sum_{n=1}^{\infty} (-1)^n \frac{x^{2n-1}}{n^2}$$

Therefore:

$$\int_0^1 \frac{\log(1+x) \log(1-x)}{x} dx=\int_0^1 \sum_{n=1}^{\infty} (-1)^n \frac{x^{2n-1}}{n^2} dx=\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \left [ \frac{x^{2n}}{2n}\right ]_0^1=\frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}$$

 

[alyafey22]

Spoiler

$$\frac{\log(1-x)\log(1+x)}{x}=\frac{- \sum_{n=1}^{\infty} \frac{x^n}{n} \cdot \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}}{x}=\frac{\sum_{n=1}^{\infty} (-1)^{n+2} \frac{x^{2n}}{n^2}}{x}=\sum_{n=1}^{\infty} (-1)^n \frac{x^{2n-1}}{n^2}$$

How did you perform the multiplication of the two series ?

 

[Euge]

In evinda's solution, there is an error made in multiplying the two series. The product of two absolutely convergent power series is given by the Cauchy product, not pointwise product.

 

[evinda]

In evinda's solution, there is an error made in multiplying the two series. The product of two absolutely convergent power series is given by the Cauchy product, not pointwise product.

I see..I am sorry! (Tmi)

 

[alyafey22]

Spoiler

$$\int^1_0 \frac{\log(1-x)\log(1+x)}{x}\,dx = -\zeta(2)\log(2)+\int^1_0 \frac{\mathrm{Li}_2(x)}{1+x}\,dx $$

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{1+x}\,dx = \int^1_0 \sum_k (-x)^k \mathrm{Li}_2(x) = \sum_{k\geq 1} (-1)^{k-1}\int^1_0 x^{k-1} \mathrm{Li}_2(x) $$

Since

$$\int^1_0 x^{k-1}\mathrm{Li}_2(x)\, dx = \sum_{n\geq 1}\frac{1}{n^2(n+k)} =\frac{1}{k}\left(\zeta(2)-\frac{H_k}{k}\right)$$

We've got

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{1+x}\,dx =\zeta(2)\sum_{k\geq 1} \frac{(-1)^{k-1}}{k}-\sum_{k\geq 1} (-1)^{k-1}\frac{H_k}{k^2} = \frac{1}{24}\left(4 \pi^2 \log(2)-15 \zeta(3)\right)$$

$$\int^1_0 \frac{\log(1-x)\log(1+x)}{x}\,dx = -\zeta(2)\log(2)+\zeta(2)\log(2)- \frac{15}{24} \zeta(3)=\frac{5}{8}\zeta(3)$$