Proving an Equality Between an Integral and an Infinite Series

In this article, we will prove the equality:

(int_0^1 frac{ln x ln (1 - x^2)}{x^2} , dx sum_{n1}^{infty} frac{1}{n (2n-1)^2})

Step-by-Step Solution

First, we will use the Taylor series expansion of (ln(1 - x^2)) around (x 0):

(ln(1 - x^2) - sum_{m1}^{infty} frac{x^{2m}}{m})

This expansion is valid for (|x| 1).

Substituting the series expansion into the integral, we get:

(int_0^1 frac{ln x ln(1 - x^2)}{x^2} , dx - int_0^1 frac{ln x}{x^2} left( - sum_{m1}^{infty} frac{x^{2m}}{m} right) , dx)

Interchanging the sum and the integral, justified by uniform convergence on ([0, 1]), we obtain:

(- sum_{m1}^{infty} frac{1}{m} int_0^1 frac{ln x x^{2m}}{x^2} , dx - sum_{m1}^{infty} frac{1}{m} int_0^1 ln x x^{2m-2} , dx)

We will evaluate the integral using integration by parts. Let:

(u ln x), so (du frac{1}{x} , dx) (dv x^{2m-2} , dx), so (v frac{x^{2m-1}}{2m-1})

Applying integration by parts:

(int ln x x^{2m-2} , dx left[ ln x cdot frac{x^{2m-1}}{2m-1} right]_0^1 - int_0^1 frac{x^{2m-1}}{2m-1} cdot frac{1}{x} , dx)

The boundary term evaluates to (0), leading to:

(- frac{1}{2m-1} int_0^1 x^{2m-2} , dx - frac{1}{(2m-1)^2})

So, we have:

(int_0^1 ln x x^{2m-2} , dx - frac{1}{(2m-1)^2})

Substituting this result back into our sum gives:

(- sum_{m1}^{infty} frac{1}{m} left( - frac{1}{(2m-1)^2} right) sum_{m1}^{infty} frac{1}{m (2m-1)^2})

We have thus shown:

(int_0^1 frac{ln x ln (1 - x^2)}{x^2} , dx sum_{m1}^{infty} frac{1}{m (2m-1)^2})

This confirms the equality:

(int_0^1 frac{ln x ln (1 - x^2)}{x^2} , dx sum_{n1}^{infty} frac{1}{n (2n-1)^2})