How to calculate the integral $\int_{0}^{1} \frac{\log(1 - x^2)}{x^2} dx$

Question

Using Wolfram Alpha, I discovered that the answer is equal to $-\log 4$. If we use the identity $1 - x^2 = (1 - x)(1 + x)$ in the logarithm, we will have two divergent integrals.This integral is equivalent to the double integral $$ \int_{0}^{1}\int_{0}^{1}\frac{1}{xy^2 - 1}dxdy $$ These two integrals arose from the sum of the series $$ \sum_{n = 1}^{\infty}\frac{1}{n(2n - 1)} $$ Therefore, I don't want you to solve the integral using the mentioned series.

Answer 1

Integrate by parts $$\int_0^1 \frac{\log(1-x^2)}{x^2}dx =\int_0^1 {\log(1-x^2)}d(1-\frac1x)\\ = \log(1 - x^2)(1-\frac1x)\bigg|_0^1 -2\int_0^1\frac1{1+x}dx=-2\ln2 $$

Answer 2

$$I=\int \frac{\log(1 - x^2)}{x^2}\, dx$$ As the other answerers, use integration by parts with $u=\log(1 - x^2)$ $$I=-\frac{\log \left(1-x^2\right)}{x}-2\int \frac{dx}{1-x^2}=-\frac{\log \left(1-x^2\right)}{x}-2 \tanh ^{-1}(x)$$ $$J=\int_0^t \frac{\log(1 - x^2)}{x^2}\, dx=-\frac{\log \left(1-t^2\right)}{t}-2 \tanh ^{-1}(t)$$ Using Taylor series $$J=-2\log(2)+\log \left(\frac{2 (1-t)}{e}\right)(t-1)+O\left((t-1)^2\right)$$

Another solution using Feynman' trick $$I(a)=\int_0^1 \frac{\log(1 - a x^2)}{x^2}\, dx$$ $$I'(a)=-\int_0^1 \frac{1}{1-a x^2}\,dx=\frac{\log \left(\frac{2}{\sqrt{a}+1}-1\right)} {2 \sqrt{a}}$$ $$I(a)=\sqrt{a} \log\left(\frac{2}{\sqrt{a}+1}-1\right)-\log (1-a)+C$$ $$I(1)=-2\log(2)+C$$ and $C=0$ because $I(0)=0$.

Answer 3

Applying the same method found here. Using integration by parts and the substitution $\ln(1-x^2) =u$.

Therefore $$ \int_0^1 \frac{\log(1-x^2)}{x^2}dx = - \int_0^1\frac{2dx}{1+x} = - \log(4) $$

The last integral can be obtained using the substitution $u=x+1$.

Answer 4

I don't want you to solve the integral using the mentioned series.

How about a different series? Replace $x\to\sqrt{1-x^2}$, so the integral is

$$I = \int_0^1 \frac{\log\left(1-x^2\right)}{x^2} \, dx = \int_0^1 \frac{2x \log x}{\left(1-x^2\right)^{3/2}} \, dx$$

Next, we differentiate the binomial series:

$$\begin{align*} f(x) &= \frac1{\sqrt{1-x^2}} = \sum_{n\ge0} \binom{2n}n \left(\frac x2\right)^{2n} \\ \frac{df}{dx} &= \frac{2x}{\left(1-x^2\right)^{3/2}} = \sum_{n\ge1} 2n \binom{2n}n \left(\frac x2\right)^{2n-1} \end{align*}$$

Now,

$$\begin{align*} I &= \sum_{n\ge1} \frac{n\binom{2n}n}{2^{2n-2}} \int_0^1 x^{2n-1} \log x \, dx \\ &= - \sum_{n\ge1} \frac{\binom{2n}n}{n2^{2n}} & \text{by parts} \\ &= - \log\frac4{1+\sqrt{1-t}} \bigg|_{t=1} = \boxed{-2\log2} \end{align*}$$

The penultimate step follows from a slight variation on the "even case" laid out in this answer.