Evaluate the integral:
$$\int_0^1 \frac{\ln x}{1+x^2}\, {\rm d}x$$
Solution
We have successively:
$$\begin{aligned}
\int_{0}^{1}\frac{\ln x}{1+x^2}\, {\rm d}x &=\int_{0}^{1}\ln x \sum_{n=0}^{\infty}(-1)^n x^n \, {\rm d}x \\
&= \sum_{n=0}^{\infty} (-1)^n \int_{0}^{1}x^n \ln x \, {\rm d}x\\
&= -\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( n+1 \right )^2}\\
&= -G
\end{aligned}$$
where $G$ is the Catalan's constant.
$$\int_0^1 \frac{\ln x}{1+x^2}\, {\rm d}x$$
Solution
We have successively:
$$\begin{aligned}
\int_{0}^{1}\frac{\ln x}{1+x^2}\, {\rm d}x &=\int_{0}^{1}\ln x \sum_{n=0}^{\infty}(-1)^n x^n \, {\rm d}x \\
&= \sum_{n=0}^{\infty} (-1)^n \int_{0}^{1}x^n \ln x \, {\rm d}x\\
&= -\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( n+1 \right )^2}\\
&= -G
\end{aligned}$$
where $G$ is the Catalan's constant.
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