# Fractional Logarithmic Integral 02

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
$$\displaystyle \int^1_0 \frac{\log (1+x)}{1+x^2} dx$$

$\log$ is the natural logarithm .

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Here is a hint
Try series expansion and harmonic sums

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Using a result that I proved here

$$\displaystyle \int^t_0 \frac{\log(1+ax)}{1+x}\, dx = - \text{Li}_2 \left( \frac{t}{t+1} \right) +\text{Li}_2 \left(\frac{t-ta}{t+1}\right)-\text{Li}_2(-at)$$

$$\displaystyle \int^{1}_0 \frac{2\log(1+x)}{1+x^2}\, dx = i \text{Li}_2 \left( \frac{1+i}{2} \right)-i \text{Li}_2 \left( \frac{1-i}{2} \right) -i\text{Li}_2 \left(i \right)+i\text{Li}_2 \left(-i \right)$$

Which is numerically equivalent to

$$\displaystyle \int^{1}_0 \frac{2\log(1+x)}{1+x^2}\, dx= \frac{\pi}{4}\log(2)$$

$$\displaystyle \int^{1}_0 \frac{\log(1+x)}{1+x^2}\, dx= \frac{\pi}{8}\log(2)$$

The proof of numerical equivalence is quite long , post it later .