MHB Fractional Logarithmic Integral 02

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The integral $$\int^1_0 \frac{\log (1+x)}{1+x^2} dx$$ is evaluated using results from previous work on fractional logarithmic integrals. A related integral, $$\int^{1}_0 \frac{2\log(1+x)}{1+x^2}\, dx$$, is shown to be numerically equivalent to $$\frac{\pi}{4}\log(2)$$. The discussion references the dilogarithm function, specifically the use of $$\text{Li}_2$$ in the evaluations. The numerical equivalence of the original integral is established as $$\frac{\pi}{8}\log(2)$$. A detailed proof for these results will be provided later.
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$$\int^1_0 \frac{\log (1+x)}{1+x^2} dx $$

$\log $ is the natural logarithm .
 
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Here is a hint
Try series expansion and harmonic sums
 
Using a result that I proved http://www.mathhelpboards.com/f10/generalized-fractional-logarithm-integral-5467/#post25055

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

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

Which is numerically equivalent to

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

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

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