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- #1

- Jan 17, 2013

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\(\displaystyle \int^1_0 \frac{\log (1+x)}{1+x^2} dx \)

$\log $ is the natural logarithm .

$\log $ is the natural logarithm .

- Thread starter ZaidAlyafey
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- Thread starter
- #1

- Jan 17, 2013

- 1,667

\(\displaystyle \int^1_0 \frac{\log (1+x)}{1+x^2} dx \)

$\log $ is the natural logarithm .

$\log $ is the natural logarithm .

- Thread starter
- #2

- Jan 17, 2013

- 1,667

Here is a hint

Try series expansion and harmonic sums

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- #3

- Jan 17, 2013

- 1,667

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

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

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 .