Interestingly, I seem to have an integral I have posted before, but I want to take a different approach to it.

$\int_{0}^{1} \frac{\ln(1+x)}{1+x^2} \,dx$

The beta function states,

$B(x,y) = \int_{0}^{1} {t}^{x-1}({1-t}^{y-1}) \,dx$

So, I was just thinking [B]if[/B] there a possible way to compute this integral using the beta function also knowing

$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$


I am thinking about a creative use of the Euler Beta function.

For now, just a symmetry trick related to the formula $\tan(\pi/4-u)=\frac{1-\tan u}{1+\tan u}$:

$$\begin{eqnarray*}I&=&\int_{0}^{1}\frac{\log(1+x)}{1+x^2}\,dx = \int_{0}^{\pi/4}\log(1+\tan\theta)\,d\theta\\&=&\int_{0}^{\pi/4}\log\left(1+\frac{1-\tan\theta}{1+\tan\theta}\right)\,d\theta=\frac{\pi}{4}\log 2-I\end{eqnarray*}$$ from which: $$ I = \color{red}{\frac{\pi}{8}\log 2}.$$

  • $\begingroup$ how did the $1+x^2$ disappear from the denominator of the fraction as you made the substitution? $\endgroup$ – Amad27 Sep 30 '14 at 13:43
  • 1
    $\begingroup$ @Amad27: $\tan'u=1+\tan^2u=\dfrac1{\cos^2u}=\sec^2u$. $\endgroup$ – Lucian Sep 30 '14 at 17:33

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