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Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} dx$

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

I am having a hard time deriving the answer, $\frac{\pi}{8} \log(2) $. I have tried Taylor expansion of both numerator and denominator, both seem too complicated and fruitless.

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Ron Gordon
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1 Answers1

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Let $x = \tan(t)$. Then we get that \begin{align} I & = \int_0^1 \dfrac{\log(1+x)}{1+x^2} dx = \int_0^{\pi/4} \dfrac{\log(1+\tan(t))}{\sec^2(t)} \sec^2(t) dt\\ & = \int_0^{\pi/4} \log(\sin(t) + \cos(t)) dt - \int_0^{\pi/4} \log(\cos(t)) dt\\ & = \int_0^{\pi/4} \log\left( \dfrac{\sin(t) + \cos(t)}{\sqrt{2}} \right) dt + \int_0^{\pi/4} \log(\sqrt{2}) dt- \int_0^{\pi/4} \log(\cos(t)) dt\\ & = \int_0^{\pi/4} \log(\cos(t-\pi/4)) dt + \int_0^{\pi/4} \log(\sqrt{2}) dt- \int_0^{\pi/4} \log(\cos(t)) dt\\ & = \underbrace{\int_{-\pi/4}^{0} \log(\cos(t)) dt}_{(t-\pi/4) \to t} + \int_0^{\pi/4} \dfrac{\log(2)}2 dt- \int_0^{\pi/4} \log(\cos(t)) dt\\ & = \underbrace{\int_{0}^{\pi/4} \log(\cos(t)) dt}_{t \to -t} + \dfrac{\pi}8 \log2- \int_0^{\pi/4} \log(\cos(t)) dt\\ & = \dfrac{\pi}8 \log 2 \end{align}

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    Very nice, I didn't see it! – Amzoti Dec 16 '12 at 07:18
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    From the second expression in the first line:

    $\displaystyle{\int_{0}^{\pi/4}\ln\left(1 + \tan\left(t\right)\right),{\rm d}t = {1 \over 2}\left[\int_{0}^{\pi/4}\ln\left(1 + \tan\left(t\right)\right),{\rm d}t + \int_{0}^{\pi/4}\ln\left(1 + \tan\left({\pi \over 4} - t\right)\right),{\rm d}t \right]={1 \over 2}\left[\int_{0}^{\pi/4}\ln\left(1 + \tan\left(t\right)\right),{\rm d}t + \int_{0}^{\pi/4}\ln\left(2 \over 1 + \tan\left(t\right)\right),{\rm d}t \right]={1 \over 2}\int_{0}^{\pi/4}\ln\left(2\right),{\rm d} t={1 \over 8},\pi\ln\left(2\right)}$.

     – Felix Marin Mar 27 '14 at 07:47