definite integral #3

**Random Variable** · February 5th 2011, 02:32 PM

$\displaystyle \int_{0}^{1} ( \arctan x )^{2} \ dx$

**TheCoffeeMachine** · February 5th 2011, 06:46 PM

Starting with integration by parts (twice), we have:

$\displaystyle \begin{aligned} \int_{0}^{1}\left(\tan^{-1}(x)\right)^2\;{dx} &= x(\tan^{-1}{x})^2\bigg|_{ 0}^{1}-\int_{0}^{1} \frac{2x\tan^{-1}{x}}{1+x^2}\;{dx} \\& = \frac{\pi^2}{16}- \int_{0}^{1} \left[\ln(x^2+1)\right]'\tan^{-1}{x}\;{dx} \\&= \frac{\pi^2}{16}-\ln(x^2+1)\tan^{-1}{x}\bigg|_{0}^{1}+\int_{0}^{1} \frac{\ln\left(x^2+1\right)}{x^2+1}\;{dx} \\& = \frac{\pi^2}{16}-\frac{\pi}{4}\ln(2)+\int_{0}^{1} \frac{\ln\left(x^2+1\right)}{x^2+1}\;{dx} \\& = \frac{\pi^2}{16}-\frac{\pi}{4}\ln(2)+I_{1}.\end{aligned}$

Where letting $u = \arctan{x} \Rightarrow \frac{du}{dx} = \frac{1}{1+x^2} \Rightarrow dx = (1+x^2)\;{du}$ gives:

$\displaystyle \begin{aligned} I_{1} & = \int_{0}^{1} \frac{\ln\left(x^2+1\right)}{x^2+1}\;{dx} = \int_{0}^{\frac{\pi}{4}}\frac{\ln(\tan^2{u}+1)(1+x ^2)}{1+x^2}\;{du} \\& = \int_{0}^{\frac{\pi}{4}}\ln(\sec^2{u})\;{du} = -2\int_{0}^{\frac{\pi}{4}}\ln(\cos{u})\;{du} \\& = -2\int_{0}^{\frac{\pi}{4}}\ln\left(2\cos{u}}\right) \;{du}+2\int_{0}^{\frac{\pi}{4}}\ln(2)\;{du} \\& = -K+\frac{\pi}{2}\ln(2). \end{aligned}$

Where $K$ is Catalan's constant.

Thus $\displaystyle \int_{0}^{1} ( \arctan x )^{2} \;{dx} = \frac{\pi^2}{16}+\frac{\pi}{4}\ln(2)-K.$

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