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Arctan integral

Prove that

    \[\int_0^{\infty}\frac{\arctan x -x e^{-x}}{x^2}\,\mathrm{d}x=1+\gamma\]

where \gamma denotes the Euler – Mascheroni constant.

Solution

Beginning by parts we have,

    \begin{align*} \int_{0}^{\infty} \frac{\arctan x- xe^{-x}}{x^2} \, \mathrm{d}x &= \left [ -\frac{\arctan x - xe^{-x}}{x} \right ]_0^\infty + \\ &\quad \quad \quad + \int_{0}^{\infty} \frac{1}{x} \left ( \frac{1}{x^2+1} +e^{-x} (x-1) \right ) \, \mathrm{d}x \\ &=\int_{0}^{\infty} \left ( \frac{1}{x\left ( x^2+1 \right )} + \frac{e^{-x}(x-1)}{x} \right ) \, \mathrm{d}x \\ &=\int_{0}^{\infty} \left ( \frac{1}{x} - \frac{x}{x^2+1} + e^{-x} - \frac{e^{-x}}{x} \right ) \, \mathrm{d}x \end{align*}

However,

\begin{aligned} \int_{\epsilon}^{M} \left ( \frac{1}{x} - \frac{x}{x^2+1} - \frac{e^{-x}}{x} \right ) \, \mathrm{d}x &= \left ( \ln M - \ln \epsilon - \frac{\ln \left ( M^2+1 \right ) - \ln \left ( \epsilon^2+1 \right ) }{2} \right ) - \\ &\quad \quad \quad - \int_{\epsilon}^{M} \frac{e^{-x}}{x} \, \mathrm{d}x \\ &=\left ( \ln M - \ln \epsilon - \frac{\ln \left ( M^2+1 \right ) - \ln \left ( \epsilon^2+1 \right ) }{2} \right ) - \\ &\quad \quad \quad - \left [ \ln x e^{-x} \right ]_\epsilon^{M}  - \int_{\epsilon}^{M} e^{-x} \ln x \, \mathrm{d}x \\ &=\left ( \ln M - \frac{\ln \left ( M^2+1 \right )}{2} \right ) + \frac{\ln \left ( \epsilon^2+1 \right )}{2} - \\ &\quad \quad \quad \quad -\ln M e^{-M} + \left (\ln \epsilon \; e^{-\epsilon} - \ln \epsilon \right ) - \\ &\quad \quad \quad \quad  -\int_{\epsilon}^{M} e^{-x} \ln x \, \mathrm{d}x \end{aligned}

The result now follows taking \epsilon \rightarrow 0^+ and M \rightarrow +\infty.

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