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\begin{aligned} \displaystyle\int_0^{ \infty} \frac{\arctan x}{x^{\alpha}} \,dx=\frac{1}{\alpha-1}\int_0^{ \infty} \frac{dx}{x^{\alpha-1}(1+x^2)}
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\begin{aligned} \displaystyle -6\pi i J + 8\pi^{3} i \int_{0}^{\infty} \frac{1}{1+x^{2}} \, dx = 2\pi i \left(Res(f(z),i)+Res(f(z),-i)) \end{aligned}
\begin{aligned} \displaystyle\int_0^{ \infty} \frac{\arctan x}{x^{\alpha}} \,dx=\frac{1}{\alpha-1}\int_0^{ \infty} \frac{dx}{x^{\alpha-1}(1+x^2)} \,dx
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