Integral via Inversions

Let us compute the following integral

(1)   \begin{align*} \int_0^{\infty} {\ln (2x) \over 1+x^2}dx \end{align*}

Performing x\to {1\over x}, this integral becomes

(2)   \begin{align*} \int_{\infty}^{0} {\ln (2x^{-1}) \over 1+x^{-2}}\bigg(-\frac{dx}{x^2}\bigg) \,=\, \int_0^{\infty} {\ln (2x^{-1}) \over 1+x^2} dx \end{align*}

Thus we have

(3)   \begin{align*} \int_0^{\infty} {\ln (2x) \over 1+x^2}dx \,&=\, {1\over 2} \int_0^{\infty} {\ln (2x) + \ln (2x^{-1}) \over 1+x^2}dx \\ \,&=\, {\ln 2} \int_0^{\infty} {dx \over 1+x^2} \\ \,&=\, {\pi \ln 2 \over 2} \end{align*}

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