# Contour method for shifted logarithm branch

Prove  $a,b,c,d >0$

$$\int^\infty_0 \frac{\log(a^2+b^2x^2)}{c^2+d^2x^2}\,dx = \frac{\pi}{cd} \log \frac{ad+bc}{d}$$

Consider the function

$$f(z) = \frac{\log(a-ibz)}{c^2+d^2z^2}$$

We need the logarithm with the branch cut $y<-\frac{a}{b} , x =0$ . Note that this corresponds to

$$\log(a+ibz) = \log\sqrt{(a+y)^2+b^2x^2}+i\theta \,\,\,\, , \, \theta \in [-\frac{\pi}{2},\frac{3\pi}{2})$$

Consider the contour that avoids the branch-cut on the negative imaginary part

$$\int_{C_{\rho}}f(z)\,dz+\int^\rho_0\frac{\log(a-ibx)}{c^2+dx^2}\,dx+\int_{-\rho}^0\frac{\log(a-ibx)}{c^2+dx^2}\,dx =2\pi i\, \mathrm{Res}\left(f,\frac{c}{d}i \right)$$

The second integral $x \to -x$

$$\int_{C_{\rho}}f(z)\,dz+\int^\rho_0\frac{\log(a-ibx)}{c^2+dx^2}\,dx+\int_{0}^{\rho}\frac{\log(a+ibx)}{c^2+dx^2}\,dx =2\pi i\, \mathrm{Res}\left(f,\frac{c}{d}i \right)$$

Note that in the x-axis we have

$$\log(a \pm bix)= \log(\sqrt{a^2+b^2x^2})\pm i\arctan\left(\frac{bx}{a} \right)$$

Using that we deduce

$$\log(a – bix)+\log(a+ibx)= \log(a^2+b^2x^2)$$

This implies to

$$\int_{C_{\rho}}f(z)\,dz+\int^\rho_0\frac{\log(a^2+b^2x^2)}{c^2+dx^2}\,dx =2\pi i\, \mathrm{Res}\left(f,\frac{c}{d}i \right)$$

For the circular part

$$\left|\int_{C_{\rho}} \frac{\log(a-ibz)}{c^2+d^2z^2} \,dz\right|\leq \pi \rho \, \frac{\log( a+ b \rho) + 2\pi }{|c^2-d^2 \rho^2|} \sim_{\infty} 0$$

Let us look at the residue

$$2\pi i\mathrm{Res}\left(f,\frac{c}{d}i \right) = 2\pi i\lim_{z \to \frac{c}{d}i} \frac{\log(a-ibz)}{2d^2 z} = \frac{\pi}{cd} \log \frac{ad+bc}{d}$$

Hence this simplifies to

$$\boxed{\int^\infty_0 \frac{\log(a^2+b^2x^2)}{c^2+dx^2}\,dx =\frac{\pi}{cd} \log \frac{ad+bc}{d}}$$

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