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I'm trying to show that $\int_0^{\infty} \frac{\log x}{(x + a)^2 + b^2} \operatorname d\!x = \frac{1}{b}\arctan \sqrt{a^2 + b^2}.$ However, I am a bit confused applying the key hole "method." I consider $f(z) = \frac{\log^2(z)}{(z+a)^2+b^2}$ and can see that there are poles at $z_{1,2} = -a + bi, -a-bi.$ Then, I'm not sure what to do from there. I know that the integral on the outer circle vanishes as $R\to \infty$ and the integral of the inner circle also goes to zero as $\epsilon \to 0.$ I'm not sure what my integrands are for the contours $C_1$ going from $R$ to $\epsilon$ and $C_2$ going from $\epsilon$ to $R.$ I also don't know how to compute their residues.

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    $\begingroup$ 1) Your integral is incorrect, should be $\frac1b\arctan\left(\frac{b}{a}\right) \log\sqrt{a^2+b^2}$. 2) The residue for $$\frac{(\log x)^2}{(x+a)^2 + b^2} = \frac{(\log x)^2}{(x + a - ib)(x + a + ib)}$$ at $x = -a + ib$ is $$\frac{\log(-a + ib)^2}{(-a + ib) + a + ib} = \frac{1}{2ib}\log(-a+ib)^2$$ If a memomorphic function $f(x)$ has a simple pole at $\alpha$ and $f(x) = \frac{g(x)}{x-\alpha}$ for $x$ near $\alpha$, the residue of $f(x)$ at $x = \alpha$ equals to $g(\alpha)$ $\endgroup$ – achille hui May 3 at 17:19
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without complex analysis$$I = \int^{\infty}_{0}\frac{\ln x}{(x+a)^2+b^2}dx$$

set $\displaystyle x = \frac{a^2+b^2}{y}$ and $\displaystyle dx=-\frac{a^2+b^2}{y^2}dy$

$$I = \int^{\infty}_{0}\frac{\ln(a^2+b^2)-\ln(y)}{(y+a)^2+b^2}dy$$

$$I = \ln(a^2+b^2)\int^{\infty}_{0}\frac{1}{(y+a)^2+b^2}dy-I$$

$$2I=\frac{\ln(a^2+b^2)}{b}\arctan\bigg(\frac{y+a}{b}\bigg)\bigg|^{\infty}_{0}$$

$$I=\frac{\ln(a^2+b^2)}{2b}\bigg[\frac{\pi}{2}-\arctan\bigg(\frac{a}{b}\bigg)\bigg]$$

$$I=\frac{\ln(a^2+b^2)}{2b}\cot^{-1}\bigg(\frac{a}{b}\bigg)=\frac{1}{b}\arctan\bigg(\frac{b}{a}\bigg)\ln\sqrt{a^2+b^2}$$

For $a,b>0$

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