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I got stuck trying to find $$\int_0^\infty\frac{\log x}{(x+a)^2+b^2}\,dx$$ using complex analysis.

My attempt is to evaluate the contour integral of $$\int_C\frac{\log z}{(z+a)^2+b^2}\,dz$$ for some nicely chosen contour $C$. But for now the best $C$ I can think of is two segments plus two semicircles, which gives the value $$\int_0^\infty\frac{\log x}{(x+a)^2+b^2}\,dx+\int_0^\infty\frac{\log x}{(x-a)^2+b^2}\,dx=\frac{\pi}{2b}\log(a^2+b^2).$$ The problem is that the integrand is not an even function. For example, $\int_0^\infty\frac{\log x}{1+x^2}\,dx$ would have been rather easy.

Please, I need some help!

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    $\begingroup$ Use a keyhole contour. An example to give you some ideas for how to calculate this can be found on Example 5 of the Wikipedia Page $\endgroup$ – Brevan Ellefsen Jun 21 at 3:56
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    $\begingroup$ See also here for a more complicated example that shows how this process can be extended to higher powers of $\log$ $\endgroup$ – Brevan Ellefsen Jun 21 at 3:59
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Based on this answer to the question "How to evaluate $ \int_0^\infty \frac{\log x}{(x^2+a^2)^2} dx $" and on this answer to the question "Evaluate $\int_0^\infty \frac {(\log x)^4dx}{(1+x)(1+x^2)}$". Instead of a function with $\log z$ on the numerator, we consider a function with $\log^2 z$. This is the very same method as that pointed to in the comments.

For $a,b>0$, this method gives the closed formula

$$ \int_{0}^{\infty }\frac{\log x}{\left( x+a\right) ^{2}+b^{2}}\,dx=\frac{1}{2b }\arctan \left( \frac{b}{a}\right) \log \left( a^{2}+b^{2}\right) ,\qquad a,b>0\tag{$\ast$}. $$

We chose the multiple-valued function $f(z)$ with branch cut $\arg z=0$ defined as

$$ \begin{align*} f(z) &=\frac{\log ^{2}z}{\left( z+a\right) ^{2}+b^{2}},\quad \text{with } 0<\arg z<2\pi ,\quad z=re^{i\theta } \\ &=\frac{\log ^{2}z}{\left( z-z_{1}\right) \left( z-z_{2}\right) }\qquad z_{1}=-a+ib,\quad z_{2}=-a-ib, \end{align*} $$

and integrate it counterclockwise around the closed contour $\Gamma $ shown in the figure. This contour is indented around the branch point $O$ and consists of the circles $\gamma _{R}$ ($\left\vert z\right\vert =R$) and $\gamma _{\varepsilon }$ ($\left\vert z\right\vert =\varepsilon $), $ 0<\varepsilon <1<R$, and the segment $\left[ \varepsilon ,R\right] $ described in the positive sense above the $x$-axis and in the negative sense below the $x$-axis.

$\qquad\qquad$enter image description here

$$\text{Closed contour } \Gamma $$

On the upper edge, $\theta =0$ ($r\in \left[ \varepsilon ,R\right] $) and

$$ f(z)=\frac{\left( \log r\right) ^{2}}{\left( r+a\right) ^{2}+b^{2}}. $$

On the lower edge, $\theta =2\pi $ ($r\in \left[ \varepsilon ,R\right] $) and

$$ f(z)=\frac{\left( \log \left( re^{i2\pi }\right) \right) ^{2}}{\left( r+a\right) ^{2}+b^{2}}=\frac{\left( \log r+i2\pi \right) ^{2}}{\left( r+a\right) ^{2}+b^{2}}. $$

As such,

$$ \begin{align*} I &=\lim_{\varepsilon \rightarrow 0,R\rightarrow \infty }\oint_{\Gamma} \frac{\left( \log z\right) ^{2}}{\left( z+a\right) ^{2}+b^{2}}\,dz, \\ &=\int_{0}^{\infty }\frac{\left( \log r\right) ^{2}}{\left( r+a\right) ^{2}+b^{2}}\,dr-\int_{0}^{\infty }\frac{\left( \log \left( re^{i2\pi }\right) \right) ^{2}}{\left( re^{i2\pi }+a\right) ^{2}+b^{2}}\,dr \\ &\quad+\lim_{R\rightarrow \infty }\int_{\gamma _{R}}\frac{\left( \log z\right) ^{2}}{\left( z+a\right) ^{2}+b^{2}}\,dz-\lim_{\varepsilon \rightarrow 0}\int_{\gamma _{\varepsilon }}\frac{\left( \log z\right) ^{2}}{\left( z+a\right) ^{2}+b^{2}}\,dz \\ &=\int_{0}^{\infty }\frac{\left( \log r\right) ^{2}-\left( \log r+i2\pi \right) ^{2}}{\left( r+a\right) ^{2}+b^{2}}\,dx \\ &=4\pi ^{2}\int_{0}^{\infty }\frac{1}{\left( r+a\right) ^{2}+b^{2}} \,dr-i4\pi \int_{0}^{\infty }\frac{\log r}{\left( r+a\right) ^{2}+b^{2}}\,dr \end{align*} $$

provided that

$$ \lim_{R\rightarrow \infty }\int_{\gamma _{R}}\frac{\left( \log z\right) ^{2} }{\left( z+a\right) ^{2}+b^{2}}\,dz=\lim_{\varepsilon \rightarrow 0}\int_{\gamma _{\varepsilon }}\frac{\left( \log z\right) ^{2}}{\left( z+a\right) ^{2}+b^{2}}\,dz=0,\quad \text{(see below).} $$

By the residue theorem,

$$ \begin{align*} I &=2\pi i\left( \operatorname{Res}_{z=z_{1}}f(z)+ \operatorname{Res}_{z=z_{2}}f(z)\right) \\ &=2\pi i\left[ \operatorname{Res}_{z=z_{1}}\frac{ \left( \log z\right) ^{2}}{\left( z-z_{1}\right) \left( z-z_{2}\right) }+ \operatorname{Res}_{z=z_{2}}\frac{\left( \log z\right) ^{2}}{\left( z-z_{1}\right) \left( z-z_{2}\right) }\right] \\ &=2\pi i\left[ \frac{\left( \log z_{1}\right) ^{2}}{z_{1}-z_{2}}+\frac{ \left( \log z_{2}\right) ^{2}}{z_{2}-z_{1}}\right] \\ &=2\pi i\left[ \frac{\left( \log \left( -a+ib\right) \right) ^{2}}{i2b}- \frac{\left( \log \left( -a-ib\right) \right) ^{2}}{i2b}\right] \\ &=\frac{\pi }{b}\left[ \log \left( -a+ib\right) \right] ^{2}-\frac{\pi }{b} \left[ \log \left( -a-ib\right) \right] ^{2} \end{align*} $$

We now assume that $a,b>0$. Then

$$ \begin{align*} I &=\frac{\pi }{b}\left[ \log \left( \left\vert -a+ib\right\vert \right) +i\left( \pi -\arctan \left( \frac{b}{a}\right) \right) \right] ^{2} \\ &\quad-\frac{\pi }{b}\left[ \log \left( \left\vert -a-ib\right\vert \right) +i\left( \pi +\arctan \left( \frac{b}{a}\right) \right) \right] ^{2} \\ &=\frac{\pi }{b}\left[ \frac{1}{2}\log \left( a^{2}+b^{2}\right) +i\left( \pi -\arctan \left( \frac{b}{a}\right) \right) \right] ^{2} \\ &\quad-\frac{\pi }{b}\left[ \frac{1}{2}\log \left( a^{2}+b^{2}\right) +i\left( \pi +\arctan \left( \frac{b}{a}\right) \right) \right] ^{2} \\ &=\frac{4\pi ^{2}}{b}\arctan \left( \frac{b}{a}\right) -i\frac{2\pi }{b} \arctan \left( \frac{b}{a}\right) \log \left( a^{2}+b^{2}\right) \end{align*} $$

because $$ \log \left( \left\vert -a+ib\right\vert \right) =\log \left( \left\vert -a-ib\right\vert \right) =\frac{1}{2}\log \left( a^{2}+b^{2}\right) . $$

Taking the imaginary part of $I$ we obtain $(\ast)$ in the form $$ \text{Im }( I )=-4\pi \int_{0}^{\infty }\frac{\log r}{\left( r+a\right) ^{2}+b^{2} }\,dr=-\frac{2\pi }{b}\arctan \left( \frac{b}{a}\right) \log \left( a^{2}+b^{2}\right) $$


Proof that $\int_{\gamma _{R}}f,\int_{\gamma _{\varepsilon }}f\rightarrow 0$. If $z$ is any point on $\gamma _{R}$,

$$ \begin{align*} \left\vert f(z)\right\vert &=\frac{\left\vert \log z\right\vert ^{2}}{ \left\vert \left( z+a\right) ^{2}+b^{2}\right\vert },\qquad z=R\,e^{i\theta },R>1,0<\theta <2\pi \\ &\leq \frac{\left( \log R+2\pi \right) ^{2}}{\left\vert z+\left( -z_{1}\right) \right\vert \left\vert z+\left( -z_{2}\right) \right\vert }, \\ &\leq \frac{\left( \log R+2\pi \right) ^{2}}{\left\vert R-\sqrt{a^{2}+b^{2}} \right\vert ^{2}}\leq M_{R} \end{align*} $$

where

$$ M_{R}=\frac{4\pi \log R+4\pi ^{2}+\log ^{2}R}{R^{2}+\left( a^{2}+b^{2}\right) -2R\sqrt{a^{2}+b^{2}}} $$

because $$ \left\vert z+\left( -z_{1}\right) \right\vert \geq \left\vert R-\left\vert z_{1}\right\vert \right\vert ,\left\vert z+\left( -z_{2}\right) \right\vert \geq \left\vert R-\left\vert z_{2}\right\vert \right\vert ,\left\vert z_{1}\right\vert =\left\vert z_{2}\right\vert =\sqrt{a^{2}+b^{2}}. $$

This means that

$$ \begin{align*} \left\vert \int_{\gamma _{R}}f(z)\,dz\right\vert &\leq M_{R}\times \,2\pi R \\ &=\frac{4\pi \log R+4\pi ^{2}+\log ^{2}R}{R^{2}+\left( a^{2}+b^{2}\right) -2R\sqrt{a^{2}+b^{2}}}\times \,2\pi R\longrightarrow 0\qquad \left( R\rightarrow \infty \right) . \end{align*} $$

Similarly, if $z$ is any point on $\gamma _{\varepsilon }$ $$ \begin{align*} \left\vert f(z)\right\vert &=\frac{\left\vert \log z\right\vert ^{2}}{ \left\vert \left( z+a\right) ^{2}+b^{2}\right\vert },\qquad z=\varepsilon \,e^{i\theta },0<\varepsilon <1,0<\theta <2\pi \\ &\leq \frac{\left( \log \varepsilon +2\pi \right) ^{2}}{\left\vert z+\left( -z_{1}\right) \right\vert \left\vert z+\left( -z_{2}\right) \right\vert } \\ &\leq \frac{\left( \log \varepsilon +2\pi \right) ^{2}}{\left\vert \varepsilon -\sqrt{a^{2}+b^{2}}\right\vert ^{2}}\leq M_{\varepsilon }, \end{align*} $$

where

$$ M_{\varepsilon }=\frac{4\pi \log \varepsilon +4\pi ^{2}+\log ^{2}\varepsilon }{\varepsilon ^{2}+\left( a^{2}+b^{2}\right) -2\varepsilon \sqrt{a^{2}+b^{2}}} $$

and

$$ \begin{align*} \left\vert \int_{\gamma _{\varepsilon }}f(z)\,dz\right\vert &\leq M_{\varepsilon }\times \,2\pi \varepsilon \qquad z=\rho \,e^{i\theta },\rho <1 \\ &\leq \frac{4\pi \log \varepsilon +4\pi ^{2}+\log ^{2}\varepsilon }{ \varepsilon ^{2}+\left( a^{2}+b^{2}\right) -2\varepsilon \sqrt{a^{2}+b^{2}}} \times \,2\pi \varepsilon \longrightarrow 0\qquad \left( \varepsilon \rightarrow 0\right) . \end{align*} $$

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