Evaluate $\int_0^\infty\frac{\ln x}{1+x^2}dx$ Evaluate $$\int_0^\infty\frac{\ln x}{1+x^2}\ dx$$
I don't know where to start with this so either the full evaluation or any hints or pushes in the right direction would be appreciated.  Thanks.
 A: Another general approach :
Consider
$$
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx.\tag1
$$
Rewrite $(1)$ as
\begin{align}
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx&=\frac1{a^2}\int_0^\infty\frac{x^{b-1}}{1+\left(\frac{x}{a}\right)^2}\ dx.\tag2
\end{align}
Putting $x=ay\;\color{blue}{\Rightarrow}\;dx=a\ dy$ yields
\begin{align}
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx&=a^{b-2}\int_0^\infty\frac{y^{b-1}}{1+y^2}\ dy,
\end{align}
where
\begin{align}
\int_0^\infty\frac{y^{b-1}}{1+y^2}\ dy=\frac{\pi}{2\sin\left(\frac{b\pi}{2}\right)}.
\end{align}
Hence
\begin{align}
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx&=\frac\pi2\cdot\frac{a^{b-2}}{\sin\left(\frac{b\pi}{2}\right)}.\tag3
\end{align}
Differentiating $(3)$ with respect to $b$ and setting $b=1$ yields
\begin{align}
\int_0^\infty\frac{\partial}{\partial b}\left[\frac{x^{b-1}}{a^2+x^2}\right]_{b=1}\ dx&=\frac\pi2\frac{\partial}{\partial b}\left[\frac{a^{b-2}}{\sin\left(\frac{b\pi}{2}\right)}\right]_{b=1}\\
\int_0^\infty\frac{\ln x}{x^2+a^2}\ dx&=\large\color{blue}{\frac{\pi\ln a}{2a}}.
\end{align}
Thus
$$
\int_0^\infty\frac{\ln x}{x^2+1}\ dx=\large\color{blue}{0}.
$$
A: In general
$$
\mathcal{I}(\alpha)=\int_0^\infty\frac{\ln x}{x^2+\alpha^2}\ dx
$$
can be evaluated by using substitution $u=\dfrac{\alpha^2}{x}\;\Rightarrow\;x=\dfrac{\alpha^2}{u}\;\Rightarrow\;dx=-\dfrac{\alpha^2}{u^2}\ du$, then
\begin{align}
\mathcal{I}(\alpha)&=\int_0^\infty\frac{\ln \left(\dfrac{\alpha^2}{u}\right)}{\left(\dfrac{\alpha^2}{u}\right)^2+\alpha^2}\cdot \dfrac{\alpha^2}{u^2}\ du\\
&=\int_0^\infty\frac{2\ln \alpha-\ln u}{\alpha^2+u^2}\ du\\
&=2\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du-\int_0^\infty\frac{\ln u}{u^2+\alpha^2}\ du\\
&=2\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du-\mathcal{I}(\alpha)\\
\mathcal{I}(\alpha)&=\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du.
\end{align}
The last integral can easily be evaluated since it is a common integral. Using substitution $u=\tan\theta$, the integral turns out to be
\begin{align}
\mathcal{I}(\alpha)&=\frac{\ln \alpha}{\alpha}\int_0^{\Large\frac\pi2} \ d\theta\\
&=\large\color{blue}{\frac{\pi\ln \alpha}{2\alpha}}.
\end{align}
Thus
$$
\mathcal{I}(1)=\int_0^\infty\frac{\ln x}{x^2+1}\ dx=\large\color{blue}{0}.
$$
A: Here is one appraoch!!
changing $x = \tan \theta$ 
$$\int_{0}^{\pi/2} \frac{\log(\tan\theta)}{\sec^2 \theta} \sec^2 \theta d\theta = \int_0^{\pi/2} \log (\sin \theta) d\theta - \int_{0}^{\pi/2} \log (\cos \theta) d\theta $$
By changing $\theta \to \pi/2 - \theta$ on the latter integrand, we get
$$\int_0^{\pi/2} \log (\sin \theta) d\theta - \int_{0}^{\pi/2} \log (\cos (\pi/2-\theta)) d\theta = \int_0^{\pi/2} \log (\sin \theta) d\theta - \int_{0}^{\pi/2} \log (\sin \theta) d\theta = 0$$
A: Here's another very short solution.
The substitution $x \to e^t$ transforms the integral to
$$\int_{-\infty}^{\infty} \frac{t}{e^t+e^{-t}}\,dt$$
which is zero by the anti-symmetry of the integrand.
A: $\newcommand{\+}{^{\dagger}}
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$$
\color{#66f}{\Large I}\equiv\ \overbrace{\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{2}}\,\dd x}
^{\ds{x \mapsto {1 \over x}}}\ =\
\int_{\infty}^{0}{\ln\pars{1/x} \over 1 + \pars{1/x}^{2}}
\,\pars{-\,{\dd x \over x^{2}}}
=-\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{2}}\,\dd x=\color{#66f}{\Large -I}
$$

$$
\imp\ \color{#66f}{\Large I} + \color{#66f}{\Large I}=0\ \imp\
\color{#66f}{\Large I\equiv\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{2}}\,\dd x
=\color{#c00000}{0}}
$$

A: Hint
Write $$\int_0^\infty\frac{\ln x}{(1+x^2)}dx=\int_0^1\frac{\ln x}{(1+x^2)}dx+\int_1^\infty\frac{\ln x}{(1+x^2)}dx$$ For the second integral make a change of variable $x=\frac{1}{y}$ and see the beauty of the result.
I am sure that you can take from here.
A: Here's another general result, consider for $|a|\le 1$:
\begin{align*}
  I(a,b) &= \int_{0}^{\infty} \, \frac{x^a}{b^2+x^2}\, dx \\
  &= \frac{b^{a-1}}{2} \int_{0}^{\infty} \, \frac{t^{(a-1)/2}}{1+t}\, dt \tag{1}\\
  &= \frac{b^{a-1}}{2}\, \mathrm{B}\left(\frac{1+a}{2},\frac{1-a}{2}\right) \tag 2\\
  &= \frac{b^{a-1}}{2}\, \frac{\pi}{\displaystyle \cos{\left(\frac{\pi}{2}a\right)}} \tag 3
\end{align*}
$(1)$ is by subst. $\displaystyle x=b\sqrt{t}$
$(2)$ is by the definition of Beta function: $\displaystyle \mathrm{B}(a,b)=\int_{0}^{\infty} \, \frac{x^{a-1}}{(1+x)^{a+b}} \, dx$
$(3)$ is by using $\displaystyle \mathrm{B}(a,b)=\frac{\Gamma{(a)}\Gamma{(b)}}{\Gamma{(a+b)}}$ and Euler's reflection formula $\displaystyle \Gamma{(a)}\Gamma{(1-a)}=\frac{\pi}{\sin{\displaystyle \left({\pi}\, a\right)}}$
Hence, 
\begin{align*}
  \int_{0}^{\infty} \, \frac{x^a \left(\log{x}\right)^n}{b^2+x^2}\, dx &= \frac{\partial^{n} }{\partial a^n} \left(\frac{b^{a-1}}{2}\, \frac{\pi}{\cos{\displaystyle \left(\frac{\pi}{2}a\right)}}\right)
\end{align*}
and when $b=1, n=1, a= 0$, the result is $0$
Update: 
An even better result:
\begin{align*}
  \int_{0}^{\infty} \, \frac{x^a \left(\log{x}\right)^n}{b^c+x^c}\, dx &= \frac{\partial^{n} }{\partial a^n} \left(\frac{b^{a+1-c}}{c}\, \frac{\pi}{\sin{\displaystyle \left(\frac{1+a}{c}\pi\right)}}\right)
\end{align*}
where $\displaystyle 0<\frac{1+a}{c}<1$
