Integrate: $ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $ without using complex analysis methods Can this integral be solved without using any complex analysis methods: $$ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $$
Thanks.
 A: So, consider the following:
$$\int_0^{\infty} dx \frac{\log{x}}{(1+x^2)^2} = \int_0^{1} dx \frac{\log{x}}{(1+x^2)^2} + \int_1^{\infty} dx \frac{\log{x}}{(1+x^2)^2}$$
Sub $x \mapsto 1/x$ in the latter integral on the RHS and get that
$$\int_0^{\infty} dx \frac{\log{x}}{(1+x^2)^2} = \int_0^{1} dx \frac{1-x^2}{(1+x^2)^2} \log{x}$$
Note that
$$\frac{1-x^2}{(1+x^2)^2} = \sum_{m=0}^{\infty} (-1)^m (2 m+1) x^{2 m}$$
So we get
$$\int_0^{\infty} dx \frac{\log{x}}{(1+x^2)^2} = \sum_{m=0}^{\infty} (-1)^m (2 m+1) \int_0^1 dx \, x^{2 m} \log{x}$$
Using the fact that
$$\int_0^1 dx \, x^{2 m} \log{x} = -\frac{1}{(2 m+1)^2}$$
we get that
$$\int_0^{\infty} dx \frac{\log{x}}{(1+x^2)^2} = -\sum_{m=0}^{\infty} \frac{(-1)^m}{2 m+1} = -\frac{\pi}{4}$$
A: Here is an approach.
$$F(s)= \int_{0}^{\infty}\frac{x^{s-1}}{(1+x^2)^2}dx= -\frac{1}{4}\,{\frac { \left( s-2 \right) \pi }{\sin \left(\pi \,s/2
 \right) }}
 $$
$$ \implies F'(s)= \int_{0}^{\infty}\frac{x^{s-1}\ln(x)}{(1+x^2)^2}dx= -\frac{1}{4}\,\frac{d}{ds}{\frac { \left( s-2 \right) \pi }{\sin \left(\pi \,s/2
 \right) }} . $$
Now, differentiate and take the limit as $s\to 1$. The answer should be $-\frac{\pi}{4} $.
Note: Here is the technique how you find $F(s)$. It is based on the $\beta$ function.
A: Here is a rather elementary method to approach this problem:
Consider $$I(a)= \int^{\infty}_0 \frac{\log(x)}{x^2+a^2} \,\mathrm{d}x $$
Realize that on differentiating under the integral sign with respect to $a$, we have
$$I'(a)= \int^{\infty}_0 -2a. \frac{\log(x)}{(x^2+a^2)^2} \,\mathrm{d}x $$
It follows that the required integral  $$\int^{\infty}_0  \frac{\log(x)}{(x^2+1)^2} \mathrm{d}x = \frac{-I'(1)}{2}.$$
Now, let us calculate $I(a)$. Set $x=a \tan(\theta)$ where $a>0$. Thus,
$$I(a)= \int^{\frac{\pi}{2}}_0 \log(a \tan\theta) \,\mathrm{d}\theta $$
$$I(a)= \int^{\frac{\pi}{2}}_0 \log(a) + \log( \tan\theta) \,\mathrm{d}\theta $$
It is trivial (by replacing $\theta$ with $\frac{\pi}{2}-\theta$) to show that $$\int^{\frac{\pi}{2}}_0  \log( \tan\theta) \,\mathrm{d}\theta = 0. $$
Thus, $$I(a)= \int^{\frac{\pi}{2}}_0 \log(a) \, \mathrm{d}\theta = \frac{\pi \log(a)}{2a}$$
And, $$I'(a)=\frac{\pi}{2a^2} (1-\log(a))$$
Finally, the required integral is $$\int^{\infty}_0  \frac{\log(x)}{(x^2+1)^2} \,\mathrm{d}x = \frac{-I'(1)}{2}=-\frac{\pi}{4}$$
And we are done.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\infty}{\ln\pars{x} \over \pars{1 + x^{2}}^{2}}\,\dd x:\
     {\large ?}}\quad$
      " without using any complex analysis methods$\ds{\ldots}$ "


*
*First, we evaluate the integral
    $\ds{\pars{~\mbox{with}\ \mu > 0~}}$:
\begin{align}
\int_{0}^{\infty}{\ln\pars{x} \over \mu + x^{2}}\,\dd x&
={1 \over \root{\mu}}
\int_{0}^{\infty}{\ln\pars{\root{\mu}\bracks{x/\root{\mu}}} \over 1 + \pars{x/\root{\mu}}^{2}}\,{\dd x \over \root{\mu}}
={1 \over \root{\mu}}
\int_{0}^{\infty}{\ln\pars{\root{\mu}x} \over 1 + x^{2}}\,\dd x
\\[3mm]&={\ln\pars{\mu} \over 2\root{\mu}}\
\underbrace{\int_{0}^{\infty}{\dd x \over 1 + x^{2}}}_{\ds{\pi \over 2}}\ +\
{1 \over \root{\mu}}\
\underbrace{\int_{0}^{\infty}{\ln\pars{x} \over 1 + x^{2}}\,\dd x}_{\ds{0}}
\qquad\qquad\qquad\pars{1}
\end{align}


*
*The first integral in $\pars{1}$ right hand side is trivially found with the change of variables $\ds{x = \tan\pars{\theta}}$.

*In the second one, we make the change $\ds{x \mapsto 1/x}$ such that the integral just changes its sign. It shows that it vanishes out.


$$
\begin{array}{|c|}\hline\\
\ds{\quad\int_{0}^{\infty}{\ln\pars{x} \over \mu + x^{2}}\,\dd x
={\ln\pars{\mu} \over \root{\mu}}\,{\pi \over 4}\quad}
\\ \\ \hline
\end{array}\tag{2}
$$



*Second, differentiate both members of $\pars{2}$ respect of $\ds{\mu}$:
$$
-\int_{0}^{\infty}{\ln\pars{x} \over \pars{\mu + x^{2}}^{2}}\,\dd x
=\bracks{{1 \over \mu^{3/2}} - {\ln\pars{\mu} \over 2\mu^{3/2}}}\,{\pi \over 4}
$$


*Then, set $\ds{\mu = 1}$ in both members:
$$\color{#66f}{\large%
\int_{0}^{\infty}{\ln\pars{x} \over \pars{1 + x^{2}}^{2}}\,\dd x
=-\,{\pi \over 4}}
$$

A: $x \to 1/x$. We then have
$$I = \int_0^{\infty}\dfrac{\log(x)}{(1+x^2)^2}dx = \int_0^1\dfrac{\log(x)}{(1+x^2)^2}dx + \int_1^{\infty}\dfrac{\log(x)}{(1+x^2)^2}dx$$
$$\int_1^{\infty}\dfrac{\log(x)}{(1+x^2)^2}dx = \int_1^0 \dfrac{x^2\log(x)}{(1+x^2)^2}dx$$
Hence,
$$I = \int_0^1 \dfrac{1-x^2}{(1+x^2)^2}\log(x)dx = \int_0^1 (1-x^2)\left(\sum_{k=0}^{\infty} (k+1)(-1)^k x^{2k} \right)\log(x)dx$$
We now have
$$\int_0^1 x^{2n} \log(x) dx = -\dfrac1{(2n+1)^2}$$
Hence,
$$I = \sum_{k=0}^{\infty}(k+1)(-1)^k\left(-\dfrac1{(2k+1)^2} + \dfrac1{(2k+3)^2}\right)= -\dfrac{\pi}4$$
To get the last step, note that
$$(k+1)\left(\dfrac1{(2k+1)^2} - \dfrac1{(2k+3)^2}\right) = \dfrac14 \left(\dfrac1{2k+1} - \dfrac1{2k+3} + \dfrac1{(2k+1)^2} + \dfrac1{(2k+3)^2} \right)$$
A: Use the Beta function (all results regarding $\beta$ are derivable by real methods)!
$$\beta (x,y)\stackrel{1}{=}\int_{0}^{\infty}\frac{t^{x-1}dt}{(1+t)^{x+y}}\stackrel{2}{=}\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
, and letting $y=2-x$ and $t=u^2$ in the first equality, 
$$\beta(x,2-x)=2\int_{0}^\infty\frac{u^{2x-1}du}{(1+u^2)^2}.$$
If we differentiate this with respect to $x$ (c.f. Leibniz integral rule), we obtain
$$\frac{d}{dx}\beta(x,2-x)=2\int_0^\infty\frac{\partial}{\partial x}\frac{u^{2x-1}du}{(1+u^2)^2}$$
$$=4\int_0^\infty  \log(x)\frac{u^{2x-1}du}{(1+u^2)^2}.$$
To kill the $u$ term in the numerator, let $x=\frac{1}{2}$, divide by four, and use the second equality:
$$I=\int_0^\infty  \frac{\log(u)du}{(1+u^2)^2}=\frac{1}{4}\frac{d}{dx} \beta\left(\frac{1}{2}, \frac{3}{2}\right)$$
$$=\frac{1}{4}\frac{d}{dx}\frac{\Gamma(x)\Gamma(2-x)}{\Gamma(2)}_{x=2}.$$
Obviously $\Gamma(2)=1!=1$, and $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$, together with $\Gamma(2-x)=(1-x)\Gamma(1-x)$, 
$$I=\frac{1}{4} \frac{d}{dx}(1-x) \frac{\pi}{\sin( \pi x)}=\frac{-\pi}{4}.$$

This method is easily applied to the integral 
$$I(a,b,c)=\int_0^\infty  \frac{\log^a(u)du}{(1+u^b)^c}, a \in \mathbb{N}, c \text{ is a nice number},$$
make the substitution $t=u^b$ in equality $1$, let $y=c-x$ and differentiate $a$ times with respect to $x$ (in fact, I strongly suspect partial derivatives could be used to drop the restriction that $a \in \mathbb{N}$, although I'm not sure whether partial derivatives of the Gamma function are well known). Then all that remains is to find a nicer form for $\Gamma(x)\Gamma(c-x)$ which you can take the $a$th derivative of, which mostly exists only for integral or some rational values of $c$, which is why I stipulated that $c$ has to be nice.
A: A really easy method using the substitution $x=\tan t$ offered by Jack D'Aurizio but followed by elementary integration by parts without using Fourier series.
$$\begin{align}
&\int_0^\frac{\pi}{2}\cos^2x\ln\tan x dx=\frac12 \int_0^\frac\pi2 \cos2x\ln\tan xdx\\
=&\frac14\sin2x\ln\tan x\bigg|^\frac{\pi}{2}_0-\frac14\int_0^\frac{\pi}2\frac{\sin 2x}{\sin x \cos x}dx=-\frac14\int_0^\frac\pi22dx=\color{blue}{-\frac\pi 4}
\end{align}$$
A: [Corrected and justified]
The change of variables $x = e^{-t}$ converts the integral to
$-\int_{-\infty}^\infty te^{-t} dt \left/(1+e^{2t})^2 \right.$.
Splitting into $\int_{-\infty}^0 + \int_0^\infty$ and combining
$t$ with $-t$ yields
$$
- \int_0^\infty \frac{t(e^{-t}-e^{-3t})dt}{(1+e^{-2t})^2}
= -\int_0^\infty t(e^{-t} - 3e^{-3t} + 5e^{-5t} - 7e^{-7t} + - \cdots) \, dt.
$$
Integrating termwise yields
$-(1 - \frac13 + \frac15 - \frac17 + - \cdots) = -\pi/4$.
Termwise integration requires some justification because the sum does not
converge absolutely, but no complex analysis is needed.
One way to justify the termwise integration in this setting is
to write the integral as the limit as $N \rightarrow \infty$ of
$$
- \int_0^\infty
 \frac{t(e^{-t}-e^{-3t}-2e^{-(4N+1)t})dt}{(1+e^{2t})^2}.
$$
Expanding $(e^{-t}-e^{-3t}-2e^{-(4N+1)t}) \left/ (1+e^{-2t})^2 \right.$
in powers of $e^{-t}$, and integrating termwise, yields
$$
S_N := -\left( 1 - \frac 13 + \frac15 - \frac17 + - \cdots \right.
\phantom{\infty\infty\infty\infty\infty\infty\infty\infty\infty\inftyinfty\infty\infty}
$$ $$
\phantom{\infty\infty\infty\infty\infty}
\left.
- \frac1{4N-1}
 + \frac{4N-1}{(4N+1)^2}
 - \frac{4N-1}{(4N+3)^2}
 + \frac{4N-1}{(4N+5)^2}
 - \frac{4N-1}{(4N+7)^2}
 + - \cdots \right),
$$
this time justified by absolute convergence.
The series $S_N$, like $-(1 - \frac13 + \frac15 - \frac17 + - \cdots)$,
is an alternating series whose terms decrease in absolute value.
The two series agree through the $1/(4N-1)$ term, and both remainders
are bounded in absolute value by $1/(4N+1)$.  Therefore as
$N \rightarrow \infty$ the new series $S_N$ approaches
$-(1 - \frac13 + \frac15 - \frac17 + - \cdots) = -\pi/4$,
and we're done.
A: Once you do the variable change in Ron Gordon's answer there's a trick that finishes it off pretty quickly. The integral is reduced to
$$\int_0^1 {1 - x^2 \over (1 + x^2)^2}\log(x)\,dx$$
$$= \int_0^1 {{1\over x^2} - 1 \over ({1 \over x}  + x)^2}\log(x)\,dx$$
Note that ${\displaystyle {{1\over x^2} - 1 \over ({1 \over x}  + x)^2}}$ is the derivative of ${\displaystyle{1 \over {1 \over x } + x}}$. So it's natural to integrate by parts in the above, and get
$$-\int_0^1{1 \over {1 \over x} + x} {1 \over x}\,dx$$
(The boundary terms here are both zero). This is the same as
$$-\int_0^1 {1 \over 1 + x^2}\,dx$$
$$= -{\pi \over 4}$$
A: An alternative approach is to enforce the substitution $x=\tan\theta$ then exploit the fact that the Fourier (cosine) series of $\log\tan\theta$ and $\cos^2\theta$ are pretty simple. Since for any $\theta\in\left(0,\frac{\pi}{2}\right)$
$$ \log\tan\theta = 2\sum_{k\geq 0}\frac{\cos((4k+2)\theta)}{(2k+1)} \tag{A}$$
$$ \cos^2\theta = \frac{1}{2}+\frac{\cos(2\theta)}{2} \tag{B}$$
$$ \int_{0}^{\pi/2}\cos(2m\theta)\cos(2n\theta)\,d\theta =\frac{\pi}{4}\delta(m,n)\tag{C}$$
the identity
$$ \int_{0}^{+\infty}\frac{\log(x)}{(1+x^2)^2}\,dx = \color{red}{-\frac{\pi}{4}} $$
is a straightforward consequence of $(C)$.

This approach also shows that $\int_{0}^{1}\frac{\log(x)}{(1+x^2)^{k+1}}\,dx$ just depends on the binomial transform of $\frac{(-1)^k}{2k+1}$. As shown here, the binomial transform of $\frac{1}{2k+1}$ is given by $\frac{4^k}{(2k+1)\binom{2k}{k}}$. With similar techniques one may prove that for any $k\in\mathbb{N}^+$
$$ \int_{0}^{+\infty}\frac{\log(x)\,dx}{(1+x^2)^{k+1}}=\color{red}{-\frac{\pi\binom{2k}{k}}{2\cdot 4^k}\sum_{m=1}^{k}\frac{1}{2m-1}}\approx -\frac{\sqrt{\pi}\log(2k)}{4\sqrt{k}}\tag{D} $$
which, of course, also follows by applying Feynman's trick to the integral representation of the Beta function.
A: Integrate by parts
\begin{align}
& \int_0^\infty \frac{\log x}{(1+x^2)^2} dx 
=\int_0^\infty \frac{\log x}{2x}d\left( \frac{x^2}{1+x^2} \right) \\
\overset{IBP}= &-\frac12 \int_0^\infty \frac{1}{1+x^2}dx+\frac12 \int_0^\infty \underset{x\to \frac1x} {\frac{\log x}{1+x^2}}dx
= -\frac12\cdot\frac\pi2+0=  -\frac{\pi}4
\end{align}
A: Using twice the result that $\int_0^{\infty} \frac{\ln x}{1+x^2} d x=0$, we have
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln x}{\left(1+x^2\right)^2} d x&=\int_0^{\infty} \frac{\left(1+x^2-x^2\right) \ln x}{\left(1+x^2\right)^2} d x \\
&=\int_0^{\infty} \frac{\ln x}{1+x^2} d x-\int_0^{\infty} \frac{x^2 \ln x}{\left(1+x^2\right)^2} d x \\
&=\frac{1}{2} \int_0^{\infty} x \ln x d\left(\frac{1}{1+x^2}\right) \\
&=\frac{1}{2} \int_0^{\infty} \frac{1-\ln x}{1+x^2} d x \\
&=\frac{1}{2} \int_0^{\infty} \frac{d x}{1+x^2} \\
&=\frac{\pi}{4}
\end{aligned}
$$
