To find definite integral using differentiation under integral sign I want to calculate
$$I=\int_{0}^{\infty}\dfrac{\ln x}{x^2+\alpha^2}\,\mathrm{d}x.$$
my first attempt:
$$I(t)=\displaystyle\int_{0}^{\infty}\dfrac{\ln {tx}}{x^2+\alpha^2}\,\mathrm{d}x$$
$$I'(t)=\dfrac{1}{t}\displaystyle\int_{0}^{\infty}\dfrac{1}{x^2+\alpha^2}\,\mathrm{d}x=\dfrac{\pi}{2\alpha t}\implies I(t)=\dfrac{\pi\ln t }{2\alpha}\ + C$$
But I'm unable to calculate $C$ because $I(0)=\infty$ 
so,please give me hint  to calculate $C$ 
or hint me to stick parameter $t$ in right place
my second attempt (but without differentiation under integral sign):
I substituted $x=\alpha \tan t$
$I=\displaystyle \int _{0}^{\dfrac{\pi}{2}} \dfrac{\ln\alpha \tan t}{\alpha^2 \sec^2 t}\alpha \sec^2 t dt$
$I=\displaystyle\dfrac{1}{\alpha} \int _{0}^{\dfrac{\pi}{2}}{\ln\alpha \tan t} dt$
$I=\displaystyle\dfrac{1}{\alpha} \int _{0}^{\dfrac{\pi}{2}}{\ln \alpha} dt +\displaystyle\dfrac{1}{\alpha} \int _{0}^{\dfrac{\pi}{2}}\ln\tan t dt $
$I=\dfrac{1}{\alpha}\dfrac{\pi}{2}\ln \alpha+0$
$I=\dfrac{\pi}{2\alpha}\ln\alpha$
but this question was given in my coaching sheet under the topic DUIS (Differentiation under integral sign) so, i want to do it by that technique only, any help will be appreciated, thank you .. 
 A: It is enough to exploit symmetry. Assuming $\alpha>0$ we have
$$ \int_{0}^{+\infty}\frac{\log x}{x^2+\alpha^2}\,dx \stackrel{x\mapsto\alpha z}{=} \frac{1}{\alpha}\int_{0}^{+\infty}\frac{\log\alpha+\log z}{z^2+1}\,dz=\color{red}{\frac{\pi\log\alpha}{2\alpha}}+\frac{1}{\alpha}\int_{0}^{+\infty}\frac{\log z}{z^2+1}\,dz$$
and the last integral equals zero by mapping $z$ into $\frac{1}{w}$.
A: Let $F(a)$ be given by the integral
$$\begin{align}
F(a)&=\int_0^\infty \frac{x^a}{x^2+\alpha^2}\,dx\\\\
&=\alpha^{a-1} \int_0^\infty \frac{x^a}{x^2+1}\,dx\tag1
\end{align}$$
Next, enforce the substitution $x\mapsto e^x$ to obtain 
$$\begin{align}
\int_0^\infty \frac{x^a}{1+x^2}\,dx&=\int_{-\infty}^\infty \frac{e^{(a+1)x}}{1+e^{2x}}\,dx\\\\
&=\int_{-\infty}^0\frac{e^{(a+1)x}}{1+e^{2x}}\,dx+\int_{0}^\infty\frac{e^{(a-1)x}}{1+e^{-2x}}\,dx\\\\
&=\sum_{n=0}^\infty (-1)^n\left(\int_{-\infty}^0 e^{(2n+1+a)x}\,dx+\int_{0}^\infty e^{-(2n+1-a)x}\,dx\right)\\\\
&=\sum_{n=0}^\infty (-1)^n \left(\frac{1}{2n+1+a}+\frac{1}{2n+1-a}\right)\\\\
&=2\sum_{n=0}^\infty (-1)^n\left(\frac{2n+1}{(2n+1)^2-a^2}\right) \\\\
&=\frac{\pi}{2}\sec\left(\frac{\pi a}{2}\right)\tag 2
\end{align}$$


See THIS ANSWER along with other answers on that page and THIS ANSWER for an alternative approaches to evaluating $\displaystyle \int_0^\infty \frac{x^a}{x^2+1}\,dx$.


Substituting $(2)$ into $(1)$ reveals
$$F(a)=\alpha^{a-1}\frac{\pi}{2}\sec\left(\frac{\pi a}{2}\right)\tag 3$$
Differentiating $(3)$ and setting $a=0$ yields
$$F'(0)=\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}\,dx=\frac{\pi \log(\alpha)}{2\alpha}$$
A: Here is an alternative complex method that still uses differentiation under the integral sign. Denote
$$I(b)=\int_0^\infty \frac{x^b}{x^2+\alpha^2}\,dx=\frac{1}{2}\int_{-\infty}^\infty \frac{x^b}{x^2+\alpha^2}\,dx$$
Assuming that $b=2\beta$, for $\beta\in\mathbb{R}$. Now,  $b$ is restricted to certain values to ensure convergence, and we will choose $b\in[0,1)$. We are then trying to compute $I'(0)$. 
We will evaluate $I(b)$ via the Residue Theorem, differentiate the resulting function, and let $b\to 0$. 
Using a semi-circle centered at the origin, its contour integral is given by the upper semi-circle, parametrized by $z=Re^{i\theta},\,\,\theta
\in[0, \pi]$ added to the real contribution, parametrized by $z=x,\,\,x\in[-R,R]$.
$$ \int_{-R}^R \frac{x^b}{x^2+\alpha^2}\,dx\, +\, \int_0^{\pi}\frac{R^be^{ib\theta}}{R^2e^{2i\theta}+\alpha^2}Rie^{i\theta}\,d\theta$$
Because of our choice of $b$, the second integral vanishes as $R\to\infty$. The first integral is equal to $2I(b)$ in the same limit. Using the Residue Theorem, 
$$\int_{-\infty}^{\infty} \frac{x^b}{x^2+\alpha^2}\,dx=2\pi i\,\text{Res}\left(z=\alpha i\right) \frac{z^b}{z^2+\alpha^2}\ $$
As the singularity of our function inside our semi circle occurrs at $z=\alpha i$. The Residue is equal to 
$$\text{Res}=\lim_{z\to\alpha i} \,\,\,(z-\alpha i)\frac{z^b}{(z-\alpha i)(z+\alpha i)}=\frac{\alpha^b i^b}{2\alpha i}$$
Thus, 
$$2I(b)=2\pi i \frac{\alpha^b i^b}{2\alpha i}$$
$$I(b)=\frac{\pi}{2\alpha}\alpha^b i^b$$
Differentiating both sides with repect to $b$, 
$$I'(b)=\frac{\pi}{2\alpha} \left(\alpha^b i^b \log \alpha +\alpha^bi^b\frac{i\pi}{2}\right)$$
Letting $b\to0$, and taking the real part of our expression, we have 
$$\int_0^{\infty}\frac{\log x}{x^2+\alpha^2}\,dx=\frac{\pi}{2\alpha}\log \alpha$$
