$f'(x)=\frac{\text{d} f(x)}{\text{d}x}=\int_{0}^{\infty} \frac{\partial }{\partial x}\left(\frac{e^{-u}-e^{-u^{\alpha}x}}{u}\right) du$
Taking, $z=u^{\alpha}x$ we get
$=\frac{x^{-1}}{\alpha}(\int_{0}^{\infty} e^{-z} dz) =\frac{x^{-1}}{\alpha}$.....(1)
Taking, $f(x)=y$
So, from (1) we get the differential equation
$\frac{\text{d} y}{\text{d}x}=\frac{x^{-1}}{\alpha}$ with the boundary condition $y(1)=\int_{0}^{\infty} \frac{e^{-u}-e^{-u^{\alpha}}}{u} du =-\frac{(\alpha-1)}{\alpha}\gamma$
Proof: Take $y_{alpha}(1)=\psi(\alpha)=\int_{0}^{\infty} \frac{e^{-u}-e^{-u^{\alpha}}}{u} du$
So, $\frac{\text{d}\psi(\alpha)}{\text{d}x}=\int_{0}^{\infty} \frac{\partial}{\partial \alpha}\left(\frac{e^{-u}-e^{-u^{\alpha}}}{u}\right) du$
$=\int_{0}^{\infty} u^{\alpha-1}\text{ln}(u)e^{-u^{\alpha}} du$
Taking $u^{\alpha}=z$
$=\frac{1}{{\alpha}^2}\int_{0}^{\infty} \text{ln}(z)e^{-z} dz=-\frac{\gamma}{{\alpha}^2}$
As $\Gamma'(1)=-\gamma$ (https://en.m.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant)
So, we get the differential equation
$\frac{\text{d}\psi}{\text{d}\alpha}=\frac{-\gamma}{{\alpha}^2}$ with the boundary condition $\psi(1)=0$ we get
$y(1)=\psi(\alpha)-\psi(1)=\psi(\alpha)=-\frac{\alpha-1}{\alpha}\gamma$
And for, all $\alpha\geq 1$, $y(1)=\frac{-(\alpha-1)\gamma}{\alpha}$.
Where, $\gamma$ is Euler- Mascheroni constant.
So, $f(x)=\frac{\text{ln}x}{\alpha} -\frac{(\alpha-1)\gamma}{\alpha}$.
So, for $\alpha=1$, $f(x)=\text{ln}x$.
Let $f \colon (0,\infty)^2 \to \mathbb{C}, \, f(a,b) = \int_{-\infty}^\infty \frac{\log(a^2+x^2)}{(x-\mathrm{i} b)^2} \, \mathrm{d} x$. We can get rid of $a$ by letting $x = a t$ and using the result $\int_{-\infty}^\infty \frac{\mathrm{d} t}{(t-\mathrm{i} c)^2} = 0$ for $c > 0$:
$$ f(a,b) = \frac{1}{a} \int \limits_{-\infty}^\infty \frac{2 \log(a) + \log(1+t^2)}{\left(t - \mathrm{i} \frac{b}{a} \right)^2} \, \mathrm{d} t = \frac{1}{a} \int \limits_{-\infty}^\infty \frac{\log(1+t^2)}{\left(t - \mathrm{i} \frac{b}{a} \right)^2} \, \mathrm{d} t \, . $$
Integration by parts then yields
\begin{align}
f(a,b) &= \frac{2}{a} \int \limits_{-\infty}^\infty \frac{t}{(1+t^2)\left(t - \mathrm{i} \frac{b}{a}\right)} \, \mathrm{d} t = \frac{2}{a} \int \limits_{-\infty}^\infty \frac{t \left(t + \mathrm{i} \frac{b}{a}\right)}{(1+t^2)\left(\frac{b^2}{a^2} + t^2\right)} \, \mathrm{d} t \\
&= \frac{2}{a} \int \limits_{-\infty}^\infty \frac{t^2}{(1+t^2)\left(\frac{b^2}{a^2} + t^2\right)} \, \mathrm{d} t \, ,
\end{align}
since the imaginary part of the integrand is an odd function. For $a \neq b$ we can now use partial fractions to compute the remaining integral, while for $a = b$ integrating by parts once more does the trick. The final result in either case is
$$ f(a,b) = \frac{2}{a} \frac{\pi}{1 + \frac{b}{a}} = \frac{2\pi}{a+b} \, . $$
Best Answer
Consider first $$\int_{-\infty}^\infty \frac{i}{y^2-x^2+i\epsilon} dy$$ for fixed $x$. We will use the identity $$\frac{1}{f+i\epsilon}=\mathrm{PV}\frac{1}{f}-i\pi\delta(f)\,,$$ where PV denotes the principal value. Noting that $$ \int_{-\infty}^{+\infty} \mathrm{PV}\frac{1}{y^2-x^2}dy= \frac{1}{2x}\int_{-\infty}^{+\infty} \mathrm{PV}\left(\frac{1}{y-x}-\frac{1}{y+x}\right)dy=0 $$ and $$ \delta(y^2-x^2)dy=\frac{1}{2x}\left( \delta(y-x)+\delta(y+x) \right)dy\,, $$ we then have $$\int_{-\infty}^\infty \frac{i}{y^2-x^2+i\epsilon} dy=\frac{\pi}{x}\,.$$ The integral over $x$ then yields $$\pi\int_{\Lambda/s}^{\Lambda}\frac{dx}{x} = \pi\log s\,. $$