Constructing improper integrals that can be solved by ${\int_{0}^{\infty}\frac{f(t)}{t}\text{d}t=\int_{0}^{\infty}\mathcal{L}\{f(t)\}\text{d}s}$ One useful formula evaluate some improper integrals is the following;
$$\boxed{\int_{0}^{\infty}\frac{f(t)}{t}\text{d}t=\int_{0}^{\infty}\mathcal{L}\{f(t)\}\text{d}s}$$
where $\mathcal{L}\{f(t)\}$ is the Laplace transform of the function $f(t)$.

I know an example of this which is
$$\int_{0}^{\infty}\frac{\sin(t)}{t}\text{d}t=\int_{0}^{\infty}\mathcal{L}\{\sin(t)\}\text{d}s=\int_{0}^{\infty}\frac{1}{s^2+1}\text{d}s=\tan^{-1}(s)|_{0}^{\infty}=\frac{\pi}{2}$$

Now I need to have some other examples to use this formula, I tried to construct (using inverse Laplace) but failed to do so.
Can you recommend me some easy examples as the one above?

Your help would be appreciated. Thanks!
 A: If you read it backward, we can combine that identity with the Frullani integral to get the formula: $$
\int_{0}^{\infty} \alpha f(\alpha s) - \beta f(\beta s)\, \mathrm{d}s = \int_{0}^{\infty} \frac{\mathcal{L}^{-1}\left\{f\right\}\left( \frac{t}{\alpha}\right) - \mathcal{L}^{-1}\left\{f\right\}\left( \frac{t}{\beta}\right)}{t}\, \mathrm{d}t = \mathcal{L}^{-1}\left\{f\right\}\Big\vert_{t=0}^{t=\infty} \ln\left\vert\frac{\beta}{\alpha}\right\vert
$$
where  $\lim\limits_{t\to 0^+} \mathcal{L}^{-1}\left\{f(s)\right\}(t)$ and $\lim\limits_{t\to \infty} \mathcal{L}^{-1}\left\{f(s)\right\}(t)$ exist.

One example you where can use this is
$$
\int_0^{\infty} \arctan\left( \frac{1}{x\pm 1}\right) - \frac{1}{\xi}\arctan\left(\frac{\xi}{x \pm \xi}\right)\, \mathrm{d}x = \ln|\xi|
$$
where two sign choices are independent of each other. This integral looks rather complicated considering the fact that the integrand can be discontinuous and that splitting the integral results into two divergent integrals. However, if you know that
$$\mathcal{L}\left\{\frac{\sin(t)}{t}e^{-t} \right\} = \int_0^{\infty}\frac{\sin(t)e^{-t(s+1)}}{t}\, \mathrm{d}t= \int_0^{\infty}\int_0^{\infty} e^{-t(s+1 +x)}\sin(t)\,\mathrm{d}t\, \mathrm{d}x=\arctan\left(\frac{1}{s+1}\right)
$$ then an immediate application of the formula with $\alpha = \pm 1$ and $\beta = \pm\frac{1}{\xi}$ gives the result.
