Proving $\ln \lambda = \int_0^\infty \frac{\mathrm dt}t e^{-\lambda t}$

Another approach is to explicitly identify what the author called an infinite constant. This amounts to finding a proper (though not unique) regularization, and in this case we can do as follows:

$$ \int_{0}^{\infty} \left( \frac{\mathrm{e}^{-\lambda t}}{t} - \frac{\mathrm{e}^{-t}}{t} \right) \, \mathrm{d}t = \log \lambda. \tag{*}$$

Now this is a quintessential example of what is called Frullani integral. The differentiating-under-the-integral-sign trick works perfectly with this regularization that leads to the identity $\text{(*)}$.

Take a look at the footnote $4$ in the paper. This is not meant to be an identity, since $e^{-\lambda t}/t \sim 1/t$ as $t \to 0$ and the integral diverges.

Note that if you differentiate the LHS, you get $\frac{1}{\lambda}$. If you take the derivative with respect to $\lambda$ inside the integral of the RHS, you also get $$-\int_0^\infty \frac{1}{t} \frac{d}{d\lambda} e^{-\lambda t} \, dt = \int_0^\infty -e^{-\lambda t} = \frac{1}{\lambda}.$$

So the "derivative" on both sides is the same, which means that both sides only "differ by an infinite constant". The way I understand it, this is just a heuristic to derive equation (1.18), which is proven rigorously later in section 2.2.