Proving $\ln \lambda = \int_0^\infty \frac{\mathrm dt}t e^{-\lambda t}$ I am currently reading this paper. On page 5, it writes:

For each positive eigenvalue $\lambda$ of the operator $D$ we may write an identity
  $$\ln \lambda = \int_0^\infty \frac{\mathrm dt}t e^{-\lambda t}.\tag{1.17}$$
  This identity is "correct"  to an infinite constant, which does not depend on $\lambda$ and, therefore, may be ignored in what follows. Now we use $\ln \det(D)=\mathrm{Tr}\ln(D)$ and extend $(1.17)$ to the

I can not prove this identity. I tried
$$\int_0^{\infty} \frac{\mathrm dt}{t}e^{-t\lambda}=\int_0^{\infty} \frac{\mathrm d(\lambda t)}{(\lambda t)}e^{-t\lambda}=\int_0^{\infty} x^{-1}e^{-x}\ \mathrm dx=\Gamma(0),$$
where fort the third equality, I used the definition of Gamma function.
 A: 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.
A: 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 $\text{(*)}$.
