I'm trying to find the canonical representation of the exponential distribution with mean $1.$ The canonical representation of an infinitely divisible random variable has the next form:
$$\phi(t)=\exp\int_{\mathbb{R}}(e^{itx}-1-itx)\frac{1}{x^{2}}\mu(dx),$$ where $\mu$ is a finite measure.
The exercise asks to prove a series of steps to do this; I'm stuck in the next because I'm not familiar with complex integration:
Show that (use the principal branch of the logarithm or else operate formally for the moment) $d(\log(\phi(t)))/dt=i\phi(t),$ where $\phi(t)$ is the characteritic function of the exponential distribution. Integrate with respect $t$ to obtain $$\frac{1}{1-it}=\exp\int_{0}^{\infty}(e^{itx}-1)\frac{e^{-x}}{x}dx.$$ Verify the previous formula after the fact by showing that the ratio of the two sides has derivative $0.$
I've computing the characteristic function of exponential distribution, without using complex variable, utilizing the functions $\frac{1}{1+t^2}e^{-x}(-\cos(tx)+t\sin(tx))$ and $\frac{1}{1+t^2}e^{-x}(-t\cos(tx)-\sin(tx))$ whose derivatives are the real and complex part of the characteristic function. So we have $$\phi(t)=\frac{1}{1+t^2}-i\frac{t}{1+t^2}=\frac{1}{1-it}.$$
I don't find a way to prove the equality requested. What about the ratio of the two sides?
Any kind of help is thanked in advanced.