Is this Expectation finite? How do I prove that
$\int_{0}^{+\infty}\text{exp}(-x)\cdot\text{log}(1+\frac{1}{x})dx$ 
is finite? (if it is)
I tried through simulation and it seems finite for large intervals. But I don't know how to prove it analytically because I don't know the closed form integral of this product.
I am actually taking expectation over exponential distribution.
Thank you

Note that $\int_{0}^{\infty}e^{-x}dx < \infty$ for $x \in \mathbb{R}$. Also note that $\log(1+\frac{1}{x})>1$ iff $x< \frac{1}{e-1}$, hence your integral $I$ can be upper-bounded by 
$$
I=\int_{0}^{\frac{1}{e-1}}e^{-x}\log (1+\frac{1}{x})dx + \int_{\frac{1}{e-1}}^{\infty}e^{-x} \log (1+\frac{1}{x})dx \\
< \int_{0}^{\frac{1}{e-1}}e^{-x}\log (1+\frac{1}{x})dx +\int_{\frac{1}{e-1}}^{\infty}e^{-x}dx
$$
Clearly both integral are finite, hence the original integral is finite too
