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

There are two things to show here:
1) The integral is finite on, say, $[1,\infty)$.
2) The integral is finite on, say, $[0,1)$.
The first part is easy: $\log(1+\frac{1}{x})$ decreases with $x$; so, for $x\geq 1$, 
$$
0\leq\log\left(1+\frac{1}{x}\right)\leq\log\left(1+\frac{1}{1}\right)=\log2;
$$
hence
$$
\int_1^{\infty}e^{-x}\log\left(1+\frac{1}{x}\right)\,dx\leq\log2\int_1^{\infty}e^{-x}\,dx,
$$
which you can show is finite.
For the second part, the $e^{-x}$ isn't helping us at all - it just approaches 1 as $x\rightarrow0$, which doesn't help offset the fact that $\log(1+\frac{1}{x})\rightarrow\infty$. So, you might as well bound $e^{-x}\leq 1$ for $x\geq 0$, so that
$$
\int_0^1e^{-x}\log\left(1+\frac{1}{x}\right)\,dx\leq\int_0^1\log\left(1+\frac{1}{x}\right)\,dx=\int_0^1\left(\log(x+1)-\log(x)\right)\,dx.
$$
Try handling it from here!
