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
 A: Note  that $\log(1+\frac{1}{x})=\log(1+x)-\log x$
We have 
$$e^{-x}\log(1+x)=_\infty o\left(\frac{1}{x^2}\right)$$
hence the integral
$$\int_0^\infty e^{-x}\log(1+x)dx\tag{1}$$
is convergent.
Moreover, since
$$e^{-x}\log(x)\sim_0\log x$$
then the integral
$$\int_0^1 e^{-x}\log x dx$$
is convergent and as in $(1)$ we prove easily that the integral
$$\int_1^\infty e^{-x}\log(x)dx$$
is convergent and finally we conclude.
A: If we write $$\int_{0}^{+\infty}=\int_0^1+\int_1^{\infty}$$ then we see that $$\lim_{x\to 0^+} x^{1/2}f(x)=0<\infty\Longrightarrow\int_0^1f(x)dx~~\text{is convergent}$$ and $$\lim_{x\to +\infty} x^{2}f(x)=0<\infty\Longrightarrow\int_1^{\infty}f(x)dx~~\text{is convergent}$$
A: 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!
A: 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
