How to prove $\int_{0}^{\infty}xe^{-x}dx$ converges without calculating the limit I am given $\int_{0}^{\infty}xe^{-x}dx$ and asked to prove that it converges and if it does, calculate the integral.
I have calculated the integral and it gives 1. However, I cannot find a way to prove that it converges before calculating the limit.
 A: Note $e^{x}\ge x^3$ for $x>M$ when $M$ large enough, and we know $\int_0^\infty xe^{-x} dx=\int_0^M xe^{-x} dx+\int_M^\infty xe^{-x} dx$.
$\int_M^\infty xe^{-x} dx\le \int_M^\infty x \dfrac{1}{x^3} dx<\infty$.
Also $\int_0^M xe^{-x} dx<\infty$ since the integrand is continuous.
A: Enforce the substitution $x\to \log(x)$ and write
$$\int_0^\infty xe^{-x}\,dx=\int_1^\infty \frac{\log(x)}{x^2}\,dx$$
Then, note that for any $\alpha>0$ and $x\ge 1$, we have
$$0\le \log(x)\le \frac{x^{\alpha}-1}{\alpha} \tag 1$$
Can you finish?

NOTE:
To establish the inequalities in $(1)$, we use the result from THIS ANSWER in which I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\frac{z-1}{z}\le \log(z)\le z-1 \tag 2$$
for $z>0$.  Now, since 
$$\log(x^\alpha)=\alpha \log(x)$$
we establish the right-hand side inequality of $(1)$ by letting $z=x^\alpha$ in $(2)$ using $(3)$, and dividing by $\alpha>0$.  
The left-hand side inequality of $(1)$ is true since $\log(x)$ is monotonically increasing with $\log(1)=0$.
A: Here is another comparison method.  We can use L'Hôpital's Rule to show that
$$
    \lim_{x\to\infty} xe^{-x/2} = 0
$$
Therefore there exists a number $M$ such that 
$$
0 < x e^{-x/2} < 1 \quad\text{when $x>M$}
$$
It follows that
$$
x e^{-x} < e^{-x/2} \quad\text{when $x>M$}
$$
So you can use the Comparison test to show that $\int_M^\infty xe^{-x}\,dx$ converges.
A: A standard inequality yields, for $x\ge 0$:
$$0\le \frac{x}{2}\le e^\frac{x}{2}\,. $$
Hence 
$$ \int_0^{X} x e^{-x}$$
being the integral of a non-negative function is increasing, and bounded above by 
$$ \int_0^{X} 2  e^{\frac{x}{2}} e^{-x} $$
which converges. Hence the integral is convergent.
