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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.

Marcel
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  • Indeed, you are done. One way to prove an improper integral converges is to evaluate it. – Matthew Leingang Mar 30 '16 at 17:44
  • The question was worded in a was that made it obvious I need to prove that it converges first. – Marcel Mar 30 '16 at 17:58
  • If this is a question from a course you're taking, get verification from your instructor that yours is the correct interpretation. I would never ask a question that way and expect separate arguments for (a) convergence and (b) the value it converges to. When I write the question this way, I want the students to either (1) show that the integral converges by evaluating it, or (2) show that the integral diverges by verifying $\lim_{b\to\infty}\int_0^b f(x),dx$ does not exist. – Matthew Leingang Mar 30 '16 at 18:53
  • My teacher assigned this right after teaching the comparison theory. I'm sure this time he wants us to use it rather than proving with limits. – Marcel Mar 30 '16 at 19:00

4 Answers4

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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.

John
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    I'm not sure the comparison to $x^3$ is obvious, but the comparison to $x^3/6$ is, and that would do the job here. – Ian Mar 30 '16 at 17:56
  • @lan Thanks for pointing out my mistake. I have modified my solution to express my original idea. – John Mar 30 '16 at 18:07
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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 $\log(1)=0$.

Community
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Mark Viola
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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.

Matthew Leingang
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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.

Laurent Duval
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