# Divergent Improper Integral

• Apr 18th 2010, 05:20 PM
kaiser0792
Divergent Improper Integral
I could use some help with an integral. The problem asks to determine whether a given integral converges or diverges. I know that the answer is "diverges" but I'm not sure why.

$\int_{-\infty}^0 xe^{-2x} dx$

I used integration by parts to get: $[xe^{-2x} - \int e^{-2x} dx] = xe^{-2x}-e^{-2x}$

Then I evaluated $\lim_{b\to-\infty}[xe^{-2x}-e^{-2x}]_{b}^{0} = (-1)-(be^{-2b}-e^{-2b})$

I understood $\lim_{b\to-\infty}be^{-2b}$ to require the use of L' Hopital's rule which I applied as follows: $\lim_{b\to-\infty}\frac{b}{e^{2b}}=\frac{1}{2e^2b}=\frac{1}{-\infty}=0$ If this were true, the improper integral would evaluate to -1 - 0 = -1 and would converge. Can anyone tell me where I went wrong?
• Apr 18th 2010, 05:46 PM
TKHunny
$\lim_{b\to -\infty}b\;=\;-\infty$

$\lim_{b\to -\infty}e^{2b}\;=\;0$

Perhaps your usage of L' Hopital was inappropriate.

Pardon the notation, but...

$\lim_{b\to -\infty}b\cdot e^{-2b}\;=\;-\infty \cdot \infty\;=\;Unbounded$

Why do you need any fancy theorem?
• Apr 18th 2010, 05:55 PM
AllanCuz
Quote:

Originally Posted by kaiser0792
I could use some help with an integral. The problem asks to determine whether a given integral converges or diverges. I know that the answer is "diverges" but I'm not sure why.

$\int_{-\infty}^0 xe^{-2x} dx$

I used integration by parts to get: $[xe^{-2x} - \int e^{-2x} dx] = xe^{-2x}-e^{-2x}$

Then I evaluated $\lim_{b\to-\infty}[xe^{-2x}-e^{-2x}]_{b}^{0} = (-1)-(be^{-2b}-e^{-2b})$

I understood $\lim_{b\to-\infty}be^{-2b}$ to require the use of L' Hopital's rule which I applied as follows: $\lim_{b\to-\infty}\frac{b}{e^{2b}}=\frac{1}{2e^2b}=\frac{1}{-\infty}=0$ If this were true, the improper integral would evaluate to -1 - 0 = -1 and would converge. Can anyone tell me where I went wrong?

Well,

$[xe^{-2x} - \int e^{-2x} dx] = xe^{-2x}-e^{-2x}$

Is not true.

$[xe^{-2x} - \int e^{-2x} dx] = xe^{-2x} + \frac {e^{-2x}} {2}$

But, just noticing and going back to your integration by parts, the above is not true as well (you did the by parts incorrectly).

$\int_{ - \infty }^0 xe^{-2x} dx = \frac { - xe^{-2x} } {2} - \frac{-1}{2} \int e^{-2x} = \frac { - xe^{-2x} } {2} - \frac{ e^{-2x}}{4}$

.........I did the limits on this before on your by parts, but since TKHunny already did this and I don't feel like doing it again with the correct equation I will leave it be. His answer satisfies why.
• Apr 18th 2010, 06:21 PM
kaiser0792
I realized after I posted that, my careless integration errors, i.e., missing the factor of 1/2.

But my real error was interpreting 1/(e^-2b) as b approaches negative infinity as resulting in 1/e^infinity rather than e^infinity. Thank you for clearing things up all, I'm okay now, I think!!