# Math Help - Improper Integrals - Convergence/Divergence

1. ## Improper Integrals - Convergence/Divergence

Evaluate the following problem and determine if it converges or diverges. If it converges, calculate what it converges to.

$\displaystyle\int^0_{-\infty} xe^x\,dx$

$\displaystyle\int^0_{-\infty} xe^x\,dx=\lim_{t\to-\infty}\int^0_t xe^x\,dx=\lim_{t\to-\infty}(xe^x-e^x)\bigg|_{0}^{t}$

$=[e^0(0-1)]-[e^{-\infty}(-\infty-1)]$

$=[-1]-[0(-\infty)]{}$

I know the answer is that it conveges to -1, but wouldn't it diverge since I end up with $-1-(0*\infty)$ ?

Thanks!

2. First of all, I don't know why you switched the order of the terminals...

Second, $\displaystyle \lim_{t \to -\infty}[e^t(t - 1)]$ is of the indeterminate form $\displaystyle 0 \times \infty$.

So rewrite it as $\displaystyle \lim_{t \to \infty}\left[\frac{t - 1}{e^{-t}}\right]$, which is now of the form $\displaystyle \frac{\infty}{\infty}$ and L'Hospital's Rule can be used.

3. Originally Posted by Prove It
First of all, I don't know why you switched the order of the terminals...

Second, $\displaystyle \lim_{t \to -\infty}[e^t(t - 1)]$ is of the indeterminate form $\displaystyle 0 \times \infty$.

So rewrite it as $\displaystyle \lim_{t \to \infty}\left[\frac{t - 1}{e^{-t}}\right]$, which is now of the form $\displaystyle \frac{\infty}{\infty}$ and L'Hospital's Rule can be used.
That was a typo, sorry....I certainly did not mean to switch the terminals. On the third line of your response, did you purposely switch the limit from $-\infty$ to $\infty$ or is that a typo?

If it is a typo, here is what I came up with...

$\displaystyle\int^0_{-\infty} xe^x\,dx=\lim_{t\to-\infty}\int^0_t xe^x\,dx=\lim_{t\to-\infty}(e^x(x-1))\bigg|^0_t=-1-\lim_{t\to-\infty}[e^t(t-1)]$

$\displaystyle =-1-\lim_{t\to-\infty}\frac{t-1}{e^{-t}}=^{LH}-1-\lim_{t\to-\infty}\frac{1}{-e^{-t}}=-1-0=converges\to -1$