$\int_{-\infty}^{0}xe^{x}$ diverge or converge

Question

How would one find whether the following improper integral converge or diverge.

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

I did the following.

$t\rightarrow\infty$

$\int_{t}^{0}xe^x$

I did the integration by parts.

$u=x$

$dv=e^x$

$xe^x-\int 1e^x$

$xe^x-1xe^x$

$(0)(e^0)-e^0(0)-te^t-te^t$

$0-\infty-\infty$

would this mean there is a divergence.

Answer 1

You've made an error in your integration by parts. $$xe^x-\int 1e^x=xe^x-e^x=(x-1)e^x$$

It can be shown that $$\lim_{x\to-\infty}(x-1)e^x=0,$$ so the improper integral converges.

As a side note, we also have that $$\lim_{x\to-\infty}xe^x=0,$$ so even if your integration by parts hadn't been erroneous, your evaluation of the limit would have been.

Answer 2

Step 1: set the limit $\lim_{t \rightarrow -\infty} $

Step 2: find integral (i got $xe^x - e^x$)

Step 3: plug in the limits and ftc, so u get $\lim_{ t-> -\infty} [(0-e^0)-(te^t-e^t)]$

*key: $e^\infty = \infty$, $e^-\infty = 0$

step 4: evaluate

answer: convergent sum $-1 $