Question

evaluate the improper integral or state that it diverges: integral from -inf to inf (2xe^-x) dx. can someone explain, thnx!!

Answer 1

OK, so it's helpful to first know what the formula for the simpler integral of (xe^-x) dx is:
\[\int\limits xe^{-x} dx = -xe^{-x}-e^{-x}\]
which can be found by integration by parts.

Answer 2

Of course we are looking for the slightly different integral
\[\int\limits_{-\infty}^{\infty} 2xe^{-x} dx\]

Answer 3

Alright, so the trick with these types of "double infinity" integrals is to break the integral into two separate pieces. We can break it anywhere, but I will choose to break the integral at x = 0, this gives
\[\int\limits_{-\infty}^{\infty} 2xe^{-x} dx = \int\limits_{-\infty}^0 2xe^{-x} dx + \int\limits_0^{\infty} 2xe^{-x} dx\]

Answer 4

Now, we see that we have two questions for the price of one! While that is a bit annoying, it is a lot easier to work with one infinity at a time.

Answer 5

Let's look at the first integral. To be precise we are really taking a limit as something goes toward - infinity so,
\[\int\limits_{-\infty}^0 2xe^{-x} dx= \lim_{a \rightarrow -\infty} \int\limits_a^0 2xe^{-x} dx \]

Answer 6

Ack! I made an error, let me fix that real quick...

Answer 7

Ah, tricky... the first integral will not converge. Notice we will have
\[2 \lim_{a \rightarrow -\infty} [-xe^{-x} - e^{-x}]_a^0 = 2[-0e^{-0} - e^{-0} - \lim_{a \rightarrow -\infty}(-ae^{-a} - e^{-a})]\]
which we can simplify to
\[2[-1 - \lim_{a \rightarrow -\infty}(-ae^{-a} - e^{-a})]\]

Answer 8

The trouble is with the limit. Observe that
\[\lim_{a \rightarrow -\infty}(-ae^{-a} - e^{-a}) = \lim_{a \rightarrow -\infty} (-a-1)e^{-a} \equiv (-[-\infty] - 1)e^{-[-\infty]} = (\infty - 1) e^\infty\]

Answer 9

But this will be infinite!

Answer 10

So, the first integral becomes
\[\int\limits_{-\infty}^0 2xe^{-x} dx = 2[-1 - \infty] = -\infty\]
Unless the second integral turns out to be positive infinity our original integral is doomed to diverge. But the second integral can be solved to get 2, so our final answer is
\[\int\limits_{-\infty}^{\infty} 2xe^{-x} dx = \int\limits_{-\infty}^0 2xe^{-x} dx + \int\limits_0^{\infty} 2xe^{-x} dx = -\infty + 2 = -\infty\]

Answer 11

The integral is divergent.