1
$\begingroup$

Evaluate $\int_{-\infty}^0xe^{-4x}dx$.

I first did integration by parts. My result is the following: $\frac{-1}{4}xe^{-4x}-\frac1{16}e^{-4x}$, which I will evaluate from $-\infty$ to $0$ using the fundamental theorem of calculas. The part that gets evaluated at zero is just $\frac{-1}{16}$. I keep having problems figuring out $\lim\limits_{b \to -\infty} \frac{xe^{-4x}}{4}+\frac{e^{-4x}}{16}$. For the first summand, I use L'Hopital's rule and keep ending up with $-\infty$. For the second summand, I get $\infty$. This is a problem because I $-\infty+\infty$ is still indeterminant, and I can't figure out if I'm messing up a sign somewhere.

I know that I can evaluate this by factoring out $e^{-4x}$ from both summands. The final result is $-\infty$. But I want to figure out how to do it without factoring as well.

  • 0
    It's from $\infty$ to $0$, not from $-\infty$ to $0.$ If it were $-\infty$ to $0$ you would rightly get $-\infty$ – 2017-01-26
  • 0
    Well, your integral evaluates from $\infty$ to $0$ and not from $-\infty$. This should restore order. – 2017-01-26
  • 0
    Sorry, it was supposed to say $-\infty$ – 2017-01-26
  • 0
    Well then the integral is $-\infty.$ $e^{-x}$ blows up in that direction. – 2017-01-26

3 Answers 3

0

The integral $\int_{-\infty}^0 xe^{-4x}\,dx$ does not converge.

The following solution is applicable to the originally posted question. After this solution was posted, the OP changed the lower limit from $\infty$ to $-\infty$.

Begin by writing

$$\int_{\infty}^0xe^{-4x\,}dx=-\int_0^{\infty}xe^{-4x}\,dx$$

Then, integrating by parts with $u=x$ and $v=-\frac14e^{-4x}$ yields

$$\begin{align} -\int_0^{\infty}xe^{-4x}\,dx&=-\left.\left(-\frac14 xe^{-4x}\right)\right|_{0}^{\infty}-\frac14\int_0^\infty e^{-4x}\,dx\\\\ &=-\frac14 \int_0^\infty e^{-4x}\,dx\\\\ &-\frac{1}{16} \end{align}$$

since using L'Hospital's Rule we have

$$\begin{align} \lim_{x\to \infty}xe^{-4x}&=\lim_{x\to \infty}\frac{x}{e^{4x}}\\\\ &=\lim_{x\to \infty}\frac{1}{4e^{4x}}\\\\ &=0 \end{align}$$

EDIT:

If the lower limit is $-\infty$, then enforcing the substitution $x\to -x$ reveals

$$\int_{-|L|}^0 xe^{-4x}\,dx=-\int_0^{|L|}xe^{4x}\,dx\le -\int_0^{|L|}x\,dx=-\frac12 |L|^2$$

Hence, $\lim_{|L|\to \infty}\int_{-|L|}^0 xe^{-4x}\,dx=-\infty$. And we are done!

  • 0
    According to my textbook, the solution should be $-\infty$ though. – 2017-01-26
  • 0
    @mmm If the lower limit is $+\infty$, then your answer $-1/16$ is correct. If the lower limit is $-\infty$, the improper integral diverges (to $-\infty$). – 2017-01-26
  • 0
    That's the part I needed clarification on. How can I show analytically that it diverges to $-\infty$? – 2017-01-26
  • 0
    @mmm I've edited by adding a section labeled $EDIT$ that proves the limit of the improper integral is $-\infty$. There is no integration by parts required. – 2017-01-26
  • 0
    @mmm Did that suffice? – 2017-01-26
0

Sometimes is useful:

Using $\displaystyle\int pe^xdx=e^x(p-p'+p''-p'''+\cdots)$ where $p$ is a polynomial, we write

$$\frac{1}{16}\int-4xe^{-4x}d(-4x)=\frac{1}{16}\int ue^{u}du=\frac{1}{16}e^u(u-1)=\frac{1}{16}e^{-4x}(-4x-1)$$

0

Indefinite integral is $-\exp{(-4x)}\left(\frac{1}{16}+\frac{x}{4}\right)$. Whether it converges or not depends on the bounds.