# Integrating an absolute value on exponential

This might be a bit rusty but hopefully it can be brushed up.

I need to integrate $$\int_{-\infty}^{\infty}xe^{-2\lambda \left | x \right |}dx$$

Recall:

$$\left | x \right |=\left\{\begin{matrix} x &x\geq 0 \\ -x& x< 0 \end{matrix}\right.$$

Then,

$$\lim_{t\rightarrow \infty}\int_{-t}^{0}xe^{-2\lambda(-x)}dx+\lim_{t\rightarrow \infty}\int_{0}^{t}xe^{-2\lambda(x)}dx$$

I would appreciate a nudge. Intuition suggest odd and even function have a role to play. Absolute values are nasty.

• As Ron Gordon explained, the answer slides out nicely given the fact the function is odd. But you should also be able to integrate $xe^{kx}$ (hint: by parts). Absolute values aren't too bad: you just need to split them up in their two cases (as you've done) and then look at each case as you would for any other normal question ^^ – bilaterus Dec 15 '15 at 11:53

Absolute value is even, so exponential is even. $x$ is odd. Odd times even is odd. Thus, an integral of an odd function about a symmetric interval is...
• Well, technically it's 0 iff the two branches converge, since this is an improper integral. Pretty sure that's the case iff $\lambda \ne 0$ – Alan Dec 15 '15 at 12:07
• @Alan: I think I know what you mean, but I do not know what a "branch" is nor how it converges. I do know, though, that if $\lambda=0$, the integral is a Cauchy principal value defined as the limit of a symmetric finite interval. – Ron Gordon Dec 15 '15 at 12:09