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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.

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  • $\begingroup$ 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 ^^ $\endgroup$ – bilaterus Dec 15 '15 at 11:53
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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...

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    $\begingroup$ What the heck. It's zero! $\endgroup$ – Mathematicing Dec 15 '15 at 11:20
  • $\begingroup$ @Mathematicing: bingo. $\endgroup$ – Ron Gordon Dec 15 '15 at 11:21
  • $\begingroup$ Well, technically it's 0 iff the two branches converge, since this is an improper integral. Pretty sure that's the case iff $\lambda \ne 0$ $\endgroup$ – Alan Dec 15 '15 at 12:07
  • $\begingroup$ @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. $\endgroup$ – Ron Gordon Dec 15 '15 at 12:09
  • $\begingroup$ @RonGordon If memory serves, "branch" refers to "branch cut". It is a concept/ method commonly employed in complex integrals. $\endgroup$ – Mathematicing Dec 15 '15 at 13:57

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