Integrate e^-a|x|: Troubleshooting & Solution

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SUMMARY

The integral of the function \( e^{-a|x|} \) over the entire real line evaluates to \( \frac{2}{a} \). This result is derived by recognizing the symmetry of the function about the x-axis, allowing the integral to be split into two parts: \( \int_{-\infty}^{0} e^{ax} \, dx \) and \( \int_{0}^{\infty} e^{-ax} \, dx \). The incorrect approach of evaluating the integral directly without considering the absolute value leads to a misleading result of zero, due to the differing behaviors of the function on either side of zero.

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Homework Statement



I'm a little confused with this integral:

[tex]\int^\infty_{-\infty}e^{-a|x|}\,dx[/tex]

Homework Equations


The Attempt at a Solution



Now, I believe the typical way to evaluate this is to say, hey, because of the |x|, this thing is symmetric about the x axis, and so we can instead evaluate:

[tex]\int^\infty_{-\infty}e^{-a|x|}\,dx = 2\int^\infty_0e^{-ax}\,dx[/tex]
[tex]= 2[-\dfrac{1}{a}e^{-ax}|^\infty_0] \, \, = \, 2[\dfrac{1}{a}] \, \, = \, \dfrac{2}{a}[/tex]

which I believe is correct. However, and this is my question, can it be evaluated without using this trick? I ran into trouble, and I'm not sure where I made my mistake, although I suspect it has to do with not really doing anything about the absolute value of x:

[tex]\int^\infty_{-\infty}e^{-a|x|}\,dx[/tex]

[tex]= \, \, [-\dfrac{1}{a}e^{-a|x|}|^\infty_{-\infty}][/tex]

[tex]= \, \, -\dfrac{1}{a}[e^{-a|\infty|} - e^{-a|-\infty|}][/tex]

[tex]= \, \, -\dfrac{1}{a}[e^{-a\infty} - e^{-a\infty}][/tex]

[tex]= \, \, -\dfrac{1}{a}[0 - 0] \, = \, 0[/tex]

And I've gotten nowhere, but I can't tell why, or what mistake I committed (if any).

Thoughts? What am I missing here? Thanks!
 
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the first way is correct it comes form the fact
[tex]e^{-a|x|} = e^{-ax},x\geq 0[/tex]
[tex]e^{-a|x|} = e^{ax},x < 0[/tex]

so the integral becomes
[tex]\int^\infty_{-\infty}e^{-a|x|}dx = \int^{0}_{-\infty}e^{ax}dx + \int^{\infty}_{0}e^{-ax}dx[/tex]

which simplifies to what you gave (use substitution u = -x in first part)

your 2nd interegral is not valid due to the different behaivour of [itex]e^{-a|x|}[/itex] either side of zero
 

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