This is one heck of an embarrassment but it is impressive exactly how these bits of subtlety gets lost in the earlier of the head after the initially year of undergraduate studies-with eexceptionally computation chucked into brianowens.tvematica.

You are watching: Integral of e^abs(x)

Here goes, I"m trying to incorporate \$int_-infty^+inftye^ x ideal e^-ikxdx\$

As usual, absolute value is somepoint to be wary of.

I spilt the integral down to the negative and positive x domajor to account for the \$|x|\$.

Then, we have

\$int_-infty^0e^-aleft ( -x appropriate )e^-ikxdx+int_0^+inftye^-axe^-ikxdx\$

or maybe, by symmeattempt,

\$2int_0^+inftye^-axe^-ikxdx\$

Have I nailed this?

integration definite-integrals improper-integrals indefinite-integrals absolute-worth
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edited Jan 4 "16 at 5:07
brianowens.tvematicing
asked Jan 4 "16 at 5:05

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This was correct prior to the edit; the expression

\$\$int_-infty^0 e^-a(-x) e^-ikx , dx + int_0^infty e^-ax e^-ikx , dx\$\$

was correct, yet it is not equal to \$2 int_0^infty e^-ax e^-ikx , dx\$. Rather, the first integral above is

eginalign*int_-infty^0 e^ax e^-ikx , dx = int_0^infty e^-ax e^+ikx , dxendalign*

Keep in mind the sign on the exponential is various.

As an error inspect, note that the Fourier transform of an also feature is actual. If you compute the integral \$int_0^infty e^-ax e^-ikx , dx\$, you carry out not acquire a actual attribute.

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answered Jan 4 "16 at 5:12
user296602user296602
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