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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## 1 Answer 1

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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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