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,
Have I nailed this?
integration definite-integrals improper-integrals indefinite-integrals absolute-worth
edited Jan 4 "16 at 5:07
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.
answered Jan 4 "16 at 5:12
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