2
$\begingroup$

This is one heck of an embarrassment but it is amazing how these bits of subtlety gets lost in the back of the head after the first year of undergraduate studies-with every computation chucked into Mathematica.

Here goes, I'm trying to integrate $\int_{-\infty}^{+\infty}e^{-a\left | x \right |}e^{-ikx}dx $

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

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

Then, we have

$\int_{-\infty}^{0} e^{-a\left ( -x \right )}e^{-ikx}dx+\int_{0}^{+\infty}e^{-ax}e^{-ikx}dx$

or perhaps, by symmetry,

$2\int_{0}^{+\infty}e^{-ax}e^{-ikx}dx$

Have I nailed this?

$\endgroup$
0

1 Answer 1

5
$\begingroup$

This was correct before the edit; the expression

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

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

\begin{align*} \int_{-\infty}^0 e^{ax} e^{-ikx} \, dx = \int_0^{\infty} e^{-ax} e^{+ikx} \, dx \end{align*}

Note the sign on the exponential is different.


As an error check, note that the Fourier transform of an even function is real. If you compute the integral $\int_0^{\infty} e^{-ax} e^{-ikx} \, dx$, you do not get a real function.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .