Question about the Fourier transform of $\exp(-a |t|)$

Question

I would like to understand how to compute the Fourier transform

$$F(x) = \int_{-\infty}^\infty f(t) \exp(-ixt) \,\Bbb dt$$

of the function $f(t) = \exp(-a |t|)$ (where $a>0$). Following this source (slide 6), I understand all the steps until

$$F(x) = \int_{-\infty}^0 e^{at} e^{-ixt} \,\Bbb dt + \int_0^\infty e^{-at} e^{-ixt} \,\Bbb dt.$$

What I do not understand, are the following equations:

$$\int_{-\infty}^0 e^{at} e^{-ixt} \,\Bbb dt = \frac{1}{a-ix}$$ and $$\int_0^\infty e^{-at} e^{-ixt} \,\Bbb dt = \frac{1}{a+ix}.$$

I assume that they used

\begin{align} \int_{-\infty}^0 e^{at} e^{-ixt} \,\Bbb dt &= \int_{-\infty}^0 e^{(a-ix)t} dt \\ &= \left[\frac{e^{(a-ix)t}}{a-ix}\right]_{t=-\infty}^0 \end{align} and similarly $$\int_0^\infty e^{-at} e^{-ixt} \,\Bbb dt = \left[\frac{e^{(-a-ix)t}}{-a-ix}\right]_{t=0}^\infty.$$

However, I think it should not be possible to determine the limits $$\lim_{t \to -\infty} \frac{e^{(a-ix)t}}{a-ix}$$ and $$\lim_{t \to \infty} \frac{e^{(-a-ix)t}}{-a-ix}$$ due to the imaginary part $e^{-ixt}$ in both terms, or am I wrong?

I would like to hear an explanation for this.

Answer 1

$|\frac{e^{(a-ix)t}}{-a-ix}|=\frac {e^{at}} {\sqrt {a^{2}+x^{2}}}\to 0$ as $t \to -\infty$.

Similarly,

$|\frac{e^{(-a-ix)t}}{-a-ix}|=\frac {e^{-at}} {\sqrt {a^{2}+x^{2}}}\to 0$ as $t \to \infty$.

Answer 2

You may restrict to real valued functions. Since the function $e^{-a|t|}$ is even while the function $t\mapsto\sin(xt)$ is odd, we get $$\int\limits_{-\infty}^\infty e^{-a|t|}e^{-ixt}\,dt= \int\limits_{-\infty}^\infty e^{-a|t|}\cos(xt)\,dt\\ =2 \int\limits_0^\infty e^{-at}\cos(xt)\,dt=:2I$$ Applying the integration by parts two times to $I$ gives $$I=-{1\over a}e^{-at}\cos(xt)\Big\vert_0^\infty-{x\over a} \int\limits_0^\infty e^{-at}\sin(xt)\,dt\\ ={1\over a}+{x\over a^2}e^{-at}\sin(xt)\Big\vert_0^\infty -{x^2\over a^2}I$$ Therefore $$ 2I={2a\over x^2+a^2}$$ Remark The application of complex variables is more convenient as we can easily determine the antiderivatives on the way.