Fourier transform of double exponential function

Question

As the title states, I'm currently trying to compute (or "prove") the Fourier transform of the double exponential function (or Laplace function), that is

$$\int_{-\infty}^{\infty}e^{i\xi x}e^{-\lambda|x|}\,dx = \int_{-\infty}^{0}e^{i\xi x}e^{\lambda x}\,dx+\int_{0}^{\infty}e^{i\xi x}e^{-\lambda x}\,dx$$

which clearly leads to the following result after a few simple integration rules:

$$\left[\frac{e^{i\xi x}e^{\lambda x}}{i\xi+\lambda}\right]_{-\infty}^{0}+\left[\frac{e^{i\xi x}e^{-\lambda x}}{i\xi-\lambda}\right]_{0}^{\infty}=\left(\frac{1}{i\xi+\lambda}-\frac{\lim_{x\to-\infty}e^{i\xi x}e^{\lambda x}}{i\xi+\lambda}\right)+\left(\frac{\lim_{x\to\infty} e^{i\xi x}e^{-\lambda x}}{i\xi-\lambda}-\frac{1}{i\xi-\lambda}\right).$$

Now, it's quite obvious that

$$\lim_{x\to-\infty}e^{\lambda x}\to 0\quad \text{and} \quad \lim_{x\to\infty} e^{-\lambda x}\to 0.$$

However, I'm struggling with interpreting what

$$\lim_{x\to-\infty}e^{i\xi x}\quad \text{and}\quad \lim_{x\to\infty}e^{i\xi x}$$

lead to since the textbook that I'm using seems to neglect this detail and simply reaches the result of

$$\frac{2\lambda}{\lambda^{2}+\xi^{2}},$$

which is naturally a Cauchy function.

Therefore, what would I do with $e^{i\xi x}$ when computing the two respective limits shown above for the definite integral? Any kind of help would be highly appreciated.

Answer 1

There is no $\lim_{x\to\infty}e^{i\xi x}$. But that doesn't matter. The function $x\mapsto e^{i\xi x}$ is bounded, so since $\lim_{x\to\infty}e^{-\lambda x}=0$ then also $\lim_{x\to\infty} e^{i\xi x}e^{-\lambda x}=0$.

The general principle is that if $\lim_{x\to a}f(x)=0$ and $g(x)$ is bounded as $x\to a$, then $\lim_{x\to a}f(x)g(x)=0$.

Answer 2

While you didn't write this, it is important to note that $\lambda$ is a non-negative real number.

The term $e^{i\xi x}$ is a purely imaginary exponential, and as such lies on the unit circle in the complex plane (by Euler's famous formula). In particular, it is bounded in absolute value. Combined with your observations, it follows that the limits at infinity vanish (because the rest of the factors being multiplied decay to zero).

As I am sure you are aware, it is the endpoints at $0$ that are more interesting (and yield the formula you seek).

Answer 3

To be a little more precise than previous answers, you have the following:

$$0 \le |e^{i\xi x}e^{\lambda x}| \le e^{\lambda x}$$

since $|e^{i\xi x}| = 1$, so if $e^{\lambda x}\to 0$, then $e^{i\xi x}e^{\lambda x} \to 0$ by the squeeze theorem. Particularly, assume that $\lambda < 0$, then

$$ 0 = \lim_{x\to\infty} 0\le \lim_{x\to\infty} |e^{i\xi x}e^{\lambda x}| \le \lim_{x\to\infty} e^{\lambda x} = 0.$$

Since $0 \le \lim_{x\to\infty} |e^{i\xi x}e^{\lambda x}| \le 0$, you have that $|e^{i\xi x}e^{\lambda x}|\to 0$ and so $e^{i\xi x}e^{\lambda x}\to 0$.