Inverse Fourier transform of a characteristic function

Let the Fourier transform of a function $f(x)$ be

$g(t) = \int_{-\infty}^{\infty} f(x) e^{itx} dx \qquad \qquad \qquad (1)$

The corresponding inverse Fourier transform would then be

$f(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} g(t) e^{-itx} dt \qquad \qquad \qquad (2)$

As an exercise, what I want to do in this note is derive the characteristic function $\phi_X(t) = E[e^{itX}]$ of an exponential random variate $X$ with probability density function

$f(x) = \lambda e^{-\lambda x} \qquad \qquad \qquad (3)$

for $x \geq 0$ . The required characteristic function is the Fourier transform of this density function. I then want to apply the inverse Fourier transform to $\phi_X(t)$ and thereby recover the density $f(x)$ .

Simply applying (1) above to (3) we easily get the characteristic function:

$\phi_X(t) = \int_0^{\infty} \lambda e^{-\lambda x} e^{itx} dx$

$= \lambda \int_0^{\infty} e^{-(\lambda-it)x} dx$

$= \bigg[-\frac{1}{(\lambda - it)} e^{-(\lambda - it)x}\bigg]_0^{\infty}$

$= \frac{\lambda}{\lambda - it}$

$= \bigg(1 - \frac{it}{\lambda}\bigg)^{-1} \qquad \qquad \qquad (4)$

(Note that $e^{-(\lambda - it)x} = e^{-\lambda} e^{itx} \rightarrow 0$ as $x \rightarrow \infty$ if $\lambda > 0$ ).

Let us now reverse the process, and apply the inverse Fourier transform to (4). Using (2) above, we get

$f(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \bigg(1 - \frac{it}{\lambda}\bigg)^{-1} e^{-itx} dt$

$= \frac{-i\lambda}{2 \pi} \int_{-\infty}^{\infty} \frac{e^{-itx}}{(t + i\lambda)} dt \qquad \qquad \qquad (5)$

(Note that $t$ is unrestricted, so the integration limits are from $-\infty$ to $\infty$ ). It quickly transpires that elementary techniques will not work easily with (5), so we try contour integration using the residue theorem. We turn the integration variable $t$ into a complex variable $z = t + iy$ , and formulate the integral in (5) as a contour integral

$\frac{-i\lambda}{2 \pi} \oint_C \frac{e^{-izx}}{(z + i\lambda)} dz \qquad \qquad \qquad (6)$

The trick will be to design the contour $C$ in such a way that part of the contour includes the integral in the form (5), while the rest of the contour vanishes, in a limiting process. We can then apply the residue theorem to solve (5).

The integrand in (6) has a pole of order 1 at $z = -i\lambda$ . To find the residue of the integrand at this pole, we multiply the integrand by $(z + i \lambda)$ , obtaining $e^{-izx}$ , and then evaluate this result at $z = -i \lambda$ , yielding $e^{-\lambda x}$ . Thus, $e^{-\lambda x}$ is the residue of the integrand at its pole. If we design the contour $C$ in such a way that it encloses the singularity at $-i \lambda$ , then Cauchy’s residue theorem tells us that the integral in (6) equals $2 \pi i$ times the residue $e^{-\lambda x}$ , so we will get the density function we are looking for:

$\frac{-i\lambda}{2 \pi} \oint_C \frac{e^{-izx}}{(z + i\lambda)} dz = \frac{-i\lambda}{2 \pi} (2 \pi i) e^{-\lambda x} = \lambda e^{-\lambda x} \qquad \qquad \qquad (7)$

Clearly then, when we use contour integration we get the required result straight away. But what contour do we use for the integral in (6)? The contour must enclose the point $-i \lambda$ , so letting $a > \lambda$ be some initial starting value, we can try the following:

Then the contour $C$ may be divided into a straight part along the real axis (which only involves the real part of $z$ , namely $t$ ), and a curved arc. Therefore the contour integral in (6) can be decomposed as

$\frac{-i\lambda}{2 \pi} \int_{-a}^{a} \frac{e^{-itx}}{(t + i\lambda)} dt - \frac{i\lambda}{2 \pi} \int_{arc} \frac{e^{-izx}}{(z + i\lambda)} dz = \lambda e^{- \lambda x} \qquad \qquad \qquad (8)$

We can now use estimations to show that the second integral on the left-hand side of (8) vanishes as $a \rightarrow \infty$ . First, noting that $\pi a$ is half the circumference of a circle of radius $a$ , we have

$\int_{arc} \frac{e^{-izx}}{(z + i\lambda)} dz \leq \pi a \sup_{arc} \bigg|\frac{e^{-izx}}{(z+i \lambda)}\bigg| \qquad \qquad \qquad (9)$

Next, we observe that

$|e^{-izx}| = |e^{-x|z|(\cos(\arg(z)) + i \sin(\arg(z)))}| = e^{-x |z| \cos(\arg(z))} = \frac{1}{e^{x |z| \cos(\arg(z))} } \leq \frac{1}{x |z| \cos(\arg(z))} \qquad \qquad \qquad (10)$