Sometimes I come across a Fourier integral that I have no idea how to attack. And then I find that I can convert it to another, more familiar integral using complex analysis. So here’s an example of such a satisfying situation.

The problem is to evaluate

$$\int_{-\infty}^{\infty} dx \, (1+i a^2 x)^{-1/2} (1+i b^2 x)^{-1/2} e^{i c x}$$

Mathematica v.9 was not able to evaluate the definite integral in closed form.

To begin, consider the following complex contour integral:

$$\oint_C dz \, (1+i a^2 z)^{-1/2} (1+i b^2 z)^{-1/2} e^{i c z}$$

where $C$ is the following contour:

$c \gt 0$, and we will assume WLOG that $a \gt b \gt 0$. Note that $C$ is a semicircular contour of radius $R$ in the upper half plane, with a detour around the branch points at $z=i/a^2$ and $z=i/b^2$. The circular detours have radii of $\epsilon$.

We first write out the contour integral in terms of various parametrization about the contour. That is, the contour integral is equal to

$$\int_{-R}^R dx \, (1+i a^2 x)^{-1/2} (1+i b^2 x)^{-1/2} e^{i c x} \\+ i R \int_0^{\pi/2} d\theta \, e^{i \theta} \left (1+i a^2 R e^{i \theta} \right )^{-1/2} \left (1+i b^2 R e^{i \theta} \right )^{-1/2} e^{i c R e^{i \theta}}\\ + i \int_R^{1/b^2 + \epsilon} dy \, (1-a^2 y)^{-1/2} (1-b^2 y)^{-1/2} e^{-c y} \\+ i \epsilon \int_{\pi/2}^{-\pi/2} d\phi \, e^{i \phi} \left [1+i a^2 \left (\frac{i}{b^2}+ \epsilon e^{i \phi}\right ) \right ]^{-1/2} \left [1+i b^2 \left (\frac{i}{b^2}+ \epsilon e^{i \phi}\right ) \right ]^{-1/2} e^{i c \left (\frac{i}{b^2}+ \epsilon e^{i \phi}\right )}\\ +i \int_{1/b^2-\epsilon}^{1/a^2+\epsilon} dy \, (1-a^2 y)^{-1/2} (1-b^2 y)^{-1/2} e^{-c y} \\ + i \epsilon \int_{\pi/2}^{-3 \pi/2} d\phi \, e^{i \phi} \left [1+i a^2 \left (\frac{i}{a^2}+ \epsilon e^{i \phi}\right ) \right ]^{-1/2} \left [1+i b^2 \left (\frac{i}{a^2}+ \epsilon e^{i \phi}\right ) \right ]^{-1/2} e^{i c \left (\frac{i}{a^2}+ \epsilon e^{i \phi}\right )}\\+ i \int_{1/a^2+\epsilon}^{1/b^2-\epsilon} dy \, (1-a^2 y)^{-1/2} (1-b^2 y)^{-1/2} e^{-c y} \\+ i \epsilon \int_{3 \pi/2}^{\pi/2} d\phi \, e^{i \phi} \left [1+i a^2 \left (\frac{i}{b^2}+ \epsilon e^{i \phi}\right ) \right ]^{-1/2} \left [1+i b^2 \left (\frac{i}{b^2}+ \epsilon e^{i \phi}\right ) \right ]^{-1/2} e^{i c \left (\frac{i}{b^2}+ \epsilon e^{i \phi}\right )}\\+ i \int_{1/b^2 + \epsilon}^R dy \, (1-a^2 y)^{-1/2} (1-b^2 y)^{-1/2} e^{-c y} \\+i R \int_{\pi/2}^{\pi} d\theta \, e^{i \theta} \left (1+i a^2 R e^{i \theta} \right )^{-1/2} \left (1+i b^2 R e^{i \theta} \right )^{-1/2} e^{i c R e^{i \theta}}$$

OK, that is a lot: $10$ contour pieces, $10$ integrals. The good news is that most of them either cancel or vanish in the limits $R \to \infty$ and $\epsilon \to 0$. Let’s split this up as follows: i) integrals over arcs as $R \to \infty$, ii) integrals over arcs as $\epsilon \to 0$, and iii) integrals along the imaginary axis.

**Integrals over arcs as $R \to \infty$**

Here we show that each such integral vanishes in this limit by bounding the magnitude. (This is essentially Jordan’s lemma.) Thus, let’s look at the magnitude of the integral over $\theta \in [0,\pi/2]$, which is bounded by

$$\begin{align} \left |i R \int_0^{\pi/2} d\theta \, e^{i \theta} \left (1+i a^2 R e^{i \theta} \right )^{-1/2} \left (1+i b^2 R e^{i \theta} \right )^{-1/2} e^{i c R e^{i \theta}} \right | &\le R (a^2 R-1)^{-1/2} (b^2 R-1)^{-1/2} \int_0^{\pi/2} d\theta \, e^{-c R \sin{\theta}}\\ & \le R (a^2 R-1)^{-1/2} (b^2 R-1)^{-1/2} \int_0^{\pi/2} d\theta \, e^{-2 c R \theta/\pi}\\ &\le \frac{\pi}{2 a b c R}\end{align}$$

which indeed vanishes in the limit as $R \to \infty$.

The integral over $\theta \in [\pi/2,\pi]$ vanishes similarly.

**Integrals over arcs as $\epsilon \to 0$
**

Let’s take as an example

$$i \epsilon \int_{\pi/2}^{-\pi/2} d\phi \, e^{i \phi} \left [1+i a^2 \left (\frac{i}{b^2}+ \epsilon e^{i \phi}\right ) \right ]^{-1/2} \left [1+i b^2 \left (\frac{i}{b^2}+ \epsilon e^{i \phi}\right ) \right ]^{-1/2} e^{i c \left (\frac{i}{b^2}+ \epsilon e^{i \phi}\right )}$$

which behaves, as $\epsilon \to 0$, as

$$i \epsilon^{1/2} \left ( 1-\frac{a^2}{b^2} \right )^{-1/2} e^{-c/b^2} \int_{\pi/2}^{-\pi/2} d\phi \, e^{-i \phi/2} = O \left ( \epsilon^{1/2} \right )$$

which vanishes. The other integrals over the circular detours vanish similarly.

Five down, five to go.

**Integrals along the imaginary axis**

Here we have to deal with the branch cuts of the square roots we have defined by the contour. Let us do that by examining the value of the integrand over the various intervals along the imaginary axis. For example, consider the term $(1-a^2 y)^{-1/2}$. Over $y \gt 1/a^2$, $1-a^2 y \le 0$, so that $(1-a^2 y)^{-1/2} = (-1)^{-1/2} (a^2 y-1)$. Note that the meaning of the term $(-1)^{-1/2}$ depends on which side of the branch cut we are on. For example, to the right of the imaginary axis, $(-1)^{-1/2}= -i$; to the left, $(-1)^{-1/2} = i$. A similar analysis may be made for the other square root in the integrand: just replace $a$ with $b$.

Now consider

$$i \int_R^{1/b^2 + \epsilon} dy \, (1-a^2 y)^{-1/2} (1-b^2 y)^{-1/2} e^{-c y}$$

in the limit as $R \to \infty$ and $\epsilon \to 0$. This integral is taken over the right side of the imaginary axis, so we write the integral as

$$i \int_{\infty}^{1/b^2} dy \, (-i) (a^2 y-1)^{-1/2} (-i) (b^2 y-1)^{-1/2} e^{-c y}$$

Analogously, on the left side of the branch cut, we have

$$i \int_{1/b^2}^{\infty} dy \, (i) (a^2 y-1)^{-1/2} (i) (b^2 y-1)^{-1/2} e^{-c y}$$

What we can conclude is that the branch cuts of each of the square roots cancel over this integration region. Thus, the sum of these two integrals is zero.

Seven down, three to go.

Now consider

$$i \int_{1/b^2}^{1/a^2} dy \, (-i) (a^2 y-1)^{-1/2} (1-b^2 y)^{-1/2} e^{-c y} $$

Note that only one of the square roots goes negative in this region; the other remains positive. Now we do the left side of the branch cut:

$$i \int_{1/a^2}^{1/b^2} dy \, (i) (a^2 y-1)^{-1/2} (1-b^2 y)^{-1/2} e^{-c y} $$

The sum of these two integrals **does not cancel**. Rather, they combine into a single integral.

Now we can write the original contour integral as a sum of two integrals (which is way better than ten). By Cauchy’s theorem, the contour integral is equal to zero because there are no poles within the contour $C$. Thus, we may now state our preliminary result:

$$\int_{-\infty}^{\infty} dx \, (1+i a^2 x)^{-1/2} (1+i b^2 x)^{-1/2} e^{i c x} = 2 \int_{1/a^2}^{1/b^2} dy \, (a^2 y-1)^{-1/2} (1-b^2 y)^{-1/2} e^{-c y} $$

**Evaluating the definite integral**

Surprisingly, Mathematica v9 has very little to say about either the integral on the left or right. However, using a straightforward substitution, we can cast the integral in a familiar form to extract a closed form expression as follows:

$$\begin{align} \int_{1/a^2}^{1/b^2} dy \, (a^2 y-1)^{-1/2} (1-b^2 y)^{-1/2} e^{-c y} &= \frac{e^{-c/a^2}}{a} \int_0^{\frac1{b^2}-\frac1{a^2}} du \, u^{-1/2} \left ( 1-\frac{b^2}{a^2} – b^2 u\right )^{-1/2} e^{-c u} \\ &= \frac{e^{-c/a^2}}{a b} \int_0^1 dv \, v^{-1/2} (1-v)^{-1/2} e^{-c \left (\frac1{b^2}-\frac1{a^2} \right ) v} \\ &= 2\frac{e^{-c/a^2}}{a b} \int_0^1 dw \, (1-w^2)^{-1/2} e^{- c \left (\frac1{b^2}-\frac1{a^2} \right ) w^2} \\ &= 2 \frac{e^{-c/a^2}}{a b} \int_0^{\pi/2} dt \, e^{- c \left (\frac1{b^2}-\frac1{a^2} \right ) \sin^2{t}} \end{align} $$

This is amazing because we know (OK, at least I do) that

$$2 \int_0^{\pi/2} dt \, e^{-\gamma \sin^2{t}} = \pi e^{-\gamma/2} I_0\left ( \frac{\gamma}{2} \right ) $$

where $I_0$ is the modified Bessel function of the first kind of zeroth order. Therefore, we have an analytical, closed form expression for the original integral. After a little algebra, we have our final result for $c \gt 0$:

>$$\int_{-\infty}^{\infty} dx \, (1+i a^2 x)^{-1/2} (1+i b^2 x)^{-1/2} e^{i c x} = \frac{2 \pi}{a b} e^{-\frac{c}{2} \left (\frac1{b^2}+\frac1{a^2} \right )} I_0 \left [\frac{c}{2} \left (\frac1{b^2}-\frac1{a^2} \right ) \right ] $$

I have verified this numerically in Mathematica v9 (albeit with some difficulty, as the numerical evaluation of the original integral is numerically problematic).

Note that the case $c \lt 0$ is simple: because the branch points are in the upper half plane for positive $a,b$, and there are no poles in the lower half plane, the integral for $c \lt 0$ is zero.

The case $c=0$ is not really problematic. Rather, it should be seen that the Fourier integral is really a principal value, so what looks like a logarithmic divergence at infinity really cancels out in the limits as $c$ goes to zero from above. The closed form result is valid for $c=0$.

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