Complex-valued Fourier integral: $ \int_{ - \infty }^{ + \infty } {\frac{{\cos (ax)}}{{{x^2} + 1}}{e^{ - ibx}}\,\mathrm dx} $

Question

I'm working on the Fourier transform, but I don't know how to evaluate the integral:

$$I = \int_{ - \infty }^{ + \infty } {\frac{{\cos (ax)}}{{{x^2} + 1}}{e^{ - ibx}}\,\mathrm dx} $$

Answer 1

Hint: Use the identity $\cos(ax)=\frac{e^{iax}+e^{-iax}}{2}$ and the fact that the Fourier transform of $\frac{1}{1+x^2}$ is

$$\int_{-\infty}^{\infty}\frac{1}{1+x^2} \rm e^{-ix w}dx=\pi\rm {e^{-|w|}}.$$

To evaluate the last integral, see the first example here. It is a worked out example.

Answer 2

The integral may be rewritten as the real part of

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

This may be evaluated using residue theory. Consider the following contour integral in the complex plane:

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

where $C$ is the semicircle of radius $R$ aligned with the real axis and extending in the upper half-plane. The contour integral is equal to

$$\int_{-R}^R dx \frac{e^{i (a-b) x}}{1+x^2} + i R \int_0^{\pi} d\phi\, e^{i \phi} \frac{e^{i (a-b)R \cos{\phi}} e^{-(a-b) R \sin{\phi}}}{1+R^2 e^{i 2 \phi}}$$

As $R \to \infty$, the second integral is bounded by

$$\frac{1}{R} \left | \int_0^{\pi} d\phi \, e^{-(a-b) R \sin{\phi}}|\right |\le \frac{2}{R} \left | \int_0^{\pi/2} d\phi \, e^{-2(a-b) R\phi /\pi}| \right |\le \frac{\pi}{(a-b)R^2}$$

where we used the inequality $\sin{\phi} \ge 2 \phi/\pi$ for $\phi \in [0,\pi/2]$. Note that this integral converges only when $a>b$. We will consider the case $a<b$ shortly. In any case, this integral vanishes as $R \to \infty$.

The contour integral is equal to $i 2 \pi$ times the sum of the residues of the poles inside $C$. The only pole inside $C$ is at $z=i$; this pole has residue

$$\frac{e^{-(a-b)}}{2 i}$$

so that we now have

$$\int_{-\infty}^{\infty} dx \frac{e^{i (a-b) x}}{1+x^2} = \pi e^{-(a-b)}$$

when $a>b$. When $a<b$, we use the semicircle in the lower half-plane. To traverse $C$ in a positive sense requires going along the real line from right-to-left, so the result for $a<b$ is

$$\int_{-\infty}^{\infty} dx \frac{e^{i (a-b) x}}{1+x^2} = \pi e^{a-b}$$

Putting this all together, and taking the real part of the integral, we finally have

$$\int_{-\infty}^{\infty} dx \frac{\cos{a x}}{1+x^2} e^{i b x} = \pi e^{-|a-b|}$$

Answer 3

Note that: \begin{split} \mathcal F(e^{-|t|})(x) = \int_{-\infty}^{\infty}e^{-|t|}e^{-ix t}\,dt &=&\int_{-\infty}^{0}e^{t}e^{-ix t}\,dt+\int_{0}^{\infty}e^{-t}e^{-ix t}\,dt\\ &=&\left[ \frac{e^{(1-ix)t}}{1-ix} \right]_{-\infty}^0-\left[\frac{e^{-(1+ix)t}}{1+ix} \right]_{0}^{\infty}\\ &=&\frac{1}{1-ix}+\frac{1}{1+ix}\\ &=&\frac{2}{x^2+1}. \end{split} Then, $$ \begin{align} e^{-|a|}=\mathcal F^{-1}\left( \frac{2}{x^2+1}\right)(a) &=\frac{1}{2\pi}\int_\Bbb R \frac{2}{x^2+1}e^{ix a}\,dx = \frac{1}{\pi}\int_\Bbb R\frac{e^{ix a}}{x^2+1}\,dx \\&=\frac{1}{\pi}\int_\Bbb R\frac{\cos a x}{x^2+1}\,dx = \frac{2}{\pi}\int_0^\infty\frac{\cos ax}{x^2+1}\,dx \end{align} $$ Given that, as $x\mapsto\sin ax $ is an old function we have, $$\int_\Bbb R \frac{\sin{a x}}{x^2+1}dx= 0.$$

Thus we have, $$ \int_0^\infty\frac{\cos ax}{x^2+1}\,dx =\frac{\pi}{2}e^{-|a|} $$ Now see that,your integral may be rewritten as the real part of

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