Principles of Fourier Analysis, 2nd Edition

20

Classical Fourier Transforms and Classically Transformable Functions

We know that

${{\mathcal{F}}_I\left[e^{-y}\text{step}\left(y\right)\right]|}_x={\displaystyle {\int }_{-\infty }^{\infty }{e}^{-y}}\text{step}\left(y\right)e^{-i2\pi xy}dy=\frac{1}{1+i2\pi x}.$

Now if ${\left(1+i2\pi x\right)}^{-1}$, the function on the right-hand side of these equations, were absolutely integrable on the real line, then its integral inverse Fourier transform would be defined by the integral formula for ${\mathcal{F}}_I^{-1}$, and the fundamental theorem on invertibility would assure us that

${{\mathcal{F}}_I^{-1}\left[\frac{1}{1+i2\pi x}\right]|}_y={\displaystyle {\int }_{-\infty }^{\infty }\frac{1}{1+i2\pi x}}{e}^{i2\pi xy}dx=e^{-y}\text{step}\left(y\right).$

But, as you verified in exercise 18.7 on page 264, ${\left(1+i2\pi x\right)}^{-1}$ is not absolutely integrable. So, we cannot invoke the fundamental theorem on invertibility to evaluate its Fourier inverse integral transform.

In fact, since ${\left(1+i2\pi x\right)}^{-1}$ is not absolutely integrable, its Fourier inverse integral ...