An Example of a Self-Reciprocal Function in Fourier Transform

Proposition

The function

$$f(t)=\frac{1}{\sqrt{|t|}}$$

is a self-reciprocal function of the Fourier transform.

In other words, it is mapped to the same function by the Fourier transform.

Proof

First, note that $f(t)$ is an even function.

We will now prove the following lemma.

Lemma

If $f(t)$ is an even function, its Fourier transform is given by

$$F(\omega )=2{\int }_0^{\mathrm{\infty }}f(t)\cos (\omega t)dt$$

$F(\omega )$ is also an even function.

Proof of the Lemma

By the definition of the Fourier transform,

$$F(\omega )={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(t)e^{-i\omega t}dt={\int }_{-\mathrm{\infty }}^{0}f(t)e^{-i\omega t}dt+{\int }_0^{\mathrm{\infty }}f(t)e^{-i\omega t}dt$$

For the first term on the right-hand side, performing the change of variables $s=-t$ gives

$${\int }_{-\mathrm{\infty }}^{0}f(t)e^{-i\omega t}dt={\int }_{\mathrm{\infty }}^{0}f(-s)e^{-i\omega (-s)}(-ds)={\int }_0^{\mathrm{\infty }}f(s)e^{i\omega s}ds={\int }_0^{\mathrm{\infty }}f(t)e^{i\omega t}dt$$

Thus,

$$F(\omega )={\int }_0^{\mathrm{\infty }}f(t)e^{i\omega t}dt+{\int }_0^{\mathrm{\infty }}f(t)e^{-i\omega t}dt={\int }_0^{\mathrm{\infty }}f(t)(e^{i\omega t}+e^{-i\omega t})dt={\int }_0^{\mathrm{\infty }}f(t)(2\cos (\omega t))dt=2{\int }_0^{\mathrm{\infty }}f(t)\cos (\omega t)dt$$

Using this expression,

$$F(-\omega )=2{\int }_0^{\mathrm{\infty }}f(t)\cos (-\omega t)dt=2{\int }_0^{\mathrm{\infty }}f(t)\cos (\omega t)dt=F(\omega )$$

It follows that $F(\omega )$ is an even function.

Proof of the Proposition

Since $f(t)$ is an even function, we can directly apply the lemma.

As $F(\omega )$ is also an even function, we first consider the range $\omega >0$.

$$F(\omega )=2{\int }_0^{\mathrm{\infty }}f(t)\cos (\omega t)dt=2{\int }_0^{\mathrm{\infty }}\frac{1}{\sqrt{|t|}}\cos (\omega t)dt=2{\int }_0^{\mathrm{\infty }}\frac{1}{\sqrt{t}}\cos (\omega t)dt$$

Performing substitution with $t=s^2$,

$$F(\omega )=2{\int }_0^{\mathrm{\infty }}\frac{1}{\sqrt{s^2}}\cos (\omega s^2)(2sds)=4{\int }_0^{\mathrm{\infty }}\frac{1}{s}\cos (\omega s^2)sds=4{\int }_0^{\mathrm{\infty }}\cos (\omega s^2)ds$$

By substitution $u=\sqrt{\omega }s$,

$$F(\omega )=4{\int }_0^{\mathrm{\infty }}\cos (u^2)(\frac{1}{\sqrt{\omega }}du)=\frac{4}{\sqrt{\omega }}{\int }_0^{\mathrm{\infty }}\cos (u^2)du$$

The right hand side is Fresnel integral. Thus,

$$F(\omega )=\frac{4}{\sqrt{\omega }}\frac{1}{2}\sqrt{\frac{\pi }{2}}=\sqrt{\frac{2\pi }{\omega }}$$

Taking the range $\omega <0$ into consideration, we have

$$F(\omega )=\sqrt{\frac{2\pi }{|\omega |}}$$

This is the same as $f(t)$ except for a constant factor of $\sqrt{2\pi }$.

If we use a definition of the Fourier transform where it is a unitary transformation (as seen in Multiple Definitions of the Fourier Transform), this becomes exactly the same.

Therefore,

$$f(t)=\frac{1}{\sqrt{|t|}}$$

is an example of a self-reciprocal function in the Fourier transform.

Additional Notes

The Gaussian function

$$f(t)=\exp [-\frac{t^2}{2}]$$

is a well-known example of a function that maps to itself under the Fourier transform (Fourier Transform of the Probability Density Function of the Normal Distribution (Gaussian Function)).

However, there are many such functions, and the current example is one of them.

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