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Assume I have a real function $f(t)$ with Fourier transform of $\hat{f}(w)$.

Can one say anything about the inverse Fourier transform of $\frac{1}{1\pm\hat{f}(w)}$?

Uri Cohen
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  • Do not worry about the down vote. Read the answer and take your time to understand it. – Mhenni Benghorbal Jul 31 '13 at 12:57
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    In summary, if the function $\frac{1}{1\pm\hat{f}(w)}$ belongs to the Schwartz class, then its inverse Fourier transform exists and vise versa. See here. – Mhenni Benghorbal Jul 31 '13 at 22:03
  • @MhenniBenghorbal Is your comment identical to your now-deleted answer? – user1729 Aug 01 '13 at 11:39
  • @user1729: Yes, it is. – Mhenni Benghorbal Aug 01 '13 at 11:43
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    Then it should be noted that there was a considerable discussion about the validity of this answer (which has been deleted along with the question). – user1729 Aug 01 '13 at 11:45
  • @user1729: The answer is correct. It is up to those who are interested in this problem to seek more details. It was the first answer to be posted. – Mhenni Benghorbal Aug 01 '13 at 12:01
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    Then it should also be noted that if the function belongs to the class of tuna casserole, then it should be heated at $200^{\circ}$ C for about an hour in a Pyrex dish before serving. This, too, is a correct answer, and about as useful as stating that if the function belongs to the Schwartz class, etc. – Ron Gordon Aug 01 '13 at 12:31
  • For discussion about Mhenni's answer, see this question – angryavian Aug 01 '13 at 12:36
  • @RonGordon: Hit the books if you want more details. The answer is correct. – Mhenni Benghorbal Aug 01 '13 at 13:14
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    @MhenniBenghorbal: I find it highly ironic that your answer to anyone that challenges you is to do more research and thinking, when it is obvious to anyone who reads carefully that you have put next to no thought into this answer (and several others, at least those I have seen). And by the way, every time you claim the answer is correct, you prove to me and everyone else reading that you have not read, nor cared about, what anyone else has said to you this entire time. This episode makes me very sad indeed. – Ron Gordon Aug 01 '13 at 13:48

2 Answers2

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Heuristically, if $\hat{f}(w)$ was a reasonably well behaved and confined transformable function (imagine e.g. a Gaussian) such that $$\lim_{\omega=\pm\infty}\hat{f}(w)=0,\tag{1}$$ then: $$\lim_{\omega=\pm\infty}\left[\hat{g}(w)=\dfrac{1}{1\pm \hat{f}(w)}\right]=1,$$ and we note that condition $(1)$ is a reasonable (if not necessary) assumption when $\hat{f}(w)$ is the spectrum of any physical system where the energy is finite. We might then write: \begin{aligned}g(t)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left[\dfrac{1}{1\pm \hat{f}(w)}-1\right]e^{i\omega t}{d\omega}+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}1e^{i\omega t}{d\omega},\\&=-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left[\dfrac{\pm \hat{f}(w)}{1\pm \hat{f}(w)}\right]e^{i\omega t}{d\omega}+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}1e^{i\omega t}{d\omega},\\&=-h(t)+\sqrt{2\pi}\delta(t),\end{aligned} and because: $$\lim_{\omega=\pm\infty}\left[\hat{h}(w)=\dfrac{\pm \hat{f}(w)}{1\pm \hat{f}(w)}\right]=0,$$ we may expect the transformed function $h(t)$ to be reasonably well behaved (some caution required of course concerning the convergence of the integral). So, what may we say about the inverse Fourier transform of $\hat{g}(w)$? We may say that we expect it to contain a $\delta$ function contribution at the origin when $(1)$ holds and $h(t)$ exists.

Can we say anything about $h(t)$? At least in one important circumstance, yes. Take the plus sign and suppose $\hat{f}(w)$ is an approximately bandlimited function such that: $$\hat{f}(w)\gg1,\,\,|\omega| \le \frac{B}{2},$$ $$\hat{f}(w)\approx 0,\,\,|\omega| \ge \frac{B}{2}, $$ then:

$$\hat{h}(w)\approx1,\,\,|\omega| \le \frac{B}{2},$$ $$h(t) \approx\frac{1}{\sqrt{2\pi}}\int_{-\frac{B}{2}}^{\frac{B}{2}}1e^{i\omega t}{d\omega}=\frac{B}{\sqrt{2\pi}}\text{sinc} \left( \dfrac{Bt}{2} \right), $$

which is a $\text{sinc}$ function who's peak value and oscillation frequency are proportional to the bandwidth of $\hat{f}(w)$. Of course we may not expect the edges of $\hat{h}(w)$ to be so sharp; allowing for a slower decay in $\hat{f}(w)$ (which may possibly not even be strictly decreasing), would allow $\hat{h}(w)$ to spread in the spectral domain which confines $h(t)$ more and encompasses the $\text{sinc}$ function in a damping envelope, but provided the approximate bandwidth of $\hat{f}(w)$ is preserved, so to is the peak value and oscillation frequency of $h(t)$, (see e.g. the Raised-cosine filter).

Graham Hesketh
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If you assume that this inverse transformation exists and $\left\|\hat{f}\right\|<1 $, you will get

$$\int\limits_{-\infty}^{+\infty}\frac{1}{1-\hat{f}(w)}e^{-iwx}dw=\int\limits_{-\infty}^{+\infty}\sum\limits_{j=0}^{+\infty}\left(\hat{f}(w)\right)^j e^{-iwx}dw=\sum\limits_{j=0}^{+\infty}\int\limits_{-\infty}^{+\infty}\left(\hat{f}(w)\right)^je^{-iwx}dw$$

There are no typical transforms of $\left(\hat{f}(w)\right)^j$ (WolframAlpha is silent, there is no such transform in "Tables of integral transforms (Bateman,Erdelyi)")

(upd)But we can say that $\left(\hat{f}\left(w\right)\right)^{j}=\hat{f}\left(w\right)\cdot\hat{f}\left(w\right)...\hat{f}\left(w\right)$, therefore,

$$ \int\limits_{-\infty}^{+\infty}\left(\hat{f}(w)\right)^je^{-iwx}dw=f^{j*}(x),$$ where $n*$ is n-th convolution of $f(x)$ with itself. So, the final answer is

$$\int\limits_{-\infty}^{+\infty}\frac{1}{1-\hat{f}(w)}e^{-iwx}dw=\sum\limits_{j=0}^{+\infty}f^{j*}(x)$$

The same you can do with

$$\int\limits_{-\infty}^{+\infty}\frac{1}{1+\hat{f}(w)}e^{-iwx}dw=\sum\limits_{j=0}^{+\infty}\left(-1\right)^{j}f^{j*}(x)$$

cool
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