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I was thinking about the Fourier inversion formula, which says

$$f(x)=\dfrac{1}{2\pi}\int_{-\infty}^\infty\left(\int_{-\infty}^\infty f(z)e^{-iyz}dz\right)e^{ixy}dy$$

I know there is an issue about changing order of integration, but I want to know algebraically why this formula should be true. So I naively change the order of integration:

$$\int_{-\infty}^\infty \left(\int_{-\infty}^\infty e^{iy(x-z)}dy \right) f(z) dz$$

The inside integral is $\dfrac{e^{iy(x-z)}}{i(x-z)}$. Plugging $y=-\infty$ and $y=\infty$ yields unknown values, since the exponent is imaginary. What can I do here?

Kunal
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  • Your formula for $f(x)$ isn't right. The inner integral is $\hat{f}(x)$, so constant in $y$. – WimC Nov 13 '13 at 19:21
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    Maybe this helps: http://math.stackexchange.com/a/123896/8157 – Giuseppe Negro Nov 13 '13 at 19:21
  • @WimC I think I've edited to correct the formula. – Kunal Nov 13 '13 at 19:24
  • @GiuseppeNegro In your link there is the line $\tag{1!} \int_{-\infty}^\infty \frac{d\xi}{2\pi} e^{i \xi(x-y)}=\delta(x-y)$. Why is that? – Kunal Nov 13 '13 at 19:31
  • Because when $x=y$ the integral is infinite and when $x\ne y$ it is zero due to cancellation. Of course it is only formal and it does not explain the role of $2\pi$. – Giuseppe Negro Nov 14 '13 at 10:09
  • A slightly better explanation is that the integral equals the Fourier anti-transform of the constant function $1$ (in a suitable generalized sense). Since $$\hat{\delta}(\xi)=\int_{-\infty}^\infty \delta(x)e^{-ix\cdot\xi},dx=1, $$ anti-transforming termwise you see that $$\delta(x)=1^\vee(x)$$ and so $\delta(x-y)=1^\vee(x-y)$, which is what we wanted to prove. – Giuseppe Negro Nov 14 '13 at 10:12

2 Answers2

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Since you're deliberately ignoring the conditions for exchanging the order of integration, you can't hope for anything that actually works here, of course.

As you have discovered, evaluating the inner integral symbolically leads to nonsense -- indeed, the improper integral never converges. So you'll have to lower your standards of rigor considerably in order to extract any intuition from this rewriting. One way which you may or may not find illuminating is this:

In the case $x=z$ the integral is just $\int_{-\infty}^\infty 1dy$ which clearly diverges to infinity in a nice restrained way.

When $x\ne z$ the integrand $e^{i(x-z)y}$ keeps circling around the origin at constant speed, and half of each round will cancel out the other half. So if one is to pretend that the integral has any value at all, it will have to be $0$ -- another way to argue this that the integrand becomes minus itself by substituting $y\mapsto y+\frac{\pi}{x-z}$, but such a translation should not change the result when we integrate over all of $\mathbb R$. Thus the integral ought to be minus itself, or in other words 0.

So the outer integral has an integrand that is zero everywhere except at $z=x$ in which case it is infinity times $f(x)$. It makes some intuitive sense that this "ought to" integrate to something proportional to $f(x)$, but the particular factor $2\pi$ needs a more careful analysis (that doesn't swap integration orders willy-nilly) to derive.

By the way, essentially these considerations, dating straight back to Fourier himself, originally led to the classical integral form for the Dirac delta function: $$ \delta(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{itx}\, dt $$ Finding a way in which this would make rigorous sense was part of the impetus behind the theory of distributions.

hmakholm left over Monica
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Try considering the discrete Fourier series: given $f:\{0,1,\dots,n-1\}\to\mathbb C$, define $$ \hat f(k) = \sum_{m=0}^{n-1} \exp(2\pi i kl/n) f(m) ,$$ $$ \check f(k) = \frac1n \sum_{m=0}^{n-1} \exp(-2\pi i kl/n) f(m) .$$ Showing that these are inverse to each other is easy algebra. Then to get intuition about the integral, think of the integral in terms of Riemann sums (and perhaps convert the sums to $\sum_{m=-n/2}^{(n/2)-1}$ if $n$ is even, and something similar if $n$ is odd).

Stephen Montgomery-Smith
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