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Plancherel's theorem says that

$f(x) = \frac{1}{2\pi} \int^\infty_{-\infty} F(k) e^{ikx} dk$

where

$F(k) = \int^\infty_{-\infty} f(x)e^{-ikx}dx$.

I'm wondering if we can prove this using Cauchy's integral formula somehow like this.

$f(x) = \frac{1}{2\pi} \int^\infty_{-\infty} \int^\infty_{-\infty} f(x')e^{-ikx'} dx' e^{ikx} dk$

$= \frac{1}{2\pi} \int^\infty_{-\infty} \int^\infty_{-\infty} e^{ik(x-x')} dk f(x') dx'$

$= \lim_{k_0\rightarrow\infty} \frac{1}{2\pi} \int^\infty_{-\infty} \frac{1}{i(x-x')} (e^{ik_0(x-x')}-e^{-ik_0(x-x')}) f(x') dx'$

$= \lim_{k_0\rightarrow\infty} -f(x)+f(x)$

$=0$

where I used Cauchy's integral formula in the next to last equality. I did contour integral over a upper half-circle and assumed f(x) goes to 0 at large x. However I got 0 instead of $f(x)$ at the last equality. I believe there are some problems in my understanding of complex analysis, so please let me know them!

user36437
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    @Batominovski has filled in the details you missed but let me caution you about one thing: going from line two to line three you made two very critical assumptions: 1) you can replace an improper integral with a Cauchy principal value integral and 2) that you can pull the $k_0$ limit outside the $x'$ integral. These are very hefty assumptions so you have not truly proved Plancherel's theorem in full generality, only in a very specific situation. That situation being when both of those very big assumptions hold. – Cameron Williams Jul 30 '15 at 04:22
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    If $F$ is $L^1$, I think the OP can assume from the first line that the $\displaystyle \int_{-\infty}^{\infty},\ldots,\text{d}k=\displaystyle \lim_{k_0\to\infty}, \int_{-k_0}^{ k_0},\ldots,\text{d}k$. Hence, the only thing to check is the integral switching, but then if $f$ is also $L^1$, then we can use Fubini's Theorem. – Batominovski Jul 30 '15 at 04:46

1 Answers1

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Observe that $$\begin{align} \frac{1}{2\pi} \int\limits^\infty_{-\infty} \int\limits^\infty_{-\infty} \textstyle f\left(x'\right)\,\exp\left({-\text{i}kx'}\right)\, \text{d}x'\, \exp({\text{i}kx})\, \text{d}k &=\lim_{k_0\rightarrow\infty}\, \frac{1}{2\pi}\, \int\limits^{+\infty}_{-\infty} \textstyle\frac{\exp\big({+\text{i}k_0\left(x-x'\right)}\big)-\exp\big({-\text{i}k_0\left(x-x'\right)}\big)}{\text{i}(x-x')} \,f\left(x'\right)\, \text{d}x' \\ &=\lim_{k_0\rightarrow\infty}\,\frac{1}{2\pi\text{i}}\,\int\limits^{+\infty}_{-\infty}\textstyle \,\frac{\exp\big({\text{i}k_0\left(x'-x\right)}\big)}{x'-x} \,\big(f\left(x'\right)+f\left(2x-x'\right)\big) \,\text{d}x'\,. \end{align}$$ Furthermore, your contour goes about the pole at $x$ half a turn. Hence, we have that $$\lim_{k_0\rightarrow\infty}\,\frac{1}{2\pi\text{i}}\,\int\limits^{+\infty}_{-\infty} \,\frac{\exp\big({\text{i}k_0\left(x'-x\right)}\big)\big)}{x'-x}\, \big(f\left(x'\right)+f\left(2x-x'\right)\big)\, \text{d}x'=\frac{1}{2}\big(f(x)+f(x)\big)=f(x)\,.$$

Batominovski
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