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Noting $\mathcal{F}^c$ the cosine transform and $\mathcal{F}^s$ the sine transform defined on real functions by:

$$\mathcal{F}^c [f (x)]=\int_0^{\infty} f(t) \cos(xt) dt $$

$$\mathcal{F}^s [f (x)]=\int_0^{\infty} f(t) \sin(xt) dt $$

What can we say about following integral ?

$$\int_0^{\infty} \mathcal{F}^c [f (x)] \mathcal{F}^s [g (x)] dx$$

(with for example $f$ and $g$ continuous functions in $L^2$)

Can we express this integral as a simple integral with original functions ? (like in Parseval's equation)

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There is no "Parseval-like" equation, in the sense of a single integral over $t$. The best you can do is a double integral over $t,t'$: $$\int_0^{\infty} \mathcal{F}^c [f (x)] \mathcal{F}^s [g (x)]\,dx=$$ $$\int_0^\infty \left(\int_0^{\infty} f(t) \cos(xt) \,dt\right)\left(\int_0^{\infty} g(t') \sin(xt')\, dt'\right)\,dx=$$ $$\frac{1}{4i}\int_0^\infty \left(\int_0^{\infty} f(t) \left[e^{ixt}+e^{-ixt}\right]\,dt\right)\left(\int_0^{\infty} g(t') \left[e^{ixt'}-e^{-ixt'}\right]\, dt'\right)\,dx=$$ $$\frac{1}{2}\int_0^\infty dt\int_0^\infty dt' f(t)g(t')\int_0^\infty {\rm Im}\,\left(e^{ix(t+t')}+e^{-ix(t-t')}\right)\,dx=^\ast$$ $$\frac{1}{2}\int_0^\infty \int_0^\infty f(t)g(t')\left(\frac{1}{t+t'}-\frac{1}{t-t'}\right)\,dtdt'$$ $$=\int_0^\infty \int_0^\infty\frac{f(t)g(t')t'}{t'^2-t^2}\,dtdt'$$


$^\ast$ Here I used the Sokhotski–Plemelj formula which gives: $$\int_0^\infty e^{ixt}dx=\pi\delta(t)+\frac{i}{t}$$ The integral over $t,t'$ is a principal value.

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    $\begingroup$ looks like an hilbert transform $\endgroup$ – T.... Mar 12 '18 at 0:33
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    $\begingroup$ I finally got to know the name of the formula for the Fourier transform of the step function (widely used in physics). $\endgroup$ – lcv Mar 12 '18 at 20:29
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No, you are one Hilbert transform away from such a formula. $\mathcal F^cf(x)=\widehat{f_e}(x)$ is the Fourier transform $\widehat{g}(x) = \int_{-\infty}^{\infty} g(t)e^{itx}\, dt$ of the even continuation $f_e(-x)=f(x)$ ($x>0$) of $f$, and similarly the sine transform is the FT of the odd continuation.

So you're looking at $$ \int_0^{\infty}\widehat{f_e}(x)\widehat{f_o}(x)\, dx = \int_{-\infty}^{\infty}\chi_{(0,\infty)}(x)\widehat{f_e}(x)\widehat{f_o}(x)\, dx , $$ and multiplication of the FT by the characteristic function is essentially the Hilbert transform.

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