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)



What gives a "Parseval like" equation mixing cosine and sine Fourier transforms ?
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 $t,t'$:
$$\int\_0^{\infty} \mathcal{F}^c [f (x)] \mathcal{F}^s [g (x)]\,dx=$$
$$\int\_0^\infty \left(\int\_0^{\infty} f(t) \cos(xt) \,dt\right)\left(\int\_0^{\infty} g(t') \sin(xt')\, dt'\right)\,dx=$$
$$\frac{1}{4i}\int\_0^\infty \left(\int\_0^{\infty} f(t) \left[e^{ixt}+e^{-ixt}\right]\,dt\right)\left(\int\_0^{\infty} g(t') \left[e^{ixt'}-e^{-ixt'}\right]\, dt'\right)\,dx=$$
$$\frac{1}{2}\int\_0^\infty dt\int\_0^\infty dt' f(t)g(t')\int\_0^\infty {\rm Im}\,\left(e^{ix(t+t')}+e^{-ix(t-t')}\right)\,dx=^\ast$$
$$\frac{1}{2}\int\_0^\infty \int\_0^\infty f(t)g(t')\left(\frac{1}{t+t'}-\frac{1}{t-t'}\right)\,dtdt'$$
$$=\int\_0^\infty \int\_0^\infty\frac{f(t)g(t')t'}{t'^2-t^2}\,dtdt'$$




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$^\ast$ Here I used the [Sokhotski–Plemelj formula](https://en.wikipedia.org/wiki/Sokhotski%E2%80%93Plemelj_theorem) which gives:
$$\int\_0^\infty e^{ixt}dx=\pi\delta(t)+\frac{i}{t}$$
The integral over $t,t'$ is a principal value.


