Proof of the "Maz identity" for solving integrals

Question

The "Maz identity" states: $$ \int_0^\infty f(x)g(x)\mathrm{d}x = \int_0^\infty \mathcal{L}\{f\}(u)\mathcal{L}^{-1}\{g\}(u)\mathrm{d}u, $$ where $\mathcal{L}$ is the Laplace transform.

I came across this identity when trying to find the Mellin transform of $\sin(x)$. The theorem turns out to be very useful, but I could not find any reference for a proof of this identity. The only references on MSE are this and this, but neither provides a derivation. This identity also appears in a recent IG post by owenmmth, for those who are interested.

PS: Although not important to the question, but the Mellin transform is defined as $$ \{\mathcal{Mf}\}(s) = \int_0^\infty x^{s-1}f(x)\mathrm{d}x .$$

Answer 1

As Zima pointed out in the comment, Wikipedia's page on the Laplace transform gives a nice proof, so I thought I would answer my own question for others who happen to search for this.

Start by showing that

$$ \int_0^\infty h(x) \{\mathcal{L}g\}(x)\mathrm{d}x = \int_0^\infty \{\mathcal{L}h\}(x)g(x)\mathrm{d}x. $$

This is obvious from the definition of the Laplace transform and an application of Fubini's theorem:

$$ \begin{align} \int_0^\infty h(x) \{\mathcal{L}g\}(x)\mathrm{d}x &= \int_0^\infty h(x) \int_0^\infty g(u)e^{-ux}\mathrm{d}u \mathrm{d}x \\ &= \int_0^\infty g(u) \int_0^\infty h(x)e^{-ux}\mathrm{d}x \mathrm{d}u \\ &= \int_0^\infty \{\mathcal{L}h\}(u)g(u)\mathrm{d}u \\ &= \int_0^\infty \{\mathcal{L}h\}(x)g(x)\mathrm{d}x . \end{align} $$

Now make the substitution $h(x) = \{\mathcal{L}^{-1}f\}(x)$, so that $\{\mathcal{L}h\}(x) = f(x)$:

$$ \int_0^\infty \{\mathcal{L}^{-1}f\}(x) \{\mathcal{L}g\}(x)\mathrm{d}x = \int_0^\infty f(x)g(x)\mathrm{d}x. $$

This proves the useful Laplace transform identity.

Answer 2

May this paper answer your question?

NB: there are typos in the last Corollary.

For those who cannot read through the paper, it might be efficient to read further. Hopefully not be too far-fetching to straight-forwardly put the main (interesting) property (probably we can call it the Frequency Convolution Theorem) of Laplace transform of the product of two functions. Let's have two functions with suitable assumptions on their behaviours of $f(x),h(x)$, then

\begin{align} \mathcal{L}[f(x)h(x)]&=\int_0^\infty\big(\!\mathcal{L}^{}f\!\big)(\xi+s)\big(\!\mathcal{L}^{-1}h\!\big)(\xi)\mathrm{d}\xi\\ &=\int_s^\infty\big(\!\mathcal{L}^{}f\!\big)(\xi)\big(\!\mathcal{L}^{-1}h\!\big)(\xi-s)\mathrm{d}\xi\\ &=\frac{1}{2\pi j}\lim\limits_{R\rightarrow\infty}\int\limits_{\sigma-iR}^{\sigma+iR}F(z)H(s-z)\mathrm{d}z,\quad\sigma>\Re\{s\in F(s)\}\;\&\; z=\sigma+jy\\ &=\frac{1}{2\pi j}F(s)*H(s)\tag{1} \end{align} where $F(s)$ and $H(s)$ are the Laplace transforms of $f(x)$ and $h(x)$, respectively. This forms a complementary to the well-known time convolution theorem $$ \mathcal{L}\left[f(t)*h(t)\right]=F(s)H(s). $$

which may not be proper to call the duality to (1). Here exposed my confusion about the duality properties (not confident to ask yet). My limited understanding of it is, taking Fourier transform for example, defined as \begin{align} \widehat{f}(\xi)&=\int_{-\infty}^\infty f(x)\ e^{- 2\pi i x \xi}\mathrm{d}x\\ f(x)&=\int_{-\infty}^\infty \widehat{f}(\xi)\ e^{2 \pi i x \xi} \mathrm{d}\xi \end{align} This definition for me is perfect as involution operation, like $\mathcal{F}[1(t)]=2\pi\delta(\omega)\iff\mathcal{F}[\delta(t)]=1(\omega)$.

Finally, the well-known integral technique OP asked about is just one special form of (1) with $s=0$. \begin{align} \int_0^\infty f(x)h(x)\mathrm{d} x=\int_0^\infty\big(\!\mathcal{L}^{}f\!\big)(\xi)\big(\!\mathcal{L}^{-1}h\!\big)(\xi)\mathrm{d}\xi. \end{align}