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I'm looking for a proof of this:

$$\int_0^\infty f(x)g(x)\,dx = \int_0^\infty \mathcal{L}\{f(x)\}\mathcal{L}^{-1}\{g(x)\}\,ds$$

Where $\mathcal{L}^{-1}\{g(x)\}$ denotes the inverse Laplace tranform of $g(x)$.

This can be used to evaluate integrals such as:

$$\int_0^\infty \frac{\sin x}{x}dx$$

with $f(x)= \sin x$ and $g(x)=\frac{1}{x}$ we get that:

$$\int_0^\infty \frac{\sin x}{x}=\int_0^\infty \mathcal{L}\{\sin x\}\mathcal{L}^{-1}\{\frac{1}{x}\}\,ds=\int_0^\infty \frac{1}{s^2+1}ds=\frac{\pi}{2}$$

My first approach would have been to write $\mathcal{L}\{f(x)\}$ and $\mathcal{L}^{-1}\{g(x)\}$ as their integral representations, but is there one for $\mathcal{L}^{-1}\{g(x)\}$?

blundered_bishop
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3 Answers3

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My first approach would have been to write $\mathcal{L}\{f(x)\}$ and $\mathcal{L}^{-1}\{g(x)\}$ as their integral representations, but is there one for $\mathcal{L}^{-1}\{g(x)\}$?

Yes, the inverse Laplace Transform is

$$\mathscr{L}^{-1}\{g\}(t)=\frac1{2\pi i}\lim_{L\to\infty}\int_{c-iL}^{c+iL}G(s)e^{st}\,ds$$

where $c$ is greater than the real part of all of the singularities of $G$.

NOTE: For a proof of the inverse Laplace Transform see THIS MSE post.

Then, we have

$$\begin{align} \int_0^\infty \mathscr{L}\{f\}(x)\mathscr{L}^{-1}\{g\}(x)\,dx&=\int_0^\infty \left(\int_0^\infty f(u)e^{-xu}\,du\right) \left(\frac1{2\pi i}\lim_{L\to\infty}\int_{c-iL}^{c+iL}g(v)e^{xv}\,dv \right)\,dx\\\\ &=\int_0^\infty f(u)\int_0^\infty e^{-xu}\left(\frac1{2\pi i}\lim_{L\to\infty}\int_{c-iL}^{c+iL}g(v)e^{xv}\,dv \right)\,dx\,du\\\\ &=\int_0^\infty f(u) g(u)\,du \end{align}$$

as was to be shown!

Mark Viola
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  • Hi Mark, hope you are doing great. I just wanted to comment on your posting. The problem I find with this is the justification of the order of integration when integrals exists in the sense of improper Riemann integrals or Henstock integrals. The wikipedia page on this issue does not give explicit conditions under which the identity in the OP works. – Mittens May 29 '23 at 23:36
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Well, we can look at the page of Laplace transform on Wikipedia to find the following:

enter image description here

Jan Eerland
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Not a complete answer but jut an attempt to justify the validity of the identity in the OP under $L_1$-conditions. The general validity of the identity, in the sense of improper Riemann integrals escapes me.

Let $f,g$ functions on $(0,\infty)$ for which the Laplace Transform is defined. Denote $\overline{f}(s)=\mathcal{L}f(s)=\int^\infty_0 e^{-st} f(t)\,dt$.

  • If $F(s,t)=e^{-st}f(t)g(s)$ is (Lebesgue) integrable on $L((0,\infty)\times(0,\infty),m_2)$, where $m_2$ is the Lebesgue measure on $\mathbb{R}^2$, an application of Fubini-Tonelli's theorem yields \begin{align} \int^\infty_0\overline{f}(s) g(s)\,ds&=\int^\infty_0\int^\infty_0e^{-st}f(t)g(s)\,dt\,ds\\ &=\int^\infty_0\int^\infty_0e^{-st}f(t)g(s)\,ds\,dt=\int^\infty_0 f(t)\overline{g}(t)\,dt \end{align}

  • If $f\in L_1(0,\infty)$, $\overline{g}(s)=\mathcal{L}(s)$ is defined for all $s>0$, and $\widetilde{g}(s;x)=\int^x_0 e^{-ys} g(y)\,dy$ is bounded as a function in $(0,\infty)\times(0,\infty)$, then \begin{align} \int^M_0\overline{f}(s)g(s)\,ds=\int^\infty_0f(t)\int^M_0e^{-st}g(s)\,ds\,dt \end{align} and so, by dominated convergence $$ \int^\infty_0\overline{f}(s) g(s)\,ds=\int^\infty_0 f(t)\overline{g}(t)\,dt$$

The identity obtained in each of the special cases outlined above is coincident with identity in the OP. The identity in the OP may not hold in more general situations, for example if the abscissas of convergence for $\overline{f}$ and $\overline{g}$ are both positive, then $\int^\infty_0\overline{f}(s)\overline{g}(s)\,ds$ may not exist (as a convergent proper integral).

Mittens
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