We use the curved arrows to point from the Laplace-transformed functions to the corresponding initial functions.

If $$f(t,\,x) \;\curvearrowleft\; F(s,\,x),$$ then one can integrate both functions with respect to the parametre $x$ between the same limits which may be also infinite provided that the integrals converge:

(1)

(2)

Proof. Using the definition of the Laplace transform, we can write $$\int_a^b\!f(t,\,x)\,dx \;\curvearrowleft\; \int_0^\infty\left(e^{-st}\int_a^b\!f(s,\,x)\,dx\right)dt \;=\; \int_0^\infty\left(\int_a^b\!e^{-st}f(s,\,x)\,dx\right)dt.$$ We change the order of integration in the last double integral and use again the definition, obtaining $$\int_a^b\!f(t,\,x)\,dx \;\curvearrowleft\; \int_a^b\left(\int_0^\infty\!e^{-st}f(s,\,x)\,dt\right)dx \;=\; \int_a^b\!F(s,\,t)\,dx,$$ Q.E.D.