Editing Laplace transform (section)

=== Evaluating integrals over the positive real axis ===
A useful property of the Laplace transform is the following:
<math display=block>\int_0^\infty f(x)g(x)\,dx = \int_0^\infty(\mathcal{L} f)(s)\cdot(\mathcal{L}^{-1}g)(s)\,ds </math>
under suitable assumptions on the behaviour of <math>f,g</math> in a right neighbourhood of <math>0</math> and on the decay rate of <math>f,g</math> in a left neighbourhood of <math>\infty</math>. The above formula is a variation of integration by parts, with the operators 
<math>\frac{d}{dx}</math> and <math>\int \,dx</math> being replaced by <math>\mathcal{L}</math> and <math>\mathcal{L}^{-1}</math>. Let us prove the equivalent formulation:
<math display=block>\int_0^\infty(\mathcal{L} f)(x)g(x)\,dx = \int_0^\infty f(s)(\mathcal{L}g)(s)\,ds. </math>

By plugging in <math>(\mathcal{L}f)(x)=\int_0^\infty f(s)e^{-sx}\,ds</math> the left-hand side turns into:
<math display=block>\int_0^\infty\int_0^\infty f(s)g(x) e^{-sx}\,ds\,dx, </math>
but assuming Fubini's theorem holds, by reversing the order of integration we get the wanted right-hand side.

This method can be used to compute integrals that would otherwise be difficult to compute using elementary methods of real calculus. For example,
<math display=block>\int_0^\infty\frac{\sin x}{x}dx = \int_0^\infty \mathcal{L}(1)(x)\sin x dx = \int_0^\infty 1 \cdot \mathcal{L}(\sin)(x)dx = \int_0^\infty \frac{dx}{x^2 + 1} = \frac{\pi}{2}. </math>