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Evaluating integrals over the positive real axis

A useful property of the Laplace transform is the following: ∫ 0 ∞ f (x) g (x) d x = ∫ 0 ∞ (L f) (s) ⋅ (L − 1 g) (s) d s {\displaystyle \int _{0}^{\infty }f(x)g(x)\,dx=\int _{0}^{\infty }({\mathcal {L}}f)(s)\cdot ({\mathcal {L}}^{-1}g)(s)\,ds} under suitable assumptions on the behaviour of f, g {\displaystyle f,g} in a right neighbourhood of 0 {\displaystyle 0} and on the decay rate of f, g {\displaystyle f,g} in a left neighbourhood of ∞ {\displaystyle \infty }. The above formula is a variation of integration by parts, with the operators d d x {\displaystyle {\frac {d}{dx}}} and ∫ d x {\displaystyle \int \,dx} being replaced by L {\displaystyle {\mathcal {L}}} and L − 1 {\displaystyle {\mathcal {L}}^{-1}}. Let us prove the equivalent formulation: ∫ 0 ∞ (L f) (x) g (x) d x = ∫ 0 ∞ f (s) (L g) (s) d s. {\displaystyle \int _{0}^{\infty }({\mathcal {L}}f)(x)g(x)\,dx=\int _{0}^{\infty }f(s)({\mathcal {L}}g)(s)\,ds.}

By plugging in (L f) (x) = ∫ 0 ∞ f (s) e − s x d s {\displaystyle ({\mathcal {L}}f)(x)=\int _{0}^{\infty }f(s)e^{-sx}\,ds} the left-hand side turns into: ∫ 0 ∞ ∫ 0 ∞ f (s) g (x) e − s x d s d x, {\displaystyle \int _{0}^{\infty }\int _{0}^{\infty }f(s)g(x)e^{-sx}\,ds\,dx,} 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, ∫ 0 ∞ sin x x d x = ∫ 0 ∞ L (1) (x) sin x d x = ∫ 0 ∞ 1 ⋅ L (sin) (x) d x = ∫ 0 ∞ d x x 2 + 1 = π 2. {\displaystyle \int _{0}^{\infty }{\frac {\sin x}{x}}dx=\int _{0}^{\infty }{\mathcal {L}}(1)(x)\sin xdx=\int _{0}^{\infty }1\cdot {\mathcal {L}}(\sin)(x)dx=\int _{0}^{\infty }{\frac {dx}{x^{2}+1}}={\frac {\pi }{2}}.}