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Under suitable regularity and growth conditions, a function with Laplace transform $F$ is recovered by the inverse contour integralwhere the vertical line lies to the right of every singularity of $F$.

If $g$ has period $T$, then

The unilateral Laplace transform is $\bar{f}(p)={\int }_0^{\infty }{e}^{-pt}f(t)dt$ in its region of absolute convergence of a Laplace transform. Under the usual inversion hypotheses, the Bromwich inversion formula isThe vertical Bromwich contour is directed upward, with $c$ inside the transform's convergence half-plane and to the right of all its singularities. At discontinuities the standard symmetric inversion gives the mean of the one-sided values.

For a piecewise continuously differentiable $f$ with an integrable transformed derivative, integration by parts gives the Laplace transform of a derivative:Here the upper endpoint vanishes by the given decay condition, and right continuity at zero supplies the lower endpoint.

For real $p>0$, substitute $t=px$ in the Laplace transform:The Gamma function integral is finite because $a>0$. For complex $p$, absolute convergence holds exactly when $\mathrm{Re}p>0$: the modulus of the integrand is $x^{a-1}{e}^{-(\mathrm{Re}p)x}$. On this half-plane the same answer holds with the branch $p^{-a}=\exp (-a\mathrm{Log}p)$ whose logarithm is real on the positive axis, by holomorphic continuation of the two sides.

If “exists” permits conditionally convergent ordinary improper integrals, there are additional boundary points when $0<a<1$: for $p=i\omega \ne 0$, the Dirichlet test gives convergence, with value $\Gamma (a)|\omega {|}^{-a}{e}^{-i\mathrm{sgn}(\omega )\pi a/2}$. For $a\ge 1$ these boundary integrals do not converge in that sense. At $p=0$, and for $\mathrm{Re}p<0$, the integral diverges. Thus $\mathrm{Re}p>0$ is the region of absolute convergence of a Laplace transform, with the stated conditional boundary qualification.