14.3: Connection to Laplace

Claim

For \(\text{Re}(z) > 1\) and \(\text{Re} (s) >0\), \(\mathcal{L} (t^{z -1}; s) = \dfrac{\Gamma (z)}{s^z}\).

Proof

By definition \(\mathcal{L} (t^{z -1}; s) = \int_{0}^{\infty} t^{z - 1} e^{-st} \ dt\). It is clear that if \(\text{Re} (z) > 1\), then the integral converges absolutely for \(\text{Re} (s) > 0\).

Let’s start by assuming that \(s > 0\) is real. Use the change of variable \(\tau = st\). The Laplace integral becomes

\[\int_{0}^{\infty} t^{z - 1} e^{-st} \ dt = \int_{0}^{\infty} (\dfrac{\tau}{s})^{z- 1} e^{-\tau} \dfrac{d \tau}{s} = \dfrac{1}{s^z} \int_{0}^{\infty} \tau ^{z - 1} e^{-\tau} = \dfrac{\Gamma (z)}{s^z} \ d\tau. \nonumber \]

This shows that \(\mathcal{L} (t^{z -1}; s) = \dfrac{\Gamma (z)}{s^z})\) for \(s\) real and positive. Since both sides of this equation are analytic on \(\text{Re} (s) > 0\), the extension to Theorem 14.2.1 guarantees they are the same.

Corollary

\(\Gamma (z) = \mathcal{L} (t^{z - 1}; 1)\). (Of course, this is also clear directly from the definition of \(\Gamma (z)\) in Equation 14.3.1.