Show that \[ B(z, w) = {\Gamma(z) \Gamma(w) \over \Gamma(z+w)} .\]
Hint: find \(\mcl(t^{z-1})\) and \(\mcl(t^{z-1}\convolve t^{w-1})\).
Solution
Let \(\operatorname{Re}(z) > 0\) and \(\operatorname{Re}(w) > 0\), let \(s>0\), and define \(f(t) = t^{z-1}\) and \(g(t) = t^{w-1}\) for \(t > 0\). Recall \(B(z, w) = \int_0^1 u^{z-1} (1 - u)^{w-1}\,du\).
1. \(\mathcal{L}\{f\}(s) = \frac{\Gamma(z)}{s^z}\) and \(\mathcal{L}\{g\}(s) = \frac{\Gamma(w)}{s^w}\).
Substituting \(u = st\), so \(t = u/s\) and \(dt = du/s\), \[\mathcal{L}\{f\}(s) = \int_0^\infty t^{z-1} e^{-st}\,dt = \int_0^\infty \left(\frac{u}{s}\right)^{z-1} e^{-u} \frac{du}{s} = \frac{1}{s^z} \int_0^\infty u^{z-1} e^{-u}\,du = \frac{\Gamma(z)}{s^z}.\] The same computation with \(w\) in place of \(z\) gives \(\mathcal{L}\{g\}\).
2. \((f * g)(x) = x^{z+w-1} B(z, w)\) for \(x > 0\).
Substituting \(t = xu\), so \(dt = x\,du\), \[(f * g)(x) = \int_0^x t^{z-1} (x - t)^{w-1}\,dt = \int_0^1 (xu)^{z-1} (x - xu)^{w-1} x\,du = x^{z+w-1} \int_0^1 u^{z-1} (1 - u)^{w-1}\,du.\]
3. \(B(z, w) \frac{\Gamma(z+w)}{s^{z+w}} = \frac{\Gamma(z)\Gamma(w)}{s^{z+w}}\).
By step 2 and step 1 with \(z+w\) in place of \(z\), \[\mathcal{L}\{f * g\}(s) = B(z, w) \int_0^\infty x^{z+w-1} e^{-sx}\,dx = B(z, w) \frac{\Gamma(z+w)}{s^{z+w}}.\] By the convolution theorem for Laplace transforms and step 1, \[\mathcal{L}\{f * g\}(s) = \mathcal{L}\{f\}(s) \cdot \mathcal{L}\{g\}(s) = \frac{\Gamma(z)\Gamma(w)}{s^{z+w}}.\]
4. Q.E.D.
Multiply step 3 by \(s^{z+w}\) and divide by \(\Gamma(z+w)\), which has no zeros.