From the result above and the substitution u=s+t, \[ P_t U_\alpha f = \int_0^\infty e^{-\alpha s} P_{s+t} f \, ds = \int_t^\infty e^{-\alpha (u - t)} P_u f \, du = e^{\alpha t} \int_t^\infty...From the result above and the substitution u=s+t, PtUαf=∫∞0e−αsPs+tfds=∫∞te−α(u−t)Pufdu=eαt∫∞te−αuPufdu Hence PtUαf−Uαft=1t[eαt∫∞te−αuPufdu−Uαf] Adding and subtracting eαuUαf and combining integrals gives \begin{align*} \frac{P_t U_\alpha f - U_\al…
