6.1 Definition and Existence
Definition 6.1.1 (Laplace transform). For \(f\) defined on \([0,\infty )\), \[\mathcal {L}\{f\}(s) = F(s) = \int ^{\infty }_{0}f(t)\,e^{-st}\,dt,\] for those \(s\) for which the integral converges.
Definition 6.1.2 (Exponential order). \(f\) is of exponential order \(\alpha \) if there are constants \(M\) and \(T\) with \(\left |f(t)\right |\leq M e^{\alpha t}\) for all \(t\geq T\).
Theorem 6.1.3 (Existence). If \(f\) is piecewise continuous on \([0,\infty )\) and of exponential order \(\alpha \), then \(\mathcal {L}\{f\}(s)\) converges absolutely for \(\operatorname {Re}s>\alpha \), and \(F(s)\to 0\) as \(s\to \infty \).
Proof. For \(t\geq T\) and \(\operatorname {Re}s = \sigma >\alpha \), \[\left |f(t)e^{-st}\right | \leq M e^{\alpha t}e^{-\sigma t} = M e^{-(\sigma -\alpha )t},\] and \(\int ^{\infty }_{T}Me^{-(\sigma -\alpha )t}dt = Me^{-(\sigma -\alpha )T}/(\sigma -\alpha )\) is finite. The integral over \([0,T]\) is finite because \(f\) is piecewise continuous there. Letting \(\sigma \to \infty \) in the same bound sends \(F(s)\) to zero. □
Note 6.1.4. The last clause of Theorem 6.1.3 is a useful check. A candidate transform that does not tend to zero at infinity — \(F(s)=1\), or \(F(s)=s/(s+1)\) — is not the transform of any function satisfying the hypotheses, and finding one in the middle of a calculation means an error earlier.
Example 6.1.5 (Powers, and where the gamma function enters). For \(f(t) = t^{a}\) with \(a>-1\), substitute \(u = st\) in the defining integral: \[\mathcal {L}\{t^{a}\}(s) = \int ^{\infty }_{0}t^{a}e^{-st}dt = \frac {1}{s^{a+1}}\int ^{\infty }_{0}u^{a}e^{-u}\,du = \frac {\Gamma (a+1)}{s^{a+1}},\] valid for \(s>0\). For integer \(a=n\) this is \(n!/s^{n+1}\).
Note 6.1.6. Example 6.1.5 is the gamma function of Chapter 2 doing the work, and it is worth noticing how little was required: a change of variable turned the Laplace integral into the gamma integral exactly. The restriction \(a>-1\) is the convergence condition established in Problem 1.6.9, arriving here for the same reason — the behaviour at the origin. Without the gamma function the transform of \(t^{1/2}\) could not be written down at all.
| \(f(t)\) | \(F(s)\) | valid for |
| \(1\) | \(1/s\) | \(s>0\) |
| \(t^{a},\ a>-1\) | \(\Gamma (a+1)/s^{a+1}\) | \(s>0\) |
| \(e^{at}\) | \(1/(s-a)\) | \(s>a\) |
| \(\sin bt\) | \(b/\left (s^{2}+b^{2}\right )\) | \(s>0\) |
| \(\cos bt\) | \(s/\left (s^{2}+b^{2}\right )\) | \(s>0\) |
| \(\sinh bt\) | \(b/\left (s^{2}-b^{2}\right )\) | \(s>|b|\) |
| \(\cosh bt\) | \(s/\left (s^{2}-b^{2}\right )\) | \(s>|b|\) |
| \(H(t-a)\) | \(e^{-as}/s\) | \(s>0\) |
| \(\delta (t-a)\) | \(e^{-as}\) | all \(s\) |
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