Explanation: First, use the identity \(\sin^2(t) = \frac{1-\cos(2t)}{2}\). So, \(f(t) = \frac{1}{2} - \frac{1}{2}\cos(2t)\). The Laplace transform is \(F(s) = \frac{1}{2s} - \frac{1}{2}\frac{s}{s^2+4}\). Now apply the first shifting theorem for \(e^{-t}f(t)\) (so \(a=-1\)), which means we replace \(s\) with \(s+1\). This gives \(\frac{1}{2(s+1)} - \frac{1}{2}\frac{s+1}{(s+1)^2+4}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 386)
Explanation: The Final Value Theorem states \(\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s)\). The poles of \(sF(s)\) are the roots of \(s^2+2s+2=0\), which are \(s = -1 \pm i\). Since these are in the left half-plane, the theorem is applicable. \(\lim_{s \to 0} s \frac{1}{s(s^2+2s+2)} = \lim_{s \to 0} \frac{1}{s^2+2s+2} = \frac{1}{2}\).
Source: Ostaszewski, A. (2012). Advanced mathematical methods. Cambridge University Press. (p. 389)
Explanation: We use the formula \(\mathcal{L}\{t^\alpha\} = \frac{\Gamma(\alpha+1)}{s^{\alpha+1}}\). We want to match \(\frac{1}{s^{3/2}}\), so \(\alpha+1 = 3/2 \Rightarrow \alpha = 1/2\). Then \(\mathcal{L}\{t^{1/2}\} = \frac{\Gamma(3/2)}{s^{3/2}} = \frac{\frac{1}{2}\Gamma(1/2)}{s^{3/2}} = \frac{\sqrt{\pi}}{2s^{3/2}}\). To get the desired inverse transform, we must have \(\mathcal{L}^{-1}\{\frac{1}{s^{3/2}}\} = \frac{2}{\sqrt{\pi}} \mathcal{L}^{-1}\{\frac{\sqrt{\pi}}{2s^{3/2}}\} = \frac{2}{\sqrt{\pi}} t^{1/2} = 2\sqrt{t/\pi}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 394)
Explanation: Taking the Laplace transform gives \(s^2Y - 2sY + Y = \frac{1}{s-1}\). So, \(Y(s)(s-1)^2 = \frac{1}{s-1} \Rightarrow Y(s) = \frac{1}{(s-1)^3}\). We know \(\mathcal{L}\{t^2\} = \frac{2}{s^3}\). So \(\mathcal{L}^{-1}\{\frac{1}{s^3}\} = \frac{t^2}{2}\). Using the first shifting theorem, \(\mathcal{L}^{-1}\{\frac{1}{(s-1)^3}\} = \frac{1}{2}t^2 e^t\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 234)
Explanation: We can use the convolution theorem. \(\mathcal{L}\{\cos(t)\} = \frac{s}{s^2+1}\) and \(\mathcal{L}\{\sin(t)\} = \frac{1}{s^2+1}\). The transform of the convolution is their product: \(\frac{s}{(s^2+1)^2}\). We know from the property \(\mathcal{L}\{tf(t)\} = -F'(s)\) that \(\mathcal{L}\{t\sin(t)\} = \frac{2s}{(s^2+1)^2}\). Therefore, the inverse transform of \(\frac{s}{(s^2+1)^2}\) is \(\frac{1}{2}t\sin(t)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 394)
Explanation: This requires the second shifting theorem. Let \(F(s) = \frac{s}{s^2+\pi^2}\), so \(f(t) = \cos(\pi t)\). The transform is \(e^{-s}F(s)\) with \(a=1\). The inverse is \(f(t-1)u(t-1) = \cos(\pi(t-1))u(t-1)\). Note that \(\cos(\pi(t-1)) = \cos(\pi t - \pi) = -\cos(\pi t)\). So an equivalent answer is \(-\cos(\pi t)u(t-1)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 386)
Explanation: This integral is \(\mathcal{L}\{t\sin(t)\}\) evaluated at \(s=3\). First, find \(\mathcal{L}\{\sin(t)\} = F(s) = \frac{1}{s^2+1}\). Then \(\mathcal{L}\{t\sin(t)\} = -F'(s) = -\frac{d}{ds}(\frac{1}{s^2+1}) = \frac{2s}{(s^2+1)^2}\). Now substitute \(s=3\): \(\frac{2(3)}{(3^2+1)^2} = \frac{6}{10^2} = \frac{6}{100} = \frac{3}{50}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 386)
Explanation: Let \(F(s) = \text{arccot}(s)\). We use the property \(\mathcal{L}\{tf(t)\} = -F'(s)\). The derivative is \(F'(s) = -\frac{1}{1+s^2}\). So, \(\mathcal{L}\{tf(t)\} = \frac{1}{1+s^2}\). The inverse transform of this is \(\sin(t)\). Therefore, \(tf(t) = \sin(t) \Rightarrow f(t) = \frac{\sin(t)}{t}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 402, Problem 12)
Explanation: True. The convolution operation is associative. This can be proven by applying the Convolution Theorem: \(\mathcal{L}\{(f*g)*h\} = \mathcal{L}\{f*g\}\mathcal{L}\{h\} = (F(s)G(s))H(s)\). And \(\mathcal{L}\{f*(g*h)\} = \mathcal{L}\{f\}\mathcal{L}\{g*h\} = F(s)(G(s)H(s))\). Since multiplication is associative, the transforms are equal, and thus the functions are equal.
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 236)
Explanation: Use the convolution theorem with \(F(s) = G(s) = \frac{s}{s^2+4}\). Then \(f(t) = g(t) = \cos(2t)\). The inverse transform is \((f*g)(t) = \int_0^t \cos(2u)\cos(2(t-u))du\). Using the identity \(\cos(A)\cos(B) = \frac{1}{2}[\cos(A-B)+\cos(A+B)]\), the integral becomes \(\frac{1}{2}\int_0^t [\cos(4u-2t) + \cos(2t)]du = \frac{1}{2}[\frac{1}{4}\sin(4u-2t) + u\cos(2t)]_0^t = \frac{1}{2}[(\frac{1}{4}\sin(2t) + t\cos(2t)) - (\frac{1}{4}\sin(-2t))] = \frac{1}{2}[\frac{1}{2}\sin(2t) + t\cos(2t)] = \frac{1}{4}t\cos(2t) + \frac{1}{8}\sin(2t)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 394)
Explanation: Use the property \(\mathcal{L}\{\frac{f(t)}{t}\} = \int_s^\infty F(u)du\). Here \(f(t) = 1 - e^{-t}\), so \(F(u) = \frac{1}{u} - \frac{1}{u+1}\). The integral is \(\int_s^\infty (\frac{1}{u} - \frac{1}{u+1})du = [\ln(u) - \ln(u+1)]_s^\infty = [\ln(\frac{u}{u+1})]_s^\infty\). As \(u \to \infty\), \(\ln(\frac{u}{u+1}) \to \ln(1) = 0\). So the result is \(0 - \ln(\frac{s}{s+1}) = \ln((\frac{s}{s+1})^{-1}) = \ln(\frac{s+1}{s}) = \ln(s+1) - \ln(s)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 402)
Explanation: Using the convolution theorem, \(\mathcal{L}\{t^p * t^q\} = \mathcal{L}\{t^p\} \mathcal{L}\{t^q\} = \frac{\Gamma(p+1)}{s^{p+1}} \frac{\Gamma(q+1)}{s^{q+1}} = \frac{\Gamma(p+1)\Gamma(q+1)}{s^{p+q+2}}\). Taking the inverse transform, we get \(t^p * t^q = \frac{\Gamma(p+1)\Gamma(q+1)}{\Gamma(p+q+2)} t^{p+q+1} = B(p+1,q+1)t^{p+q+1}\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 240)
Explanation: Use partial fractions for \(s^2\). Let \(x=s^2\). \(\frac{1}{(x+1)(x+4)} = \frac{A}{x+1} + \frac{B}{x+4}\). This gives \(A=1/3, B=-1/3\). So the transform is \(\frac{1}{3}\frac{1}{s^2+1} - \frac{1}{3}\frac{1}{s^2+4}\). The inverse transform is \(\frac{1}{3}\sin(t) - \frac{1}{3} \frac{1}{2}\sin(2t) = \frac{1}{3}\sin(t) - \frac{1}{6}\sin(2t)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 330)
Explanation: For a periodic function with period \(T\), the Laplace transform is given by the integral over one period divided by \(1-e^{-sT}\). This accounts for all subsequent periods of the function.
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 329)
Explanation: We calculate \(\int_0^2 e^{-st}f(t)dt = \int_0^1 e^{-st}(1)dt + \int_1^2 e^{-st}(-1)dt = [-\frac{1}{s}e^{-st}]_0^1 - [-\frac{1}{s}e^{-st}]_1^2 = (\frac{1-e^{-s}}{s}) - (\frac{e^{-s}-e^{-2s}}{s}) = \frac{1-2e^{-s}+e^{-2s}}{s} = \frac{(1-e^{-s})^2}{s}\). Then we divide by \((1-e^{-2s})\). This can be simplified to \(\frac{1}{s}\tanh(s/2)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 398)
Explanation: This is a standard transform pair, often derived using the multiplication by \(t\) property. We know \(\mathcal{L}\{\cos(at)\} = \frac{s}{s^2+a^2}\). Differentiating with respect to \(s\) gives \(\frac{d}{ds}(\frac{s}{s^2+a^2}) = \frac{(s^2+a^2)-s(2s)}{(s^2+a^2)^2} = \frac{a^2-s^2}{(s^2+a^2)^2}\). Since \(\mathcal{L}\{t\cos(at)\} = -\frac{d}{ds}F(s)\), we have \(\mathcal{L}\{t\cos(at)\} = -\frac{a^2-s^2}{(s^2+a^2)^2} = \frac{s^2-a^2}{(s^2+a^2)^2}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 329, Table 12.1)
Explanation: Let \(x = e^{-t}\), so \(dx = -e^{-t}dt\) and \(\ln x = -t\). When \(x=1, t=0\). When \(x \to 0, t \to \infty\). The integral becomes \(\int_\infty^0 (-t)^3 (-e^{-t})dt = \int_0^\infty (-t^3)(-e^{-t})dt = \int_0^\infty t^3 e^{-t} dt = \Gamma(4) = 3! = 6\). Wait, there is a sign error. \(\int_\infty^0 (-t)^3 (-e^{-t})dt = -\int_0^\infty (-t)^3 e^{-t} dt = \int_0^\infty t^3 e^{-t} dt = \Gamma(4) = 6\). Let me re-check. The substitution is correct. \(\int_0^1 (\ln x)^3 dx = [x(\ln x)^3]_0^1 - \int_0^1 x \cdot 3(\ln x)^2 \frac{1}{x} dx = 0 - 3\int_0^1 (\ln x)^2 dx\). Repeating this gives \(-3[-2\int_0^1 \ln x dx] = 6[x\ln x - x]_0^1 = 6(0-1) = -6\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 393)
Explanation: First, factor the denominator using Sophie Germain's identity: \(s^4+4 = (s^2+2s+2)(s^2-2s+2)\). Then use partial fractions: \(\frac{s^3}{s^4+4} = \frac{As+B}{s^2+2s+2} + \frac{Cs+D}{s^2-2s+2}\). This is complicated. A standard result is \(\mathcal{L}\{\cos(at)\cosh(at)\} = \frac{s^3}{s^4+4a^4}\). For \(a=1\), we get the desired transform.
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (Table of Laplace Transforms)
Explanation: The error function is defined as \(\text{erf}(t) = \frac{2}{\sqrt{\pi}}\int_0^t e^{-u^2}du\). A standard, but non-trivial, Laplace transform pair is \(\mathcal{L}\{\text{erf}(\sqrt{t})\} = \frac{1}{s\sqrt{s+1}}\). This is often found in tables.
Source: Standard Laplace Transform Tables.
Explanation: Take the Laplace transform of both equations: \(sX(s) - x(0) = -Y(s)\) and \(sY(s) - y(0) = X(s)\). Substitute initial conditions: \(sX(s) - 1 = -Y(s)\) and \(sY(s) = X(s)\). Substitute the second into the first: \(s(sY(s)) - 1 = -Y(s) \Rightarrow s^2Y(s) + Y(s) = 1 \Rightarrow Y(s) = \frac{1}{s^2+1}\). So, \(y(t) = \sin(t)\). Then \(X(s) = sY(s) = \frac{s}{s^2+1}\), so \(x(t) = \cos(t)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 400)
Explanation: Let \(s+1 = u\), so \(s = u-1\). The transform becomes \(\frac{u-1}{u^5} = \frac{1}{u^4} - \frac{1}{u^5}\). The inverse transform in terms of \(u\) is \(\frac{t^3}{3!} - \frac{t^4}{4!} = \frac{t^3}{6} - \frac{t^4}{24}\). Now apply the shifting theorem with \(a=-1\) to get \(e^{-t}(\frac{t^3}{6} - \frac{t^4}{24})\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 386)
Explanation: We can use the Beta function. Let \(t = \sin^2(x)\). The integral is \(2\int_0^{\pi/2} \sin^4(x) dx\). Using the formula \(\int_0^{\pi/2} \sin^{2m-1}(x)\cos^{2n-1}(x)dx = \frac{1}{2}B(m,n)\), we have \(2m-1=4 \Rightarrow m=5/2\) and \(2n-1=0 \Rightarrow n=1/2\). So the integral is \(2 \cdot \frac{1}{2} B(5/2, 1/2) = \frac{\Gamma(5/2)\Gamma(1/2)}{\Gamma(3)} = \frac{(\frac{3}{2}\frac{1}{2}\sqrt{\pi})\sqrt{\pi}}{2!} = \frac{3\pi/4}{2} = \frac{3\pi}{8}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 392)
Explanation: This is the change of scale theorem. \(\mathcal{L}\{f(at)\} = \int_0^\infty e^{-st}f(at)dt\). Let \(u=at\), so \(t=u/a\) and \(dt = du/a\). The integral becomes \(\int_0^\infty e^{-s(u/a)}f(u)\frac{du}{a} = \frac{1}{a}\int_0^\infty e^{-(s/a)u}f(u)du = \frac{1}{a}F(\frac{s}{a})\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 228)
Explanation: Complete the square in the denominator: \(s^2+s+1 = (s+1/2)^2 + 3/4\). The transform is \(\frac{1}{(s+1/2)^2 + (\sqrt{3}/2)^2}\). This matches the form for a shifted sine. We need to adjust the numerator: \(\frac{2}{\sqrt{3}} \frac{\sqrt{3}/2}{(s+1/2)^2 + (\sqrt{3}/2)^2}\). The inverse transform is \(\frac{2}{\sqrt{3}}e^{-t/2}\sin(\frac{\sqrt{3}}{2}t)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 330)
Explanation: Let \(g(t) = \frac{\sin(t)}{t}\). We know \(\mathcal{L}\{g(t)\} = G(s) = \text{arccot}(s)\). The given function is \(\int_0^t g(x)dx\). Using the transform of an integral property, \(\mathcal{L}\{\int_0^t g(x)dx\} = \frac{G(s)}{s} = \frac{\text{arccot}(s)}{s}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 402)
Explanation: Using the sifting property of the delta function, \(\mathcal{L}\{\delta(t)\} = \int_0^\infty e^{-st}\delta(t)dt = e^{-s(0)} = 1\). This is a fundamental result.
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 396)
Explanation: Taking the Laplace transform gives \(sY(s) + Y(s) = e^{-s}\). So \(Y(s) = \frac{e^{-s}}{s+1}\). This is of the form \(e^{-as}F(s)\) with \(a=1\) and \(F(s) = \frac{1}{s+1}\). The inverse of \(F(s)\) is \(f(t) = e^{-t}\). By the second shifting theorem, the solution is \(y(t) = f(t-1)u(t-1) = e^{-(t-1)}u(t-1)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 396)
Explanation: The Initial Value Theorem states \(\lim_{t \to 0^+} f(t) = \lim_{s \to \infty} sF(s)\). We have \(\lim_{s \to \infty} s \frac{2s+3}{s^2+5s+4} = \lim_{s \to \infty} \frac{2s^2+3s}{s^2+5s+4}\). Dividing numerator and denominator by \(s^2\) gives \(\lim_{s \to \infty} \frac{2+3/s}{1+5/s+4/s^2} = 2\).
Source: Ostaszewski, A. (2012). Advanced mathematical methods. Cambridge University Press. (p. 390)
Explanation: Let \(f(t) = \cos(at) - \cos(bt)\). Then \(F(u) = \frac{u}{u^2+a^2} - \frac{u}{u^2+b^2}\). We compute \(\int_s^\infty F(u)du = [\frac{1}{2}\ln(u^2+a^2) - \frac{1}{2}\ln(u^2+b^2)]_s^\infty = \frac{1}{2}[\ln(\frac{u^2+a^2}{u^2+b^2})]_s^\infty\). As \(u \to \infty\), the term goes to \(\ln(1)=0\). So the result is \(-\frac{1}{2}\ln(\frac{s^2+a^2}{s^2+b^2})) = \frac{1}{2}\ln(\frac{s^2+b^2}{s^2+a^2})\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 402)
Explanation: Let \(F(s) = \ln(1+\frac{1}{s^2}) = \ln(\frac{s^2+1}{s^2}) = \ln(s^2+1) - 2\ln(s)\). Let \(f(t) = \mathcal{L}^{-1}\{F(s)\}\). Then \(\mathcal{L}\{tf(t)\} = -F'(s) = -(\frac{2s}{s^2+1} - \frac{2}{s}) = \frac{2}{s} - \frac{2s}{s^2+1}\). The inverse transform of this is \(2 - 2\cos(t)\). So, \(tf(t) = 2(1-\cos t)\), which means \(f(t) = \frac{2(1-\cos t)}{t}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 402)
Explanation: Use \(\mathcal{L}\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n}F(s)\) with \(n=2\) and \(f(t)=\sin(t)\). So \(F(s) = \frac{1}{s^2+1}\). We need \((-1)^2 \frac{d^2}{ds^2}(\frac{1}{s^2+1})\). The first derivative is \(\frac{-2s}{(s^2+1)^2}\). The second derivative is \(\frac{-2(s^2+1)^2 - (-2s)2(s^2+1)(2s)}{(s^2+1)^4} = \frac{-2(s^2+1) + 8s^2}{(s^2+1)^3} = \frac{6s^2-2}{(s^2+1)^3} = \frac{2(3s^2-1)}{(s^2+1)^3}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 386)
Explanation: This is the integral form of a Laplace transform. Specifically, \(\mathcal{L}\{\frac{f(t)}{t}\} = \int_s^\infty F(u)du\). The given integral is \(\mathcal{L}\{\frac{e^{-t} - e^{-3t}}{t}\}|_{s=0}\). Let \(g(t) = e^{-t} - e^{-3t}\), so \(G(u) = \frac{1}{u+1} - \frac{1}{u+3}\). The transform is \(\int_s^\infty (\frac{1}{u+1} - \frac{1}{u+3})du = [\ln(u+1)-\ln(u+3)]_s^\infty = [\ln(\frac{u+1}{u+3})]_s^\infty = 0 - \ln(\frac{s+1}{s+3}) = \ln(\frac{s+3}{s+1})\). Evaluating at \(s=0\) gives \(\ln(3)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 402)
Explanation: Use partial fractions: \(\frac{1}{s^4-1} = \frac{1}{(s^2-1)(s^2+1)} = \frac{A}{s^2-1} + \frac{B}{s^2+1}\). This gives \(A=1/2, B=-1/2\). So we have \(\frac{1}{2}\frac{1}{s^2-1} - \frac{1}{2}\frac{1}{s^2+1}\). The inverse transform is \(\frac{1}{2}\sinh(t) - \frac{1}{2}\sin(t) = \frac{1}{2}(\sinh(t) - \sin(t))\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 330)
Explanation: Use \(\mathcal{L}\{t^\alpha\} = \frac{\Gamma(\alpha+1)}{s^{\alpha+1}}\). Here \(\alpha = 5/2\). We need \(\Gamma(5/2+1) = \Gamma(7/2) = \frac{5}{2}\Gamma(5/2) = \frac{5}{2}\frac{3}{2}\frac{1}{2}\Gamma(1/2) = \frac{15}{8}\sqrt{\pi}\). So the transform is \(\frac{15\sqrt{\pi}/8}{s^{7/2}} = \frac{15\sqrt{\pi}}{8s^{7/2}}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 394)
Explanation: Use the second shifting theorem. \(F(s) = \frac{1}{s^2-1}\), so \(f(t) = \sinh(t)\). The transform is \(e^{-3s}F(s)\) with \(a=3\). The inverse is \(f(t-3)u(t-3) = \sinh(t-3)u(t-3)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 386)
Explanation: First, evaluate the integral: \(\int_0^t e^x dx = [e^x]_0^t = e^t - 1\). Now find the Laplace transform of this result: \(\mathcal{L}\{e^t - 1\} = \frac{1}{s-1} - \frac{1}{s} = \frac{s - (s-1)}{s(s-1)} = \frac{1}{s(s-1)}\). Alternatively, use the transform of an integral property with \(f(t)=e^t\) and \(F(s)=\frac{1}{s-1}\). The transform is \(\frac{F(s)}{s} = \frac{1}{s(s-1)}\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 226)
Explanation: Split the fraction: \(\frac{s+2}{s^3} = \frac{s}{s^3} + \frac{2}{s^3} = \frac{1}{s^2} + \frac{2}{s^3}\). The inverse transform is \(t + 2(\frac{t^2}{2!}) = t+t^2\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 330)
Explanation: Let \(f(t) = \cosh(3t)\), so \(F(s) = \frac{s}{s^2-9}\). Then \(\mathcal{L}\{tf(t)\} = -F'(s) = -\frac{(s^2-9) - s(2s)}{(s^2-9)^2} = -\frac{-s^2-9}{(s^2-9)^2} = \frac{s^2+9}{(s^2-9)^2}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 386)
Explanation: Use partial fractions: \(\frac{1}{s(s+1)^2} = \frac{A}{s} + \frac{B}{s+1} + \frac{C}{(s+1)^2}\). Solving gives \(A=1, B=-1, C=-1\). The transform is \(\frac{1}{s} - \frac{1}{s+1} - \frac{1}{(s+1)^2}\). The inverse transform is \(1 - e^{-t} - te^{-t}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 330)
Explanation: We need to write the function in the form \(f(t-a)u(t-a)\). Here \(a=\pi\). We have \(\sin(t) = \sin(t-\pi+\pi) = \sin(t-\pi)\cos(\pi) + \cos(t-\pi)\sin(\pi) = -\sin(t-\pi)\). So we need \(\mathcal{L}\{-\sin(t-\pi)u(t-\pi)\}\). The transform of \(-\sin(t)\) is \(-\frac{1}{s^2+1}\). Applying the second shifting theorem gives \(-e^{-\pi s} \frac{1}{s^2+1}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 386)
Explanation: This is a known, but complex, transform pair. A standard result from tables is \(\mathcal{L}\{\frac{1}{2a}(\sin(at)\cosh(at) + \cos(at)\sinh(at))\} = \frac{s^2}{s^4+4a^4}\). For \(a=1\), we get the desired result.
Source: Standard Laplace Transform Tables.
Explanation: Using the formula \(B(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}\). We have \(B(1/2, 1/2) = \frac{\Gamma(1/2)\Gamma(1/2)}{\Gamma(1)} = \frac{\sqrt{\pi}\sqrt{\pi}}{1} = \pi\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 240)
Explanation: Use partial fractions: \(\frac{1}{s^2(s+1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1}\). Solving gives \(A=-1, B=1, C=1\). The transform is \(-\frac{1}{s} + \frac{1}{s^2} + \frac{1}{s+1}\). The inverse transform is \(-1 + t + e^{-t}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 330)
Explanation: First, \(\mathcal{L}\{\cosh(t)\} = \frac{s}{s^2-1}\). Now apply the first shifting theorem with \(a=4\), replacing \(s\) with \(s-4\). This gives \(\frac{s-4}{(s-4)^2-1}\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 227)
Explanation: Factor the denominator: \(s^4+10s^2+9 = (s^2+1)(s^2+9)\). Use partial fractions: \(\frac{6}{(s^2+1)(s^2+9)} = \frac{A}{s^2+1} + \frac{B}{s^2+9}\). This gives \(A=3/4, B=-3/4\). So we have \(\frac{3}{4}\frac{1}{s^2+1} - \frac{3}{4}\frac{1}{s^2+9}\). The inverse transform is \(\frac{3}{4}\sin(t) - \frac{3}{4}\frac{1}{3}\sin(3t) = \frac{3}{4}\sin(t) - \frac{1}{4}\sin(3t)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 330)
Explanation: Use \(\mathcal{L}\{t^\alpha\} = \frac{\Gamma(\alpha+1)}{s^{\alpha+1}}\). Here \(\alpha = -1/2\). We need \(\Gamma(-1/2+1) = \Gamma(1/2) = \sqrt{\pi}\). So the transform is \(\frac{\sqrt{\pi}}{s^{1/2}} = \sqrt{\frac{\pi}{s}}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 394)
Explanation: Take the Laplace transform: \(s^2Y - sy(0) - y'(0) + 2(sY-y(0)) + 2Y = 0\). Substitute initial conditions: \(s^2Y - 1 + 2sY + 2Y = 0 \Rightarrow (s^2+2s+2)Y = 1\). So \(Y(s) = \frac{1}{s^2+2s+2} = \frac{1}{(s+1)^2+1}\). This is the transform of \(e^{-t}\sin(t)\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 234)
Explanation: This is a standard transform pair, usually found in tables. \(\mathcal{L}\{\frac{1}{2a^2}\sin(at)\sinh(at)\} = \frac{s}{s^4+4a^4}\).
Source: Standard Laplace Transform Tables.
Explanation: Let \(f(t) = \sinh(t)\), so \(F(u) = \frac{1}{u^2-1}\). We compute \(\int_s^\infty \frac{1}{u^2-1}du = \frac{1}{2}\int_s^\infty (\frac{1}{u-1} - \frac{1}{u+1})du = \frac{1}{2}[\ln(u-1)-\ln(u+1)]_s^\infty = \frac{1}{2}[\ln(\frac{u-1}{u+1})]_s^\infty\). As \(u \to \infty\), the term goes to \(\ln(1)=0\). So the result is \(-\frac{1}{2}\ln(\frac{s-1}{s+1})) = \frac{1}{2}\ln(\frac{s+1}{s-1})\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 402)
Explanation: This is a standard transform pair, often derived using the multiplication by \(t\) property. We know \(\mathcal{L}\{\cos(at)\} = \frac{s}{s^2+a^2}\). Differentiating with respect to \(s\) gives \(\frac{d}{ds}(\frac{s}{s^2+a^2}) = \frac{(s^2+a^2)-s(2s)}{(s^2+a^2)^2} = \frac{a^2-s^2}{(s^2+a^2)^2}\). Since \(\mathcal{L}\{t\cos(at)\} = -\frac{d}{ds}F(s)\), we have \(\mathcal{L}\{t\cos(at)\} = -\frac{a^2-s^2}{(s^2+a^2)^2} = \frac{s^2-a^2}{(s^2+a^2)^2}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 329, Table 12.1)