Explanation: The Laplace transform of a function \(f(t)\) is defined as \( F(s) = \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) dt \). This integral transforms a function of a real variable \(t\) (often time) to a function of a complex variable \(s\) (frequency).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 221)
Explanation: A function \(f(t)\) is of exponential growth if there exist constants \(M > 0\) and \(\gamma\) such that \(|f(t)| \le Me^{\gamma t}\) for all \(t \ge 0\). This condition ensures that the integral for the Laplace transform converges for \(s > \gamma\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 223)
Explanation: Using the definition, \(\mathcal{L}\{c\} = \int_0^\infty c e^{-st} dt = c [-\frac{1}{s}e^{-st}]_0^\infty = c(0 - (-\frac{1}{s})) = \frac{c}{s}\), for \(s>0\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 329, Table 12.1)
Explanation: Using integration by parts: \(\mathcal{L}\{t\} = \int_0^\infty t e^{-st} dt = [t(-\frac{e^{-st}}{s})]_0^\infty - \int_0^\infty -\frac{e^{-st}}{s} dt = 0 + \frac{1}{s} \int_0^\infty e^{-st} dt = \frac{1}{s} (\frac{1}{s}) = \frac{1}{s^2}\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 242, Solution to Activity 7.1)
Explanation: This is a standard result, often derived using integration by parts recursively or using the Gamma function. \(\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}\) for \(s>0\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 225)
Explanation: \(\mathcal{L}\{e^{at}\} = \int_0^\infty e^{at}e^{-st} dt = \int_0^\infty e^{-(s-a)t} dt = [-\frac{1}{s-a}e^{-(s-a)t}]_0^\infty = \frac{1}{s-a}\) for \(s>a\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 222)
Explanation: The Laplace transform of \(\sin(at)\) is given by \(\mathcal{L}\{\sin(at)\} = \frac{a}{s^2 + a^2}\). This can be derived using integration by parts twice or by using Euler's formula for \(\sin(at)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 329, Table 12.1)
Explanation: The Laplace transform of \(\cos(at)\) is given by \(\mathcal{L}\{\cos(at)\} = \frac{s}{s^2 + a^2}\). This can also be derived using integration by parts or Euler's formula.
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 329, Table 12.1)
Explanation: The first shifting theorem is a fundamental property. \(\mathcal{L}\{e^{at}f(t)\} = \int_0^\infty e^{at}f(t)e^{-st}dt = \int_0^\infty f(t)e^{-(s-a)t}dt = F(s-a)\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 227)
Explanation: We know that \( \mathcal{L}\{t^2\} = \frac{2!}{s^{2+1}} = \frac{2}{s^3}\). By the first shifting theorem, with \(a=3\), we have \(\mathcal{L}\{e^{3t}t^2\} = F(s-3) = \frac{2}{(s-3)^3}\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 229, Example 7.5)
Explanation: Since \(\cosh(at) = \frac{e^{at} + e^{-at}}{2}\), we can use the linearity of the Laplace transform: \(\mathcal{L}\{\cosh(at)\} = \frac{1}{2}(\mathcal{L}\{e^{at}\} + \mathcal{L}\{e^{-at}\}) = \frac{1}{2}(\frac{1}{s-a} + \frac{1}{s+a}) = \frac{1}{2}(\frac{s+a+s-a}{s^2-a^2}) = \frac{s}{s^2-a^2}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 329, Table 12.1)
Explanation: The transform of a derivative is given by \(\mathcal{L}\{f'(t)\} = s\mathcal{L}\{f(t)\} - f(0) = sF(s) - f(0)\). This property is key to solving differential equations.
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 224, Equation 7.4)
Explanation: Applying the derivative rule twice: \(\mathcal{L}\{f''(t)\} = s\mathcal{L}\{f'(t)\} - f'(0) = s(sF(s) - f(0)) - f'(0) = s^2F(s) - sf(0) - f'(0)\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 225, Equation 7.5)
Explanation: This is the standard transform pair for an exponential function. Since \(\mathcal{L}\{e^{at}\} = \frac{1}{s-a}\), the inverse transform is \(\mathcal{L}^{-1}\{\frac{1}{s-5}\} = e^{5t}\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 222)
Explanation: We know \(\mathcal{L}\{\sin(at)\} = \frac{a}{s^2+a^2}\). Here, \(a^2=9\) so \(a=3\). The given transform is \(\frac{1}{s^2+9}\). We need to adjust for the constant 'a' in the numerator. \(\mathcal{L}^{-1}\{\frac{1}{s^2+9}\} = \frac{1}{3}\mathcal{L}^{-1}\{\frac{3}{s^2+9}\} = \frac{1}{3}\sin(3t)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 329, Table 12.1)
Explanation: We can split the fraction: \(F(s) = \frac{s}{s^2-4} + \frac{1}{s^2-4}\). The inverse transform of the first term is \(\cosh(2t)\). The inverse transform of the second term is \(\frac{1}{2}\sinh(2t)\). By linearity, the result is \(\cosh(2t) + \frac{1}{2}\sinh(2t)\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 230, Example 7.6)
Explanation: Transforming the equation gives \(sY(s) - y(0) + Y(s) = \frac{1}{s}\). With \(y(0)=0\), we get \((s+1)Y(s) = \frac{1}{s}\), so \(Y(s) = \frac{1}{s(s+1)}\). Using partial fractions, \(Y(s) = \frac{1}{s} - \frac{1}{s+1}\). The inverse transform is \(y(t) = 1 - e^{-t}\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 233)
Explanation: Transforming gives \(s^2Y(s) - sy(0) - y'(0) + 4Y(s) = 0\). Substituting initial conditions: \(s^2Y(s) - s + 4Y(s) = 0 \Rightarrow (s^2+4)Y(s) = s \Rightarrow Y(s) = \frac{s}{s^2+4}\). The inverse transform is \(y(t) = \cos(2t)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 330)
Explanation: The convolution of \(f\) and \(g\) is defined as \((f*g)(t) = \int_0^t f(u)g(t-u)du\). It is a special kind of integral that arises in many areas of mathematics and engineering.
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 236)
Explanation: The Convolution Theorem is a powerful tool that states the Laplace transform of a convolution of two functions is the product of their individual Laplace transforms: \(\mathcal{L}\{(f*g)(t)\} = \mathcal{L}\{f(t)\} \mathcal{L}\{g(t)\} = F(s)G(s)\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 236, Theorem 7.1)
Explanation: Let \(H(s) = F(s)G(s)\) where \(F(s) = 1/s\) and \(G(s) = 1/(s-1)\). Then \(f(t) = 1\) and \(g(t) = e^t\). The inverse transform is their convolution: \(h(t) = (f*g)(t) = \int_0^t 1 \cdot e^{t-u} du = e^t \int_0^t e^{-u} du = e^t [-e^{-u}]_0^t = e^t(-e^{-t} + 1) = e^t - 1\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 237, Example 7.9)
Explanation: The Gamma function is defined by the improper integral \(\Gamma(\alpha) = \int_0^\infty t^{\alpha-1} e^{-t} dt\). This integral converges for \(\alpha > 0\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 231)
Explanation: Using integration by parts on the definition of \(\Gamma(\alpha+1)\), we find the recurrence relation \(\Gamma(\alpha+1) = \alpha\Gamma(\alpha)\). This is a generalization of the factorial function.
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 232, Equation 7.10)
Explanation: Using the property \(\Gamma(\alpha+1) = \alpha\Gamma(\alpha)\) and the fact that \(\Gamma(1) = \int_0^\infty e^{-t} dt = 1\), we can show by induction that \(\Gamma(n) = (n-1)!\) for any positive integer \(n\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 232, Equation 7.11)
Explanation: The value of \(\Gamma(1/2)\) is a famous result. It is equal to \(\sqrt{\pi}\). This can be shown by evaluating the Gaussian integral.
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 231, Activity 7.10)
Explanation: The general formula for the Laplace transform of a power of \(t\) is \(\mathcal{L}\{t^\alpha\} = \frac{\Gamma(\alpha+1)}{s^{\alpha+1}}\). This generalizes the result for integer powers.
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 232, Equation 7.12)
Explanation: The Beta function has several equivalent definitions. The standard integral definition is \(\int_0^1 t^{p-1}(1-t)^{q-1} dt\). It can also be expressed in terms of Gamma functions as \(\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}\). The second option is another valid integral representation.
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 239-240)
Explanation: The Beta function \(B(p,q)\) can be seen as the convolution of \(f(t)=t^{p-1}\) and \(g(t)=t^{q-1}\) evaluated at \(t=1\). Specifically, \(B(p,q) = (t^{p-1} * t^{q-1})|_{t=1}\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 240)
Explanation: We know \(\mathcal{L}\{t\} = 1/s^2\). Using the first shifting theorem with \(a=-2\), we get \(\mathcal{L}\{t e^{-2t}\} = F(s-(-2)) = F(s+2) = \frac{1}{(s+2)^2}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 386, Table 21.7)
Explanation: We recognize the form \(F(s-a)\) with \(a=2\). The base transform is \(1/s^4\). We know \(\mathcal{L}^{-1}\{\frac{n!}{s^{n+1}}\} = t^n\). So, \(\mathcal{L}^{-1}\{\frac{1}{s^4}\} = \frac{1}{3!}\mathcal{L}^{-1}\{\frac{3!}{s^4}\} = \frac{1}{6}t^3\). Applying the shifting theorem, \(\mathcal{L}^{-1}\{\frac{1}{(s-2)^4}\} = e^{2t} \frac{t^3}{6}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 386, Table 21.7)
Explanation: This is the 'multiplication by t' property. Differentiating \(F(s) = \int_0^\infty e^{-st}f(t)dt\) with respect to \(s\) gives \(F'(s) = \int_0^\infty -t e^{-st}f(t)dt = -\mathcal{L}\{tf(t)\}\). Therefore, \(\mathcal{L}\{tf(t)\} = -F'(s)\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 228, Equation 7.9)
Explanation: Let \(f(t) = \sin(t)\), so \(F(s) = \mathcal{L}\{\sin(t)\} = \frac{1}{s^2+1}\). Then \(\mathcal{L}\{t\sin(t)\} = -F'(s) = -\frac{d}{ds}(\frac{1}{s^2+1}) = -(-1)(s^2+1)^{-2}(2s) = \frac{2s}{(s^2+1)^2}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 386)
Explanation: By definition, \(\mathcal{L}\{u(t-a)\} = \int_0^\infty u(t-a)e^{-st}dt = \int_a^\infty e^{-st}dt = [-\frac{1}{s}e^{-st}]_a^\infty = 0 - (-\frac{1}{s}e^{-as}) = \frac{e^{-as}}{s}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 329)
Explanation: The second shifting theorem is used for functions that are shifted in time. It states that \(\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s)\), where \(F(s)\) is the transform of the unshifted function \(f(t)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 386)
Explanation: This transform has the form \(e^{-as}F(s)\) with \(a=2\) and \(F(s) = 1/s^2\). The inverse transform of \(F(s)\) is \(f(t) = t\). By the second shifting theorem, \(\mathcal{L}^{-1}\{e^{-2s}F(s)\} = f(t-2)u(t-2) = (t-2)u(t-2)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 386)
Explanation: Using the integral definition, \(B(1,1) = \int_0^1 t^{1-1}(1-t)^{1-1} dt = \int_0^1 1 dt = [t]_0^1 = 1\). Alternatively, using the Gamma function relation, \(B(1,1) = \frac{\Gamma(1)\Gamma(1)}{\Gamma(2)} = \frac{1 \cdot 1}{1!} = 1\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 240)
Explanation: The integral is a convolution \(y(t) * \sin(t)\). Taking the Laplace transform of the entire equation gives \(Y(s) = \frac{1}{s^2} + Y(s)\frac{1}{s^2+1}\). Solving for \(Y(s)\): \(Y(s)(1 - \frac{1}{s^2+1}) = \frac{1}{s^2} \Rightarrow Y(s)(\frac{s^2}{s^2+1}) = \frac{1}{s^2} \Rightarrow Y(s) = \frac{s^2+1}{s^4} = \frac{1}{s^2} + \frac{1}{s^4}\). Taking the inverse transform is \(y(t) = t + \frac{t^3}{3!} = t + \frac{t^3}{6}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 402, Problem 8)
Explanation: The property \(B(p,q) = B(q,p)\) is called the symmetry property of the Beta function. It can be proven by a simple substitution \(u=1-t\) in the integral definition.
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 392)
Explanation: We use the property \(\mathcal{L}\{\frac{f(t)}{t}\} = \int_s^\infty F(u)du\). Here \(f(t) = \sin t\), so \(F(u) = \frac{1}{u^2+1}\). Then \(\mathcal{L}\{\frac{\sin t}{t}\} = \int_s^\infty \frac{1}{u^2+1}du = [\arctan(u)]_s^\infty = \frac{\pi}{2} - \arctan(s)\). This is also equal to \(\text{arccot}(s)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 402, Problem 12)
Explanation: Transforming the ODE: \([s^2Y-sy(0)-y'(0)] - 3[sY-y(0)] + 2Y = \frac{4}{s-3}\). Substituting initial conditions: \(s^2Y-s+1 - 3sY+3 + 2Y = \frac{4}{s-3}\). Grouping terms: \((s^2-3s+2)Y -s+4 = \frac{4}{s-3}\). Solving for Y(s): \(Y(s) = \frac{s-4}{s^2-3s+2} + \frac{4}{(s-3)(s^2-3s+2)}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 382, Example 21.4)
Explanation: By definition, \((t*t)(t) = \int_0^t u(t-u)du = \int_0^t (tu - u^2)du = [t\frac{u^2}{2} - \frac{u^3}{3}]_0^t = \frac{t^3}{2} - \frac{t^3}{3} = \frac{t^3}{6}\).
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 236)
Explanation: This can be solved using the convolution theorem or the multiplication by t property in reverse. Let \(F(s) = G(s) = \frac{s}{s^2+1}\). Then \(f(t)=g(t)=\cos(t)\). The inverse transform is \((f*g)(t) = \int_0^t \cos(u)\cos(t-u)du = \frac{1}{2}t\sin(t) + \frac{1}{2}\cos(t)\sin(t)\). A simpler method is to note that \(\mathcal{L}\{t\sin(t)\} = \frac{2s}{(s^2+1)^2}\). Therefore, \(\mathcal{L}^{-1}\{\frac{s}{(s^2+1)^2}\} = \frac{1}{2}t\sin(t)\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 394)
Explanation: The integral defining \(\Gamma(\alpha)\) diverges if \(\alpha \le 0\). The recurrence relation \(\Gamma(\alpha) = \frac{\Gamma(\alpha+1)}{\alpha}\) can be used to extend the definition to negative non-integers, but it reveals that \(\Gamma(\alpha)\) has simple poles at \(\alpha = 0, -1, -2, ...\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 389)
Explanation: This integral is in the form of a Laplace transform. It is \(\mathcal{L}\{t^3\}\) evaluated at \(s=2\). We know \(\mathcal{L}\{t^3\} = \frac{3!}{s^4} = \frac{6}{s^4}\). Substituting \(s=2\) gives \(\frac{6}{2^4} = \frac{6}{16} = \frac{3}{8}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 384)
Explanation: Let \(F(s) = \ln(s+a) - \ln(s+b)\). We use the property \(\mathcal{L}\{tf(t)\} = -F'(s)\). Here, \(-F'(s) = -(\frac{1}{s+a} - \frac{1}{s+b}) = \frac{1}{s+b} - \frac{1}{s+a}\). The inverse transform of this is \(e^{-bt} - e^{-at}\). So, \(tf(t) = e^{-bt} - e^{-at}\), which means \(f(t) = \frac{e^{-bt} - e^{-at}}{t}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 402, Problem 12)
Explanation: The Final Value Theorem allows finding the steady-state value of a function from its transform. It states \(\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s)\), provided the limit exists and the stability condition on the poles of \(sF(s)\) is met.
Source: Ostaszewski, A. (2012). Advanced mathematical methods. Cambridge University Press. (p. 389)
Explanation: The Initial Value Theorem allows finding the initial value of a function from its transform. It states \(\lim_{t \to 0^+} f(t) = \lim_{s \to \infty} sF(s)\).
Source: Ostaszewski, A. (2012). Advanced mathematical methods. Cambridge University Press. (p. 390)
Explanation: We use the property \(\mathcal{L}\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n}F(s)\). Here \(n=2\) and \(f(t)=\cos(t)\), so \(F(s) = \frac{s}{s^2+1}\). We need to compute \(\frac{d^2}{ds^2}(\frac{s}{s^2+1})\). The first derivative is \(\frac{(s^2+1)-s(2s)}{(s^2+1)^2} = \frac{1-s^2}{(s^2+1)^2}\). The second derivative is \(\frac{-2s(s^2+1)^2 - (1-s^2)2(s^2+1)(2s)}{(s^2+1)^4} = \frac{-2s(s^2+1) - 4s(1-s^2)}{(s^2+1)^3} = \frac{-2s^3-2s-4s+4s^3}{(s^2+1)^3} = \frac{2s^3-6s}{(s^2+1)^3} = \frac{2s(s^2-3)}{(s^2+1)^3}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 386)
Explanation: Using the formula \(B(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}\). For integers, \(\Gamma(n)=(n-1)!\). We have \(B(4,5) = \frac{\Gamma(4)\Gamma(5)}{\Gamma(9)} = \frac{3! \cdot 4!}{8!} = \frac{6 \cdot 24}{40320} = \frac{144}{40320} = \frac{1}{280}\).
Source: Wrede, R., & Spiegel, M. R. (2010). Schaum's outline of Advanced Calculus. McGraw-Hill. (p. 392)
Explanation: True. The Laplace transform satisfies the property of linearity, which means \(\mathcal{L}\{\alpha f(t) + \beta g(t)\} = \alpha \mathcal{L}\{f(t)\} + \beta \mathcal{L}\{g(t)\}\). This follows directly from the linearity of integration.
Source: Ostaszewski, A. (2012). Further calculus. University of London. (p. 223)