# Tag Info

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$$L( \frac{1}{a}cos( \frac{bt}{a} ) ) = \frac{as^2}{a^2s^2 + b^2}$$ and hence,your reasoning must surely be wrong. Think about change of variable from $s$ to $s/a$ .

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$\hat{h}(s) = \frac{s}{a^2s^2+b^2} = {1 \over a^2} {s \over s^2 + ({b \over a})^2 }$. From this we get $h(t) = {1 \over a^2} \cos ({b \over a} t)$.

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The generating function of Catalan's numbers $y_n=\frac 1{n+1}\binom{2n}n$ is $$\sum_{n\geqslant 0 }\frac1 {n+1}\binom{2n}nx^n=\frac{1-\sqrt{1-4x}}{2x}$$ Multplying by $x$ and differentitating gives that $$\sum_{n\geqslant 0}\binom {2n}nx^n=\frac{d}{dx}\frac{1-\sqrt{1-4x}}{2}=\frac{1}{\sqrt{1-4x}}$$ One can also try to prove this directly, by noting that ...

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Here is one way: Let $f(t) = \int_0^t (H(\tau) - H(\tau-2)) d \tau$. Then ${\cal L} f(s) = {1 \over s} ({\cal L} H(s)) (1-e^{-2s}) = {1 \over s^2} (1-e^{-2s})$.

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Note that, $$f(t) = t[H(t) - H(t - 2)] + 2H(t - 2)$$ and

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$2 t e^{-t} \sin(2t)$ is correct. Why do you think that's not the answer? If you have been given a different answer, it may have been expressed in a different form (e.g. expanding $\sin(2t)$ as $2 \sin(t)\cos(t)$).

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Your approach is good. Using Laplace Transforms followed by Partial Fractions is probably the best way to solve this problem. (The next easiest way would be to evaluate $\int_{0}^{t}(t-\tau)^2e^{-2\tau}\,d\tau$). When doing the partial fractions, you should have gotten: $\dfrac{2}{s^3(s+2)} = \dfrac{\tfrac{1}{4}}{s} - \dfrac{\tfrac{1}{2}}{s^2} + ... 1 You already have the correct answer - since$\sin(-x)=-\sin(x)$, your answer and the "correct" answer are one and the same. 1 I am guessing that you would like to know$\lim_{s \to 0+} f(s) = \lim_{s \to 0+} \int_0^\infty a(t) e^{-st}dt$. Let$(s_n)$be any sequence of positive numbers converging to$0$and set$f_n(t) = a(t)e^{-s_nt} \to a(t)$as$n \to \infty$. Moreover,$|f_n(t)| \leq |a(t)|$so if$a(t)$is Lebesgue integrable, then the limit equals$\int_0^\infty a(t) dt... 1 An alternative to convolution would be the following approach. You need to know the following Laplace transform pairs: $$\mathcal{L}\{\sin(\omega T)\}=\frac{\omega}{s^2+\omega^2}\\ \mathcal{L}\{\cos(\omega T)\}=\frac{s}{s^2+\omega^2}\\ \mathcal{L}\{tf(t)\}=-F'(s)\\ \mathcal{L}\{f'(t)\}=sF(s)$$ From this it follows that $$\mathcal{L}\{t\sin(\omega ... 1 One possible hint may be to use the following idea (assuming you know the convolution of two functions):$$\mathcal{L}\left(\int_0^{\infty}f(t)g(x-t)dt\right)=F(s)G(s).$$where \mathcal{L}(f(x))=F(s) and \mathcal{L}(g(x))=G(s). Note that we have here$$\frac{a}{(a^2+s^2)^2}=\frac{1}a\cdot \frac{a}{a^2+s^2}\cdot\frac{a}{a^2+s^2}$$and of course you know ... 1 Differentiation is a good short-cut for this case:$$ \mathscr{L}\left\{\sin(wt)\right\} = \frac{w}{s^{2}+w^{2}}. $$Differentiate with respect to w:$$ \begin{align} \mathscr{L}\{t\cos(wt)\} = \frac{d}{dw}\frac{w}{s^{2}+w^{2}} &= -\frac{2w^{2}}{(s^{2}+w^{2})^{2}}+\frac{1}{s^{2}+w^{2}} \\ & = ... 0 Hint: Use\frac{\omega^2}{(s^2+\omega^2)^2}=\frac{1}{2}\big[\frac{s^2+\omega^2}{(s^2+\omega^2)^2}-\frac{s^2-\omega^2}{(s^2+\omega^2)^2}\big]=\frac{1}{2}\big[\frac{1}{s^2+\omega^2}+\frac{d}{ds}\big(\frac{s}{s^2+\omega^2}\big)\big]$Now use$\mathscr{L}^{-1}(\frac{\omega}{s^2+\omega^2})=\sin(\omega t)$and ... 0 Mathematica gives me$ \dfrac{-(t w \cos(w t)) + \sin(w t)}{2 w} $Hint:$w/(s^2+w^2)$is the transform of$sin(wt)$. CAn you now use the convolution theorem? 0 Following on from my comment: The left-hand side, namely$sX(s)-X(t=0)$, is the Laplace transform of the derivative$X'(t)$. You should be able to find it in most Laplace transform tables. I'll provide a link to the 'proof' and use of it (as there is too much to possibly write here). http://www.math.uah.edu/howell/DEtext/Part4/Laplace_derivatives.pdf 0 Look at it more carefully. That '$s$' is also present on the left hand side. This is the argument of the Laplace transform$F$of$f$, which itself is a function. Instead, you could also write anything else in place of$s$, e.g. $$F(x)=\int_0^\infty f(t)e^{-xt}dt\,.$$ 0 Give you more a counterexample:$\mathcal{L}_{x\to s}\biggl\{\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{4^n(n!)^2}\biggr\}=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2s^{2n+1}}$The former is suitable for$x\in\mathbb{C}$but the latter is only suitable for$|s|\geq1$. 0$\mathcal{L}^{-1}_{s\to t}\left\{e^{-\left(\frac{b}{b+s}\right)^k}\right\}=\mathcal{L}^{-1}_{s\to t}\left\{\sum\limits_{n=0}^\infty\dfrac{(-1)^nb^{kn}}{n!(b+s)^{kn}}\right\}=\mathcal{L}^{-1}_{s\to t}\left\{1+\sum\limits_{n=1}^\infty\dfrac{(-1)^nb^{kn}}{n!(b+s)^{kn}}\right\}$... 1$\mathcal{L}_{t\to\lambda}\left\{\int_0^tx(u)x(t-u)~du\right\}=\mathcal{L}_{t\to\lambda}\{\sin t\}(X(\lambda))^2=\dfrac{1}{\lambda^2+1}X(\lambda)=\pm\dfrac{1}{\sqrt{\lambda^2+1}}x(t)=\mathcal{L}^{-1}_{\lambda\to t}\biggl\{\pm\dfrac{1}{\sqrt{\lambda^2+1}}\biggr\}=\mathcal{L}^{-1}_{\lambda\to ...

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For the special case $k=1$ you should have the answer $${{\rm e}^{-bt}} \left( i\sqrt {{\frac {b}{t}}} {\rm I_{1} \left(\,2\,\sqrt {-bt}\right)}+\delta \left( t \right) \right) ,$$ where $I_n(x)$ is the modified Bessel function of the first kind and $\delta(x)$ is the dirac delta function.

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So you need to compute $$\frac1{2\pi i}\int_{-i\infty}^{i\infty}e^{xs-\left(\tfrac{b}{b+s}\right)^k}ds=e^{-bx} \frac1{2\pi i}\int_{-i\infty}^{i\infty}e^{x(b+s)-\left(\tfrac{b}{b+s}\right)^k}ds.$$ For $x$ positive and $k$ positive integer this is equal to $$e^{-bx}\operatorname{res}_{z=0}e^{xz-\left(\tfrac{b}{z}\right)^k}.$$ The residue is the ...

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By the uniqueness of the Laplace transform it suffices to show that: $$\mathcal{L}\left(\int_{0}^t x(u)x(t-u)du\right)=\frac{1}{\lambda^2+1} \tag{1}$$ To do so, insert the sum definition: $$x \left( t \right) =\sum _{n=0}^{\infty }{\frac { \left( -1 \right) ^{ n} \left(\dfrac{t}{2} \right) ^{2\,n}}{ \left( n! \right) ^{2}}} \tag{2}$$ into the integral and ...

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Here is an approach. The integral in consideration is $$\int_{0}^{\infty}t^{-3/2} e^{-\frac{a^2}{4t}}e^{-st}dt.$$ Taking the Mellin transform w.r.t. $a$ (see note 1), we need to change the order of integration, gives $$\int_{0}^{\infty}t^{-3/2}e^{-st} \int_{0}^{\infty} a^{w-1}e^{-\frac{a^2}{4t}} da\,dt = \frac{4^{-w/2}}{2}\,\Gamma\left( ... 0 simplification of the above st = u s dt = du dt = du/s 1/s integral(0- infinity ) e^-u (u/s)^-1/2 du s^1/2 / s^1 = 1/s^1/2 therefore the answer is root pi by s since the multiplying term is s^1/2 again. 0 Consider f(t) = \sum_{j=0}^\infty X_j(t) where X_j(t) = j! for j < t < j + 1/(j!)^2 and 0 otherwise. This is not exponentially bounded: j! \exp(-sj) \to \infty as j \to \infty for every real s. But for s > 0, {\mathscr L}(f)(s) \le \sum_{j=0}^\infty \exp(-sj)/j! = \exp(\exp(-s)) converges. 2 Hint: \mathscr{L^{-1}}\left(\frac{b}{(s-a)^2+b^2}\right) = e^{at}\sin{(bt)}; in your case, a=-1 and b=1. You can prove it by the integral definition of Laplace transformation. More info here. Another hint:, you can also find the inverse of \frac{s}{((s+1)^2+1)} by smart usage of the \mathscr{L^{-1}} of e^{at}\cos{(bt)} in the link above. (That ... 0 Here is one way to think of it: if we multiply throughout by \frac{2}{2} = 1 we get$$\frac{1}{2}\cdot\frac{2s^2 + 6s + 6}{2s^2 + 7s + 7}= \frac{1}{2}\cdot\frac{(2s^2 + 7s + 7) - s - 1}{2s^2 + 7s + 7}= \frac{1}{2}\cdot\left(1 + \frac{-s-1}{2s^2 + 7s + 7}\right)= \frac{1}{2}\cdot\left(1 - \frac{s + 1}{2s^2 + 7s + 7}\right)= \frac{1}{2} ...

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Consider the Laplace transform of $e^{x^{2}}$. \begin{align} e^{t^{2}} &\doteqdot \int_{0}^{\infty} e^{-st + t^{2}} dt \\ &\doteqdot \int_{0}^{\infty} e^{(t-s/2)^{2}- s^{2}/4} \ dt \\ &\doteqdot e^{- s^{2}/4} \ \int_{0}^{\infty} e^{(t-s/2)^{2}} \ dt \\ &\doteqdot e^{- s^{2}/4} \ \int_{-s/2}^{\infty} e^{u^{2}} du = e^{-s^{2}/4} \left[ ...

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$\textbf{Hint:}$ You can start by using $L\big(u(t-\pi)(2\cos t-3\sin 3t)\big)=e^{-\pi s}L\big(2\cos(t+\pi)-3\sin(3(t+\pi)\big)=e^{-\pi s}L\big(-2\cos t+3\sin3t\big)$

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By the formula $L(u(t-a)f(t))=e^{-as}Lf(t+a)$ we have \begin{aligned} L(U(t-2)(2t^2-6t+5)) &=e^{-2s}L((2(t+2)^2-6(t+2)+5)) \\ &=e^{-2s}(\frac{4}{s^3}+\frac{2}{s^3}+\frac{1}{s}) \\ \end{aligned}

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Since $F\left( {\alpha ,\beta ,\delta ;t} \right) = \sum {\frac{{\left( \alpha \right)_n \left( \beta \right)_n }}{{n!\left( \delta \right)_n }}t^n }$. Assuming the uniform convergence of the series, then term by term integration yields that \begin{align} \int_0^\infty {e^{ - st} } t^{\gamma - 1} F\left( {\alpha ,\beta ,\delta ;t} \right)dt \\ ...

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