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Well he is wrong, the fact that he couldn't say why is a reason to mistrust his statement. You could come up with that result by using timescaling rule: $\mathcal{L} f(a\cdot) = F(s/a)/a$ which with $a=-1$ would lead to the result that your teacher claims, but this rule assumes that $a>0$ so it can't be used. The reason the rule doesn't hold is that you ...

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I realize that this is an older question, but I'll answer it anyway in case other people are looking for the answer. As an electrical engineer, a Laplace transform usually references the two-sided Laplace transform. Therefore the function being transformed does not have to be causal, meaning that the function can equal something other than $0$ for $t < ... 1 $$\mathcal{L}\left(\frac{1}{\sqrt{1+t^2}}\right) = \frac{\pi}{2}\left(H_0(s)-Y_0(s)\right)\tag{1}$$ where$H_0$is a Struve function and$Y_0$is a Bessel function of the second kind. 0 Are you familiar with the Hamilton-Jacobi equations? In optimal control case they are $$\dot{x} = \frac{\partial H}{\partial \lambda}, \dot{\lambda} = -\frac{\partial H}{\partial x}, \frac{\partial H}{\partial u} = 0$$ So, what is done is finding$x, \lambda, u$functions that satisfy the above equations. That is where$\dot{\lambda} = a \lambda$comes ... 0 You want$−0.0744 v^22/T^2$to be$-s t$. Let us get rid of the square in$v^2$. We can take$u=v^2$, so$du=2v\,dv$. We substitute, and now we have $$\int_0^\infty\,e^{−0.0744 u/T^2}\,\left(\frac1{2\sqrt u}\,p(\sqrt u)\right)\,du.$$ This is the Laplace Transform of the function in brackets at the point$s=0.0744/T^2$. So your equation is now $$... 0 This is an application of the Laplace Convolution Formula. Let h(t) = f(x,t). Let L[h] = \int_0^\infty h(t)e^{-st} dt, the Laplace transform of h. Then the Laplace Convolution Formula states that: L[h]L[g] = L[h * g], where h * g = \int_o^t h(x)g(t - x) dx. Hence L[h * g] = \int_0^\infty e^{-st}\int_o^t h(x)g(t - x) dx dt This is exactly ... 0 The Laplace transform of a function has the following bound: take s=\sigma+it, \sigma, t real$$ \lvert \mathcal{L}(f)(\sigma+it) \rvert = \left\lvert \int_0^{\infty} e^{-\sigma x-itx} f(x) \, dx \right\rvert \\ \leqslant \int_0^{\infty} \lvert e^{-\sigma x}e^{-itx} \rvert \lvert f(x) \rvert \, dx \\ = \int_0^{\infty} e^{-\sigma x} \lvert f(x) \rvert ... 1 The Laplace Transform of a function$f$is $$F(s)=\int_0^\infty\,f(t)e^{-st}\,dt.$$ The imaginary part of$s$bears no influence in whether the integral converges. And one can show that if the integral does not converge for a certain$s$, the it doesn't converge for all$s$with smaller real part. In other words, the ROC is always of the form ... 0 Hint. If$s>|a|$then$\exp((a-s)b)$tends to$0$, not$\infty$, if$b\to\infty$. 0 Hint: You can proceed by direct convolution, using $$\mathcal{L}(e^{ax})=\frac1{x-a},$$and $$e^{ax}*e^{bx}=\int_{t=0}^x e^{a(x-t)}e^{bt}dt=\left.e^{ax}\frac{e^{(b-a)t}}{b-a}\right|_{t=0}^x=\frac{e^{bx}-e^{ax}}{b-a}.$$ 1 You right, first thing here is a convolution $$x(t) = \mathcal{L}^{-1} \left( \frac{1}{(s+\mu_1 + \mu_2) (s + \hat{\lambda}_2) (s + \lambda_1 +\lambda_2 )}\right)= \mathcal{L}^{-1} \left( \frac{1}{s+\mu_1 + \mu_2 }\right)*\mathcal{L}^{-1} \left( \frac{1}{ s + \hat{\lambda}_2 }\right) *\mathcal{L}^{-1} \left( \frac{1}{s + \lambda_1 +\lambda_2 }\right) = ... 1 By the residue theorem:$$ \frac{1}{(x-a)(x-b)(x-c)}=\frac{1}{(a-b)(a-c)(x-a)}+\frac{1}{(b-a)(b-c)(x-b)}+\frac{1}{(c-a)(c-b)(x-c)}$$and \mathcal{L}^{-1}\left(\frac{1}{x-c}\right)=e^{cs}. 1 Hint :$$\lim_{t\to 0} f(t) = \lim_{s\to \infty}sF(s)$$This is the Initial value theorem. Edit: Expanding on this, since the limit we're seeking for is known to exist from the hypothesis, you may proceed as such :$$\lim_{t\to 0} \frac{f(t)}{g(t)} = \frac{\lim_{t\to 0} f(t)}{\lim_{t\to 0}g(t)}= \frac{\lim_{s\to \infty}sF(s)}{\lim_{s\to ... 0 I had evaluated the convolution integral incorrectly. The answer is $$q(t) = \frac{1}{\omega_n} \int_{0}^{t} \cos(\omega(t-\tau)) \sin(\omega_n \tau) \,d \tau = \frac{\cos(\omega t) - \cos(\omega_n t)}{\omega_n^2 - \omega^2}$$ The limit as$\omega \rightarrow \omega_n$can be obtained L'Hopital's rule, $$\lim_{\omega \rightarrow \omega_n} q(t) = ... 0 Hint: Use ~\displaystyle\int_0^\infty\exp\Big(-\sqrt[n]x\Big)~dx~=~n!~\iff~\int_0^\infty e^{-x^n}~dx~=~\Gamma\bigg(1+\frac1n\bigg)~ in conjunction with Euler's formula. 1 If p>1,$$ I(p)=\int_{0}^{+\infty}\sin(x^p)\,dx = \frac{1}{p}\int_{0}^{+\infty}x^{\frac{1}{p}-1}\sin(x)\,dx \tag{1}$$but since \mathcal{L}(\sin(x))=\frac{1}{s^2+1} and \mathcal{L}^{-1}\left(x^{1/p-1}\right)=\frac{s^{-1/p}}{\Gamma\left(1-\frac{1}{p}\right)} we have:$$ ... 2 The integral evaluates to $$\int_0^{\infty}\sin x^a\ dx=\Gamma\left(1+\frac{1}{a}\right)\sin\frac{\pi}{2a},$$ but the way I know uses complex analysis. Added by request: We will integrate the function$\exp(-x^a)$around the circular wedge of radius$R$and opening angle$\pi/(2a)$, for$a>1$. By the Residue Theorem, $$0=\int_0^R\exp(-x^a)\ ... 2 If A,B>0$$\int_{0}^{+\infty}\exp\left(-A^2 x^2-\frac{B^2}{x^2}\right)\,dx =\sqrt{\frac{B}{A}}\int_{0}^{+\infty}\exp\left(-ABx^2-\frac{AB}{x^2}\right)\,dx\tag{1}$$hence it is enough to compute:$$ I(C)=\int_{0}^{+\infty}\exp\left(-C^2 x^2-\frac{C^2}{x^2}\right)\,dx. \tag{2}$$By splitting the integration range as (0,1)\cup(1,+\infty) and setting ... 2 A very general converse of Watson's lemma is due to Feller [1]. Let \mu be a measure concentrated on [0,\infty) such that$$ \int_0^\infty e^{-s t}\,d\mu(t) $$exists for all s > 0. Let L be a real-valued function defined for large x which satisfies$$ \lim_{x \to \infty} \frac{L(ax)}{L(x)} \to 1 $$for all fixed a > 0 and let ... 3 We have:$$ I(s)=\int_{0}^{+\infty}\exp\left(-sx-\frac{1}{x}\right)\,dx = \frac{2}{\sqrt{s}}\,K_1(2\sqrt{s})\tag{1}$$and your integral can be computed by differentiating both sides of (1) n times with respect to s. Exploiting the Bessel differential equation:$$ \int_{0}^{+\infty} x^n \exp\left(-sx-\frac{1}{x}\right)\,dx = 2 ... 1 I think you played a little fast and loose with the contributions over$E$and$G$. In considering the contour integral over$E$, I let$z=e^{i \pi} x$and, over$G$, I let$z=e^{-i \pi} x$. The integrals I get, after taking appropriate limits, is $$e^{i \pi} \int_1^{-1} dx \frac{e^{i b \sqrt{1-x^2}}}{-(1-x^2)} e^{-x t} + e^{-i \pi} \int_{-1}^1 dx ... 3 No slowly varying function L can be integrable on [0,\infty). It follows from the Potter bounds (e.g., see Theorem 1.5.6 of the book Regular Variation by Bingham, Goldie, and Teugel), that for any \delta>0, one has$$ L(x) \ge C_\delta x^{-\delta} $$for x sufficiently large, where C_\delta>0. This bound is also not difficult to prove ... 0 Unfortunately I can't think of an example with the extra requirements, but there certainly exists such L that are integrable. Take L(x) =\log(x)^b, where b is a real number. and let's show using L'Hopital that this function indeed is slowly varying.$$ \lim_{x \to \infty} \frac{\log(tx)^b}{\log(x)^b} = ... -1 Hint: the partial fraction decomposition could begin with something like: $$\frac{s}{(s+1)^2(s+2)} = \frac{2s+1}{(s+1)^2} + \frac{-2}{s+2}$$ Now you continue from here. 3 Hint: recall that: $$\mathcal{L}^{-1}\left(\frac{1}{s+a}\right)=e^{-ax},\qquad \mathcal{L}^{-1}\left(\frac{1}{(s+a)^2}\right) = xe^{-ax}$$ and apply a partial fraction decomposition. 0 You posted three changes to my equations. I do agree with two of changes. Here is what I now think: \begin{eqnarray*} s^2W(s) - s + Y(s) + Z(s) &=& -\frac{1}{s} \,\,\,\, eq1 \\ W(s) + s^2Y(s) - Z(s) &=& 0 \,\,\,\,\,\,\, eq2 \\ -sW(s) - sY(s) + s^2Z(s) + s - 1 &=& 0 \,\,\,\,\, eq3 \end{eqnarray*} It is not$ -s - 1$because$z(0) = ...

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A possible approach may be the following one. Despite its appearance, the RHS is an entire function: $$\begin{eqnarray*}6\left(\frac{\cosh(t)}{t^2}-\frac{\sinh(t)}{t^3}\right)&=&6\left(\sum_{n\geq 0}\frac{t^{2n-2}}{(2n)!}-\sum_{n\geq 0}\frac{t^{2n-2}}{(2n+1)!}\right)\\&=&6\sum_{n\geq 0}\frac{2n t^{2n-2}}{(2n+1)!}\\&=&12\sum_{n\geq ... 0 Your inverse laplace transform is wrong. Notice that the numerator should also have the s-1 term to write the solution you got. The correct solution would be: Y=  2s\over{(s-1)^2+4} =2(s-1)\over{(s-1)^2+4}+2\over{(s-1)^2+4} Now Taking the inverse laplace transform you get the required result 0 You've computed the inverse Laplace transform incorrectly:$$ \mathcal{L}^{-1}\left[\frac{2s}{(s-1)^2 + 4}\right](t) = e^t\mathcal{L}^{-1}\left[\frac{2(u+1)}{u^2 + 4}\right](t) = e^t\left( 2\mathcal{L}^{-1}\left[\frac{u}{u^2 + 4}\right](t) + 2\mathcal{L}^{-1}\left[\frac{1}{u^2 + 4}\right](t) \right) = e^t \left( \mathcal{L}^{-1}\left[\frac{u/2}{(u/2)^2 + ...

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Check your Eqs.1-3. Do you agree with this ? :

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I'm understanding that $f(s)=\frac{1}{As^2+w^2}$, with $A,w$ real constants. Well, this would be my roadmap: Write $f(s)$ as the sum of two fractions Use the linearity of the Laplace transform Compute the inverse transform of each simple fraction (should be a familiar type) Put together the pieces

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We have for the Laplace transform of $f(t-a)u(t-a)$ \begin{align} \mathscr{L}(f(t-a)u(t-a))(s)&=\int_0^{\infty}f(t-a)u(t-a)e^{-st}\,dt\\\\ &=\int_{-a}^{\infty}f(t)u(t)e^{-s(t+a)}\,dt\\\\ &=e^{-sa}\int_0^{\infty}f(t)u(t)e^{-st}\,dt\\\\ &=e^{-sa}\mathscr{L}(f(t)u(t))(s) \tag 1 \end{align} For the problem of interest, we have $a=1$ in ...

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This is an inverse Laplace transform. We assume $t \gt 0$; the ILT is zero when $t \lt 0$. To evaluate, we consider the contour integral $$\oint_C \frac{dz}{z} \frac{e^{-\sqrt{a z}}}{c+\sqrt{z}} \cos{\sqrt{b z}} e^{t z}$$ where $C$ is a Bromwich contour which is detoured above and below the negative real axis and around the origin, as we define a ...

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The inverse Laplace transform of $s$ is $\delta'$ (the derivative of Dirac's delta "function"). To understand what that means, you need to know a little about distribution theory. (There is no way around this: if in your rational function $\frac{P}{Q}$, you have that $\deg P \ge \deg Q$, the inverse transform will contain $\delta$:s and their derivatives.)

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We have $$\int_0^{\infty} e^{-st} \frac{\cos{\sqrt{t}}}{\sqrt{t}} \, dt.$$ Clearly we need $s>0$ for this to converge. Set $u^2=st$, and then $2/\sqrt{s} du = dt/\sqrt{t}$, so the integral becomes $$\frac{2}{\sqrt{s}}\int_0^{\infty} e^{-u^2} \cos{\left( \frac{u}{s} \right)} \, du.$$ To do this integral, define $$I(a) = 2\int_0^{\infty} e^{-u^2} ... 1 I give you some hints along the way. First let u=\sqrt{t}. This transforms your integral to$$ \frac{1}{2}\int_0^{+\infty}e^{-su^2}\cos u\,du=\int_{-\infty}^{+\infty}e^{-su^2}\cos u\,du, $$where we in the second step used that the function is even. Now,$$ \cos u=\frac{1}{2}\bigl(e^{iu}+e^{-iu}\bigr). $$We get$$ \frac{1}{2}\int_{-\infty}^{+\infty} ...

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Using this, $$L^{-1}F'(s)=-t\cdot f(t)$$ Let $$\dfrac{d\left(\dfrac{as+b}{s^2+2s+20}\right)}{ds}=\dfrac{3s+8}{(s^2+2s+20)^2}$$ Find $a,b$ Now use this, $$A\cdot L^{-1}\dfrac{s-a}{(s-a)^2+b^2}+B\cdot L^{-1}\dfrac b{(s-a)^2+b^2}=A\cdot e^{at}\cos bt+B\cdot e^{at}\sin bt$$

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Utilizing the property \begin{align} \mathcal{L}\{f(t)\} = \frac{\int_{0}^{T} e^{- s u} \, f(u) \, du}{1 - e^{-s T}} \end{align} where $f(t+T) = f(t)$ then \begin{align} \mathcal{L}^{-1}\{\frac{1}{1- e^{-s}}\} = \frac{\int_{0}^{1} e^{-su} \, \delta(u) \, du}{1- e^{-s}} \end{align} for which $f(t) = \delta(t)$ with the periodic property $f(t+1) = f(t)$. ...

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To answer your second question, which largely seems to be about how to evaluate an integral of this type. By the integral's domain, $s \in \{ \alpha + x'i : x' \in [-x, x]\}$ Set $s' = (s-\alpha)i = -x', \textrm{d}s' = i\textrm{d}s = -\textrm{d}x'$ Transforming the integral, you end up with \underset{{x\rightarrow \infty}}{\lim} ...

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Yes, $\alpha$ is a real number, such that $\alpha + it$ is in the half-plane of convergence (strip if you are working with the two-sided Laplace transform). The second question seems meaningless. What do you mean by $\infty i$?

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Maybe it helps to consider the complex Laplace transform. Thus $$\hat{f}(z)=\int_{0}^{\infty }dt\exp [izt]f(t),\;{Im}z>0$$ This object will exist for $f(t)$ absolutely integrable or square integrable. Now put $z=\omega +i\delta$. Then ($\theta (t)$ is the Heaviside step function) $$\hat{f}(\omega +i\delta )=\int_{0}^{\infty }dt\exp [i(\omega +i\delta ... 1 The only mistake I can see is that the factor of 4 in u_{xx} =4u_t has become a factor of \frac 14. So the line$$\mathcal{L}(u'')=\mathcal{L}(\dot u) \rightarrow U''(x,s)=\frac{s}{4}U(x,s)-\frac{1}{4}u(x,0)$$should instead be$$\mathcal{L}(u'')=\mathcal{L}(\dot u) \rightarrow U''(x,s)=4sU(x,s)-4u(x,0). Following through with this correction ...

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