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You probably also know how what will happen when you differentiate a Laplace transform. Hint: what is the derivative of $$\frac{1}{s^2+5}\:?$$

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The inverse Laplace transform of $\frac1{s^2+1}$ is $\sin(t)$. Therefore we have $$\mathcal L^{-1}\left\{\frac1{(s+1)^2+1}\right\}=\mathrm e^{-t}\sin(t)=\frac1{2\mathrm i}\left(\mathrm e^{(-1+\mathrm i)t}-\mathrm e^{(-1-\mathrm i)t}\right)$$ Therefore $$\begin{split}\mathcal L^{-1}\left\{\frac1s\frac1{(s+1)^2+1}\right\}&=\int_0^t\mathrm ... 3 For the first question, take something like$$ f(t) = \sin(e^{t^2}). $$Then f is bounded (so in particular of exponential growth), but$$ f'(t) = 2t e^{t^2} \cos( e^{t^2} ) $$is not of exponential growth. (Look at the values for z=2\pi k, k \in \mathbb{Z}.) For the second question, see Jyrki's answer. 3 Answering the second question with a "standard" example of a function that is not of exponential order, but does have a Laplace transform in the region \Re s>0. Build a function out of spiky triangles$$ \Delta_{H,A}(x)= \begin{cases} H-\frac{H^2}{A}|x|,&\ \text{if $|x|\le A/H$, and}\\ 0,&\ \text{otherwise.} \end{cases} $$Here H>0,A>0 ... 2 It would be the situation in B: you would deform around the pole. It works as follows. The inverse Laplace transform is given by Cauchy's theorem. I present the parametrization of each piece of the contour, assuming that the radius of the semicircular detour about the pole z=-1 and the branch point z=0 is \epsilon:$$\int_{c-i \infty}^{c+i \infty} ...

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EDIT: This proves the wrong thing; ignore this answer. If $f'$ is of exponential order, so is $f$. Consider the integral $$\int_a^x f'(t) dt$$ From the fundamental theorem of calculus, this differs from $f$ by at most a constant. If $|f(x)| ≤ |g(x)|$ for sufficiently large $x$, $\int_a^x |f(t)| dt ≤ \int_a^x |g(t)| dt$ for sufficiently large $x$ and $a$. $$... 0 I have a copy of G.E.Roberts and H.Kaufman "Table of Laplace Transforms", 1966. On page 252, Item 3.2.55 is a more complex example but might apply if v = 0. Inverse of (s + (s^2 - a^2)^{1/2})^v \exp(-b\sqrt{s^2-a^2})/\sqrt{s^2 - a^2} is given as: 0 for 0 < t < b and a^v ((t-b)/(t+b))^{v/2} I_v(a\sqrt{t^2-b^2}) for t > b ... 0 HINTS:$$\mathscr{L}\left(\sin (wt)u(t)\right)(p)=\frac{w^2}{p^2+w^2}\mathscr{L}(f(t-0.5)u(t-0.5))(p)=e^{-0.5p}\mathscr{L}(f(t)u(t))(p)$$0 The reason for the shifting is so that the initial values will shift to y(0) and y'(0). Since the Laplace transform of the y' and y'' are dependent on y(0) and y'(0), this makes solving the problem easier. The shift gives the equivalent IVP$$ \overline{y}'' + 2\overline{y}' + \overline{y} = 50\overline{t}, \quad \overline{y}(0) = -4, \quad ...

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Define $B_n(y) := E[f(S_n)](n/y)$, $Y_n := |f(S_n) - f(y)|$ and $Z_n := |S_n - y|$ Since f is bounded and continuous on $[0,\infty)$, it is bounded and continuous on $[0,M] \ \forall M > 0$ and hence uniformly continuous in $[0,M]$. $\to \forall \epsilon > 0, \exists \delta > 0 \forall x, y \in [0,M]$ s.t. $$Z_n \le \delta \to Y_n < ... 2 What you did is OK, you get$$ \mathcal{L}(\cos^2(\omega t))=\frac{1}{2}\left(\frac{1}{s} + \mathcal{L}(\cos(2\omega t))\right)=\frac{1}{2 s}+\frac{s}{2 \left(s^2+4 w^2\right)} $$where we have used$$\mathcal{L}(\cos\omega t) = \frac{s}{s^2 + \omega^2}.$$1 By the definition of the \Gamma function,$$\begin{eqnarray*}\mathcal{L}\left(t^{n-\frac{1}{2}}\right)&=&\int_{0}^{+\infty} t^{n-\frac{1}{2}}e^{-st}\,dt = ...

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The Laplace Transform of $t^{n-1/2}$ is given by $$\mathscr{L}\left(t^{n-1/2}\right)(s)=\int_0^\infty t^{n-1/2}\,e^{-st}\,dt$$ Now, enforce the substitution $st\to t$ reveals \begin{align} \mathscr{L}\left(t^{n-1/2}\right)(s)&=\left(\frac{1}{s}\right)^{n+1/2}\int_0^\infty t^{n-1/2}\,e^{-t}\,dt\\\\ ... 1 Using \mathcal{L}_{t}\left[e^{at}\sin(\omega t)\right]_{(s)}=\frac{\omega}{(s-a)^2+\omega^2}:\mathcal{L}_{s}^{-1}\left[\frac{1}{s^2-6s+10}\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[\frac{1}{(s-3)^2+1}\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[\frac{1}{(s-3)^2+1^2}\right]_{(t)}=e^{3t}\sin(t)$$Using \mathcal{L}_{t}\left[e^{at}\cos(\omega ... 2 Hint. You may observe that$$\mathcal{L}(e^{-at}\cos\omega t) = \frac{s + a}{(s+a)^2 + \omega^2},\mathcal{L}(e^{-at}\sin\omega t) = \frac{\omega}{(s+a)^2 + \omega^2}.$$Then apply it to$$ F(s)=\frac{s+\color{red}{2}}{(s+\color{red}{2})^{2}+1^2}+6\:\times\frac{\color{blue}{1}}{(s+2)^{2}+\color{blue}{1}^2} . $$1 Divide it as s+2 and 6. Hope that helps. 0 I'm assuming: f is not a Borel set but a Borel-measurable function \mathbb R \to \mathbb R. \mathcal{E} = \mathscr{M}(E) \ or \ \mathscr{B}(E)$$\mathbb{E}[\prod_{i=1}^n e^{-f(X_i)}]= \prod_{i=1}^n \mathbb{E}[e^{-f(X_i)}]$$Now$$\mathbb{E}[e^{-f(X_i)}] = \int_{\Omega} e^{-f(X_i)} d\mathbb P= \int_{E} e^{-f(x_i)} d ...

3

Hint: Notice that \begin{align} \frac{s^3}{(s^2+4)^2}&=\frac{s(s^2+4)-4s}{(s^2+4)^2}=\frac{s}{s^2+4}-\frac{4s}{(s^2+4)^2} \end{align} Also notice $\mathscr{L}\left\{\sin (2t)\right\}=\frac{2}{s^2+4}\,\,$ and $\,\,\frac{\mathrm d}{\mathrm ds}\left(\frac{2}{s^2+4}\right)=-\frac{4s}{s^2+4}$. Then use the fact that ...

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I just joined so I can't add a comment to Ron's answer but wanted to clarify that information on y for any t can be used to find C, not just y(0).

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Using the Laplace transform is useless and ill-advised here, it creates a lot of confusion. Basically the first line is a second order differential equation with constant coefficients: $$K\frac{\mathrm d^2T_0}{\mathrm dx^2}-MT_0=-MT_a-Q_m.\tag{1}$$ The solutions of the homogeneous equation ($K\frac{\mathrm d^2T_0}{\mathrm dx^2}-MT_0=0$) are of the form ...

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I think yes if $$\nabla[k \nabla u(\underline x,t)] = \nabla \cdot [k \nabla u(\underline x,t)]:$$ By one of the product rules (), $$\nabla[k \nabla u(\underline x,t)]$$ $$= \nabla k \nabla u(\underline x,t) + k \nabla \nabla u(\underline x,t)]$$ $$= (0) \nabla u(\underline x,t) + k \nabla \nabla u(\underline x,t)]$$ $$= (0) \nabla u(\underline x,t) + ... 0 I also worked on this problem several days before. My point of view: (1) Yes we are allowed to. Because \partial_{\lambda}^{n-1}(e^{-\lambda x}f(x)) exists and it is continuous. (2) By weak law$$\frac{S_n}{n}\to \mathbb{E}(X_1)=\frac{1}{\lambda}~~\text{in probability.}$$In other words,$$S_n \to \frac{n}{\lambda}=:y~~\text{in probability.}$$... 3 Thank you for the interesting question. Here is a rather brute force solution, which may add a few steps to Jan Erland's solution. First, let us recall that, if$$ \mathcal L(f) = \int_0^\infty f(t) \, e^{-st} \, dt = F(s) $$Then$$ \mathcal L(f') = \int_0^\infty f'(t) \, e^{-st} \, dt = s \, F(s) - f(0). $$In our case, F(s) = e^{-\sqrt{s+2}}/s, so, ... 1 Mathematica 10.0 gives this output:$$\mathcal{L}_{s}^{-1}\left[\frac{e^{-\sqrt{s+2}}}{s}\right]_{(t)}=\frac{1}{2}e^{-\sqrt{2}}\left(\text{erfc}\left(\frac{1-2t\sqrt{2}}{2\sqrt{t}}\right)+e^{e\sqrt{2}}\space\text{erfc}\left(\frac{1+2t\sqrt{2}}{2\sqrt{t}}\right)\right)0 First, \begin{align} \mathscr{L}\{y'\} & = \int_{0}^{\infty}e^{-st}y'(t)dt \\ & = e^{-st}y(t)|_{t=0}^{\infty}+s\int_{0}^{\infty}e^{-st}y(t)dt \\ & = -y(0)+s\mathscr{L}\{y\} = s\mathscr{L}\{y\} \end{align} Then, using the above, the transform of the second derivative is \begin{align} \mathscr{L}\{y''\} & ... 1 Here I will assume that \sigma, S, B are all constants as you've not stated otherwise. This is a fairly simple ODE to solve as it's first order, and it's linear. Because it's linear it has a superposition principle which we can use to get the solution in the form of $$u(\tau) = u_c(\tau) + u_p(\tau).$$ Where u_c(\tau), ... 1 Starting from Mario G's suggestion in the comments...\frac{x+\frac{3}{2}}{(1-x)(2+x+x^2)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+2}$$Multiply through by (x-1)(2+x+x^2),$$x+\frac{3}{2}=A(x^2+x+2)+(Bx+C)(x-1)$$Collect powers of x on the RHS.$$x+\frac{3}{2}=(A+B)x^2+(A-B+C)x+(2A-C)$$Now due to the linear independence of the powers of x, we can ... 2 \begin{equation*} f(x)=\frac{x+\frac{3}{2}}{(x-1)(2+x+x^{2})}=\frac{A}{x-1}+\frac{Bx+C}{% x^{2}+x+2} \end{equation*} cover-up method leads to \begin{equation*} A=\left. f(x)\left( x-1\right) \right\vert _{x=1}=\left. \frac{x+\frac{3}{2}% }{(2+x+x^{2})}\right\vert _{x=1}=\frac{\frac{5}{2}}{4}=\frac{5}{8}. \end{equation*} generalized cover-up method leads to ... 1$$L^{-1} \left\{ \frac{2s^2 - 3s - 1}{s(s + 1)(s - 1)} \right\}= L^{-1}\left\{\frac{1}{s}\right\}+L^{-1}\left\{\frac{2}{s+1}\right\}-L^{-1}\left\{\frac{1}{s-1}\right\} =1+2e^{-t}-e^t$$2 Observe that$$\frac{e^{-st}}{t} = \int_s^{\infty} e^{-xt} dx.$$Then$$\int_0^{\infty} e^{-st} \frac{F(t)}{t} dt = \int_0^{\infty} \int_s^{\infty} e^{-xt} dx F(t) dt = \int_s^{\infty} \int_0^{\infty} e^{-xt} F(t) dt dx = \int_s^{\infty}f(x)dx, where in the second equality I have assumed that F is nice enough so that you can exchange the order of ... 2 Using standard Laplace properties from tables, \begin{align*}% %TCIMACRO{\tciLaplace}% %BeginExpansion \mathcal{L}% %EndExpansion \left( ty\left( t\right) \right) & =-Y^{\prime}\left( s\right) \\% %TCIMACRO{\tciLaplace}% %BeginExpansion \mathcal{L}% %EndExpansion \left( y^{\prime}\left( t\right) \right) & =sY\left( s\right) \\% ... 2F(s)=\frac{s^2}{(s^2+4)^2}=s\frac{d}{ds}((\frac{1}{s^2+4})\times\frac1{2})f(t)=\frac{d}{dt}(t\times(\frac{\sin(2t)}{2}\times\frac1{2}))=\frac1{4}(\sin(2t)+2t\cos(2t))$$2$$\frac{(s+1)^3}{s^4} = \frac 1s + \frac 3{s^2} + \frac 3{s^3} + \frac 1{s^4} and the inverse Laplace transform of each of those terms should be standard to you. After you've found it, it may be possible to simplify the answer! (If the inverse transform of these terms are not in your head, go back to your notes, text or this nice MIT lecture on the topic: ...

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I think I got it. hint: if you derivate the left part, the derivative is alot simpler. Then you can probably reintegrate, and fit the integration constant, thus getting a simpler equivalent equation to solve.

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I consulted my professor yesterday and he managed to provide me with a rather elementary but very clever solution. Consider the sequence of functions $\left\{f_{n}\right\}_{n\geq 1}$ defined by $f_{n}(t)=\frac{1}{\sqrt{t}}\chi_{[\frac{1}{n},n]}(t)$ and notice that $f_{n}\in \mathscr{L}^{2}(\mathbb{R}^{+})$ for each fixed $n\geq 1$ with ...

2

The norm is exactly $\sqrt\pi$. You can find a prof in this paper.

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Hint Solving with out using Laplace transformation (if you are allowed): we have $x''(t)=-y'(t)=-x(t)$ and so your equation is $x''(t)+x(t)=0$ It is a lnear homogenous differential equation second order with constant coeffitients. You can use this site to find the solution.

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That definition divides the plane into three parts: $y < y_0$ (below green line in image below) $y_0 \le y < y_1$ $y \ge y_1$ (above pink line in image below) For each part slopes $a$, $b$ and $c$ are required. Here is a sketch of the situation for $a=1/2, b=2, c=1$: For this situation it is possible to pick a continous $y(t)$, if one point ...

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