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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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4

Integral is a linear function ;)

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Hint: $$\int_X \alpha f + \beta g \, \mathrm{d} \mu = \alpha \int_X f \, \mathrm{d} \mu + \beta \int_X g \, \mathrm{d} \mu$$

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