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$L\{f'(t)\}=\int_0^\infty e^{-st}f'(t)dt$ Integrating by parts we have, $L\{f'(t)\}=e^{-st}f(t)|_0^\infty+s\int_0^\infty e^{-st}f(t)dt$ $L\{f'(t)\}=e^{-s(\infty)}f(\infty)-e^{-s(0)}f(0)+sL\{f(t)\}$ If $e^{-st}$ grows more rapidly than $f(t)$, we have $e^{-st}f(t)\to0$ when $t\to\infty$ $L\{f'(t)\}=sL\{f(t)\}-f(0)$ Since $f(0)=0$, this reduces to ...

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In Laplace's method, we recognize that the main contribution to the integral is in the neighborhood of the minimum of the exponent, here at $t=0$. It appears that the slowly varying part of the integrand has a singularity there, but this singularity is integrable. Thus, we may approximate as follows, noting that $\sinh{t} \sim t$ and $\cosh{t} \sim ... 2 Plot integrand for few values of$x$: It is apparent that the maximum shifts closer to the origin as$x$grows. Let's rewrite the integrand as follows: $$\int_0^{\pi/2} \sqrt{\sin(t)} \exp\left(-x \sin^4(t)\right) \mathrm{d}t = \int_0^{\pi/2} \exp\left(\frac{1}{2} \log(\sin(t))-x \sin^4(t)\right) \mathrm{d}t$$ The maximum of the integrand is ... 1 You can verify your result from the following: For laplace transform: http://www.wolframalpha.com/input/?i=laplace+transform For laplace inverse: http://www.wolframalpha.com/input/?i=laplace+inverse$1)F^{-1}\left(\frac{1}{p-2}\right)= e^{2s}$is correct$2)F^{-1}\left(\frac{e^{-2p}}{p^2}\right)=\frac{2}{s^3(s+2)}$is incorrect Correction: ... 1 This is not as difficult as it might look at first. The calculation is long, but when you don't forget anything, it works fine. First, you need to find all the derivatives in Eq. (3). It's important to remember, that also$X$and$Y$are functions of$z$, so that the$z$derivative gives$i XYe^{iZ}A_G\partial_z Z + XYe^{iZ}\partial_z A_G + ...

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This is fairly straightforward to check for say a $3\times 3$ matrix, but the details become messy to do this in general. This should give you enough idea of how to do the general case to convince you it is true though: Suppose $N=3$. Suppose $v_i=\left(\begin{matrix} a_{i1} \\ a_{i2} \\ a_{i3}\end{matrix}\right)$. Then \$B_{11}=det(e_1, v_2, ...

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