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The integral $$I(k) = \int_{-\infty}^\infty \frac{s e^{isr}}{(s-k)(s+k)} ds \tag{1}$$ where $k$ is real and the integration is for real $s$, is not really well-defined. This is precisely because the integrand has singularities on the integration domain. However consider if $k$ is a complex number $k = k_r + ik_i$ with $k_i >0$. Then the integrand is ...


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Integration operates on functions, correct? No. Integration dates back to ca. 1670. The notion of a function gradually evolved and didn't get put into its modern form until ca. 1830. Let's look at your expression $$ \int_{p_1}^{p_2} dp .$$ This is Leibniz's notation, and what he means by it is a sum of infinitely many terms $dp$. Each of these terms ...


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What does it mean to integrate $\frac{d\mathbf p}{dt}dt$? First, and in scalar form, recall from elementary calculus that $$\int_{x_1}^{x_2} dx = x_2 - x_1 $$ Second, recall that $$f(x + dx) = f(x) + f'(x)dx$$ where $$f'(x) = \frac{df(x)}{dx} $$ Denoting the differential of $f$ as $$df = f(x + dx) - f(x)$$ we have $$df = f'(x)dx$$ Since ...


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While in some contexts, $$ \frac{d\textbf{p}}{dt}dt = d\textbf{p} $$ is correct and is mathematically rigorous, there is a straightforward way to derive impulse as the change in momentum. Consider a function $f(x)$ where $f:\mathbb{R}^m \to \mathbb{R}^n$ and $f \in C^1$. Suppose its derivative is $g = f'(x)$. We can consider integrating $g(x)$ over some ...


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Lubos Motl has already provided a correct answer. This answer uses a different approach in the spirit of perturbation theory with sources: $$\int \! d^nx ~f(x)~e^{-\frac{1}{2}x^TAx +j^Tx} ~=~f\left(\frac{\partial}{\partial j}\right) \int \! d^nx ~e^{-\frac{1}{2}x^TAx +j^Tx} ~~\stackrel{\begin{matrix}\text{Gauss.}\\ \text{int.}\end{matrix}}{=}~~C~ ...


4

The case with the linear term is obtained from the original one by a simple shift, i.e. the substitution $$ x = X + A^{-1} B $$ Substitute it to the exponent in your more general integral: $$ -\frac 12 x^T A x + B^T x = -\frac 12 (X^T+B^T A^{-1}) A(X+A^{-1}B)+B^T (X+A^{-1}B)=\dots $$ I used $A=A^T$. Now, all the terms that are schematically $BX$ i.e. linear ...


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Hints: Recall that the Gamma function $\Gamma(z)$ has poles at non-positive integers $z\in -\mathbb{N}_0$. Use Euler's reflection formula to replace the factor $$ \Gamma(\omega+1)^2\Gamma(-\omega) ~=~ \frac{\pi^2}{\sin^2(\pi\omega)} \frac{1}{\Gamma(-\omega)} $$ in eq. (3.4). Now the only simple poles in the negative $\omega$ halfplane of the rewritten ...


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how should I notate my bounds of integration? I guess you mean for this integral, passing to gauge pressure \begin{align} \int_{P_s(o,t)}^{P_s(L,t)}(P_s(x,t)+b) \ dP_s(x,t). \end{align} Since $P_s=P_g+P_a$, all you need to do is that change of variables. Besides, $dP_s=dP_g$ because $P_a$ is a constant. Also, there is no need to say $dP_s(x,t)$, since ...


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Using accelerometers as a basis for determining the position of an Earth-based object is a non-trivial exercise. Using a simple numerical quadrature technique such as trapezium almost certainly is not going to cut it. Here are some of the challenges: Accelerometers report acceleration in terms of the accelerometer case frame. It is invalid to compare, for ...


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I think you have made some mistake in taking the initial condition. $$v(t)=u(0)+\int_{0}^{t}a(t)dt$$ and $$s(t)=s(0)+\int_{0}^{t}v(t)dt$$. I think you have not used $u(0)$ and $s(0)$ properly.


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You need to use recursive integration by parts to deal with the second derivatives. See, for example, equation (18) here. The last term in (18) has a multiplying factor of $(-1)^n$, where $n$ is the order of the derivative. In your case, $n=2$ and the minus sign vanishes.


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You're halfway there. The quickest way that I can see is via explicit Cartesian components and coordinates. You start by assuming that $\mathbf B$ is uniform, so that you can pull it out of the integral (which you've already used implicitly for dealing with $Y$. That means that you're interested in \begin{align} \frac{\boldsymbol \tau}{I} =\oint_L ...



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