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It is exactly because we have a factor of $\frac 1 2$ in the area formula of a triangle. To understand what I'm saying, consider what is the $v(t)$ graph of a particle under constant acceleration. Some say, a good plot is worth a million words! :)

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The result you've got would be better known as this: $$\int_0^t\biggl(\int_0^{t'} a\mathrm{d}t''\biggr)\mathrm{d}t' = \frac{1}{2}at^2$$ In other words, it's a derivation of the formula for uniformly accelerated motion. This derivation, or something like it, is one of the first things students in a good calculus-based introductory physics class learn. The ...

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It is, in fact, a double integral! The first notation used $$\varPhi_E = \oint_S \vec{E} \cdot \mathrm{d}\vec{A} = \oint_S \vec{E} \cdot \hat{n} \ \mathrm{d}A$$ is simply a more compact notation. It's much easier to write $\mathrm{d} \vec{A}$ instead of, say, $r \ \mathrm{d}r \ \mathrm{d}\theta$ all the time. Furthermore, it's more general, as $\mathrm{d} ... 15 Take a look at the notes on lectures 1 and 2 of Geometric Numerical Integration found here. Quoting from Lecture 2 A numerical one-step method$y_{n+1} = \Phi_h(y_n)$is called symplectic if, when applied to a Hamiltonian system, the discrete flow$y \mapsto \Phi_h(y)$is a symplectic transformation for all sufficiently small step sizes. From your ... 14 IMHO, the notation$\int_a^b\mathrm{d}x\,f(x)$is much cleaner than$\int_a^b f(x)\,\mathrm{d}x$, because the integration variable ($x$) and its associated integral range$(\int_a^b$) are kept together. This is particularly important in lengthy and multi-dimensional integrals. Consider $$\Upsilon_{pq}(k)= \int_0^\infty\mathrm{d}x ... 13 No, it cannot be enough. Stokes' theorem says that the volume (\Omega) integral of d\omega, a form that is the exterior derivative of another one (of \omega), may be written as a surface integral. But it doesn't allow us to rewrite the volume integral of a general integrand (which isn't the exterior derivative of anything) such as the Lagrangian ... 12 It's not just QFT literature. Physicists, especially adult research physicists, find this notation sensible and popular – even though it may be more popular among particle physicists than elsewhere. Formally, dx\,f(x) is a product of two factors and \int is a form of a sum. Because product is commutative, it doesn't hurt when the order is interchanged. ... 12 In German, this property is known as the Transformationssatz, but I do not know any appropriate translation for it. This is, however, a special case of coordinate tranformations changing the measure by the determinant of their Jacobian, since obviously \frac{\partial y_i}{\partial x_j} = A_{ij}. That it is the determinant that plays a role in the ... 11 If the functional derivative$$\tag{1} \frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)} $$exists (wrt. to a certain choice of boundary conditions), it obeys infinitesimally$$\tag{2}\delta F ~:=~ F[\phi+\delta\phi]- F[\phi] ~=~\int_M \!dx\sum_{\alpha\in J} \frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)}\delta\phi^{\alpha}(x). $$OP's functional integral ... 10 Here we will assume that OP is not questioning the fundamental physical principles/postulates/axioms of quantum mechanics, such as, e.g., the need to have a Hilbert space H in the first place, etc; and that OP is only pondering the role of L^2-spaces (as opposed to, e.g., L^1-spaces). Let us for concreteness and simplicity consider the 3-dimensional ... 10 The \delta function is not continuous, so it's a priori not differentiable. In fact, it's not even well-defined as an ordinary real-valued function, but can be made so in terms of distributions - linear maps on a space of test functions given by f\mapsto\int\delta f=f(a). It's possible to sensibly define derivatives of distributions by looking at ... 10 An important example in quantum mechanics is e.g. the Hilbert space$$H~=~L^2(\mathbb{R}^3)$$of Lebesgue square integrable wave functions \psi in the position space \mathbb{R}^3. The Lebesgue square integrable functions (as opposed to just the Riemann square integrable functions) are needed to complete the Hilbert space with respect to the square ... 9 The short answer is that the two principal value definitions agree on sufficiently well-behaved functions, but may disagree on sufficiently singular functions. For instance, on one hand$$\lim_{\epsilon\searrow 0} \int_{\mathbb{R}\backslash[-\epsilon,\epsilon]} \frac{\mathrm{d}x}{x^3}~=~0$$is zero, while on the other hand$$\lim_{\epsilon\searrow 0} ... 9 A simple reference problem Suppose we want to analyse the problem of a forced harmonic oscillator. Denote as$\phi(t)$the time dependent position of the oscillator. The oscillator experiences two forces, the spring force$-k\phi(t)$and an external force$F_{\text{ext}}(t). Newton's law says \begin{align} F(t) &= m a(t) \\ -k \phi(t) + ... 9 When the velocity is not constant you have:x(t)=\overline{v(t)} t + x_i$$where \overline{v(t)} is the average velocity from 0 to t. When you have constant acceleration the average velocity is$$\overline{v(t)}=\frac{v(0)+v(t)}{2}=\frac{at}{2} + v_i$$which will give the correct result. If the acceleration is non constant you will have to do the ... 8 Besides the reasons listed in Lubos Motl's answer, here is another reason for the \int \!dx ~f(x) notation: By writing the integral sign \int_a^b and dx next to each other in multiple nested integrations, it becomes more easy to trace which limits belong to which integration. This becomes particularly handy when changing the orders of integration. ... 8 In order to express the position as a function of the velocity you have to integrate with respect to time. When the velocity is constant this integral is simple, namely vt+C. However once the velocity becomes a function of time this integral will change and will in general not be equal to v(t)t+C. You actually have to integrate v(t) with respect to t ... 8 There are two that I know of in the context of state estimation. The first is for estimating the mean of P and is a Metropolis-Hasting MCMC algorithm here: Optimal, reliable estimation of quantum states. The second is also mainly for computing the mean (but can do other functions -- including the characteristic function of the region you are interested ... 8 a_x \Delta t = \Delta v_x = v_{xf} - v_{xi} \Delta x = v_{x,average}\Delta t = v_{xi}\Delta t + \dfrac{1}{2}a_x (\Delta t)^2 \Rightarrow v_{x,average} = v_{xi} + \dfrac{1}{2}a_x \Delta t = v_{xi} + \dfrac{1}{2}(v_{xf} - v_{xi}) = \dfrac{v_{xf}+ v_{xi}}{2} Is there a geometric interpretation or does it just work out mathematically? 7 Technically, the equation$$d = \frac{\mathrm{d}x}{\mathrm{d}t}t + \frac{\mathrm{d}^2x}{\mathrm{d}t^2}\frac{t^2}{2}$$is not right. Instead, for constant acceleration, you need$$d = \left(\left.\frac{\mathrm{d}x}{\mathrm{d}t}\right|_0\right) t + \left(\left.\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\right|_0\right) \frac{t^2}{2}$$In other words, a quantity ... 7 It's an integral over a closed line (e.g. a circle), see line integral. In particular, it is used in complex analysis for contour integrals (i.e closed lines on a complex plane), see e.g. example pointed out by Lubos. Also, it is used in real space, e.g. in electromagnetism, in Faraday's law of induction (part of the Maxwell equations, written in an ... 6 1) As OP basically notes, an n-dimensional delta function transforms under change of variables f:\mathbb{R}^n \to \mathbb{R}^n with (the absolute value of) an inverse Jacobian$$ \tag{1} \delta^n(f(x))~=~ \sum_{x_{(0)},f(x_{(0)})=0 }\frac{1}{|\det(\partial f(x_{(0)}))|} \delta^n(x-x_{(0)}), $$where the sum \sum is over all zeroes x_{(0)} of f, ... 6 It depends what you want to calculate. As you rightly note, delta functions are not dimensionless, so that including one in your integral will change its dimensionality: you will be calculating something rather different! Most of the time this won't matter if you do it right, but you do need to think about what you want to calculate. The integral \int ... 6 In theory: Lebesgue integrable functions form a Banach space, whereas Riemann integrable functions do not. This causes problems in, e.g., Quantum mechanics if we try to work with Riemann integrable functions instead of Lebesgue integrable functions... (We want the functions to form a Banach space so we can use good old fashioned linear algebra to solve ... 6 WARNING: The function is not absolutely integrable for n>1, so the integral strongly depends on how you decide to compute it if you break the integration into iterated integrals. Use instead cylindric coordinates. k = (z, \vec{r}), where \vec{r} \in \mathbb R^{n-1} and z\in \mathbb R. You have this way, assuming that x is directed along z: ... 6 That's equivalent simply to c\int dx/x. Switch to the Euclidean spacetime, k_0=ik_4 where (k_1,\dots k_4) is k_E; i.e. analytically continue in k_0 (Wick rotation). The integral is$$\int \frac{i\cdot d^4 k_E}{(2\pi)^4} \frac{1}{(k_E^2)^2} \exp(ik\cdot \epsilon)$So it's proportional to the Fourier transform of$1/k_E^4$. The original function is ... 6 This is not an equality, strictly speaking. Looks like your lecturer used spherical coordinates. If the integrand is spherically symmetric, i.e. it only depends on the magnitude of$\mathbf{p}$, then the integration over the angular coordinates is trivial and just gives you the solid angle subtended by a sphere,$4\pi$. 6 Integrate the jerk 3 times then using starting conditions to work out the integration constants. 6 One may only talk about a discrete sum over$k^\mu$vectors if all the spacetime directions are compact. In that case,$k^\mu$is quantized. If the spacetime is a periodic box with periodicities$L_x,L_y,L_z,L_t$, then$V=L_x L_y L_z$and$T=L_t$. The component$k^\mu$in such a spacetime is a multiple of$2\pi \hbar / L_\mu$(I added$\hbar$to allow any ... 6 There's really two questions here, one about the definition of distribution functions and one about the derivation of the BBGKY hierarchy. I will address them in turn. Definitions Let's define, for convenience,$\mathbf{r}^n = \mathbf{r}_1,\dots,\mathbf{r}_n$and$\mathbf{r}^{(N-n)} = \mathbf{r}_{n+1},\dots,\mathbf{r}_N\$. Next, let's denote the ...

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