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1

With the help from the comments this now makes sense. \begin{equation} \int \delta(E^2-p^2-m^2)dE \end{equation} With $$E_p^2-p^2-m^2=0$$ Use substitutions \begin{equation} f(E)=E^2-p^2-m^2\quad df=2EdE \end{equation} \begin{equation} \int \delta(f)\frac{df}{2E(f)}=\frac{1}{2E_p} \end{equation} $E(f)$ is easily found by inverting $f$ Thanks!


1

Clearly $a$ has the same dimension of $x$ (see the argument of root or of $\sin^{-1}$) so the left member is dimensionless (ratio between dimension of x: remember that differential dx count in dimensional calculus!), and the second member too has to be dimensionless: so n=0.


1

Integration is finding the area under a curve that isn't necessarily straight. If you have a velocity time graph and find the area under it, this gives you the distance travailed. If you have a acceleration-time graph the area under it is the change in velocity. There are several techniques to integration, which I will not go into here. As mentioned in the ...


2

1) On integrating dt on the RHS we get a +c(constant of integration) but why is there no +c on the LHS while integrating dv? If we start with: $$ dv = adt $$ and integrate both sides then we can indeed have a constant of integration on both sides: $$ v + C_1 = at + C_2 $$ but we can just subtract $C_1$ from both sides to get: $$ v = at + (C_2 - C_1) = ...


0

In the first step ( $\int dv=a\int dt$), we get $v+c_1=a(t+c_2)$. and therefore $v=at+(ac_2-c_1)$. The term $ac_2-c_1$ is constant (because $a$ ,here, is constant). Instead of tediously writing the integration constants for every step, we can write one at the ending, since the last constant ($c$, in this case) absorbs all the other constants which have ...


2

Use fig 13.2 of [2] as reference. Taking the example Qmechanic uses, the idea is that $I(\omega) = \int_{\zeta_a}^{\zeta_b}\frac{Z(\omega)}{\zeta - \xi(\omega)} d\zeta$ $I(\omega)$ needs to be analytically continued from $\omega_1 \rightarrow \omega_2$. The pole of the integrand travels from $\zeta = \xi(\omega_1) \rightarrow \zeta = \xi(\omega_2)$ in ...


1

I) Let there be given a meromorphic function $\zeta \mapsto F_{w}(\zeta)$ in the $\zeta$-plane with a single (not necessarily simple) pole at the position $\zeta=\xi(w)$, where $\xi$ is a holomorphic function, and $w\in \mathbb{C} $ is an external parameter. Ref. 1 is considering the contour integral $$\tag{A}I_{\Gamma,w} ~=~ \int_{\Gamma} \! d\zeta ...


0

In general, you will run into some problems with "functions" like this one. Consider, for example, U(0), which is $$ U(0)=\frac{1}{(2\pi)^3}\int d^3 p e^{i\vec p \cdot \vec {\Delta x}} $$ $$ =\delta(\vec {\Delta x}) $$ I.e., a three dimension "delta function". This is not really a "function" in the conventional sense, but rather a "functional". So, okay, ...


2

Knowing only "jerk" (third derivative of position), you cannot determine the distance traveled. To get distance traveled (or equivalently, position as a function of time) from jerk, you need to integrate three times. Each integration produces a constant of integration representing an initial value; your final equation looks something like this: $$p(t) = ...


6

Integrate the jerk 3 times then using starting conditions to work out the integration constants.


1

Your integration by parts is incorrect. You are integrating over space, so you can only move the spacial derivative rather than time derivative onto the $\psi^*$ factor. Or put another way, what you are calling the boundary term and throwing away is actually the expression you started with. You are replacing ...



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