# Tag Info

## New answers tagged integration

1

With the help from the comments this now makes sense. $$\int \delta(E^2-p^2-m^2)dE$$ With $$E_p^2-p^2-m^2=0$$ Use substitutions $$f(E)=E^2-p^2-m^2\quad df=2EdE$$ $$\int \delta(f)\frac{df}{2E(f)}=\frac{1}{2E_p}$$ $E(f)$ 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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