Sure! However, as you probably know, an antiderivative to ##f(x) = x\sin(x)## is ##F(x) = \sin(x) - x\cos(x)##. Thus, the original improper integral is
\begin{align}
I &= \int_0^\infty x\sin(x)\,dx \\ &= \lim_{a\rightarrow\infty}\Big[\sin(x) - x\cos(x)\big].
\end{align}
This limit doesn't exist...
How sure are you that the equation reads $$\ddot{\xi} = -b\xi + \cos(\omega t)(a-c\xi^2)$$ and not instead $$\dot{\xi} = -b\xi + \cos(\omega t)(a-c\xi^2).$$ I only ask because the latter is an Ricatti equation and thus exactly solvable.
If you only have calculated the pertubation series upto...
There are an infinite number of trivial solutions to this ODE.
Supose ##f:\Omega\rightarrow\mathbb{C}## is a holomorphic function, then ##\big(y(x), z(x)\big) = \big(f(x),\pm f(x)\big)## are trivial solutions to your original ODE $$(y^\prime)^2 - (z^\prime)^2 + 2m^2(y^2 - z^2) = 0.$$
So to...
In the case of a simple harmonic oscillator (SHO); your intuition is right, it is more likely to encounter it near its endpoints of oscillation.
If we choose our coordinate frame such that the origin is at the equilibrium point for the SHO and starts the oscilation at the displacement ##A##...
No, this is generally not true for indistinguishable particles as the following minimal example will demonstrate.
Consider a system consisting of two particles, each of which can be in one out of two possible one-particle-states with the energies ##E_1 = 0## and ##E_2 = \epsilon## respectively...
In general, $$Z=z^N$$ is the partition function for a system consiting of ##N## distinguishable and noninteracting particles with single particle partition function ##z## (prove this from first principle, good exercise).
Thus, in your case where the single particle partition function is given...