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the small-$\rho$ asymptotics of $$I(\rho)=-\int_{1.1}^{\infty}\frac{\sin(k\rho)}{\rho k^{1.9}\ln k}\,dk$$ is governed by the large-$k$ behavior of the integrand, which gives $I(\rho)\propto \rho^{1.9- …

since the integral for large $b$ is dominated by $x$ near zero, where it oscillates least rapidly, we can omit the fraction and approximate the integral by the sine integral, $$I(a,b)=\int_{-\infty}^ …

The first term in Watson's expansion amounts to the saddlepoint approximation (expansion of the exponent to second order around $t=1$), which gives $$I_{sp}=\frac{1}{2}e^{-x}\int_{-\infty}^\infty e^{- …

If we disregard the positivity constraint, this is not true in general, the leading order term can be of order $n-1$ rather than of order 1. The problem is treated in Laurent expansion of the inverse …

left(a,b;c,d;z\right)=\frac{\Gamma (c) \Gamma (d)}{\Gamma (a) \Gamma (b)}e^z z^{a+b-c-d}\left(1+{\cal O}(z^{-1})\right)$$ As an example, the plot shows $_{2}F_{2}\left(a,b;c,d;z\right)$ (blue) and the asymptotics …

You seek a solution $\rho$ of the equation $f'(\rho)=0$, hence $$\rho^2=1+w^{-2}\rho^{2p}.$$ The solution should remain $>0$ when $w\rightarrow\infty$. The OP says the solution should vanish as $1/w$, …

The saddle point $t^\ast$ is obtained by solving $f'(t)=0$ for $f(t)=-t+n\ln t-(n+1)\ln(t+i)$, for large $n$ we find $f(t^\ast)=-2\sqrt{in}$, so we arrive at the approximation for the integral $I_n\ap …

The strategy is as follows: put everything in the integrand in the exponent $e^{f(z)}$, with $f(z)=\zeta z-z^p/p$; calculate the saddle point, the (possibly complex) number $z_0$ where $f'(z)=0$; thi …

$$P(s)=2\int_{-\infty}^{\infty}{\rm d}p\int_{-\infty}^{\infty}{\rm d}q\ \delta\left(\frac{p^2+(p^2-q^2)^2}{p^2+4p^4}-s\right)e^{-\left(p^2+q^2\right)/a^2}$$ $$\qquad=2\int_{0}^{\infty}\frac{{\rm d}x}{ …

The integral $$I_n=\int_0^\infty e^{f(n,t)}\,dt,\;\;f(n,t)=n\ln t-n\ln(1+t)-t,$$ has a saddle point at $t^*$ where $\partial f(n,t)/\partial t=0$, $$t^\ast=-\tfrac{1}{2}+\tfrac{1}{2}\sqrt{1+4n}.$$ For …

Use the identity $$_3F_2\left(\frac{1}{2},x,x;x+\frac{1}{2},x+\frac{1}{2} \bigg|1\right)=\frac{\sqrt{\pi }\, \Gamma \left(x+\frac{1}{2}\right)}{\Gamma (x)}\,_3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1} …

Yes, there is a central limit theorem that guarantees convergence in probability to the Wigner semicircle law, see for example Central limit theorem for linear eigenvalue statistics of random matrices …

This approximate solution can be obtained by transforming the infinite sum to a contour integral and then making an asymptotic expansion. This is worked out in Application of Sommerfeld-Watson Transfo …

It may be helpful to consider an example that can be solved exactly; Using the Special-case closed form of the Baker-Campbell-Hausdorff formula one finds that if the commutator $[X,Y]$ evaluates to $$ …

We could just take the large-$n$ asymptotic of $J_n(z)\rightarrow (2\pi n)^{-1/2}(ez/2n)^n$, and then $$f_n(\rho)\rightarrow \frac{1}{n}(\tfrac{1}{2}e\rho/n)^n.$$ This seems to be quite reasonable i …