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In addition to the Carlo Beenakker's answer. The following asymptotic expansion was proved in https://www.sciencedirect.com/science/article/pii/0041555365901345?via%3Dihub (Asymptotic formulae for leg …

This relation is a special case of a more general one: $$L_\nu(z)=\frac{4}{\sqrt{\pi}\,\Gamma\left(\nu+\frac{1}{2}\right)}\sum_{n=0}^\infty\frac{(-1)^n\,(2n+\nu+1)\,\Gamma(n+\nu+1)}{n!\,(2n+1)(2n+2\nu …

Let $n$, $m$ and $k$ be some (positive) integers such that $(k+3/2)-(n+m/2)<0$. Can the hypergeometric function $$F\left (n+\frac{m}{2},n+\frac{m+1}{2};k+\frac{3}{2};-\tan^2{\phi}\right) \tag{1}$$ be …

In L.G. Afanasyev, A.V. Tarasov, Breakup of relativistic pi+pi- atoms in matter, Phys. At. Nucl. 59 (1996) 2130 the following identity is given for the spherical Bessel functions $j_n(z)=\sqrt{\frac{\ …

In the answer to Clausen’s identity for associated Legendre polynomials the following result was indicated: $$ \small{\left(P_n^m(\cos\theta)\right)^2=(\sin{\theta})^{2m}\frac{(m+n)!}{(n-m)!}\sum_{k=0 …

While calculating an integral in a quantum mechanical problem by two different methods, I came across the following identity $$\sum_{k=0}^n\sum_{m=0}^{2k}(-2)^m\binom{2(n-k)}{n-k}\binom{2k}{k}\binom{2 …

If we expand $\cos{(2y\sqrt{t})}$ into Taylor series and integrate term by term, we get $$\phi_1(y,\lambda)=\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n)!}\frac{\Gamma\left(n+\frac{1}{2}-\frac{\lambda}{ …

In a quantum mechanical problem I encountered the integral $$I_k=\int_0^\infty x^{2(l+1)-k}j_k(\sigma x)e^{-x}[L_{n-l-1}^{2l+1}(x)]^2 dx,$$ where $j_k(x)$ is a spherical Bessel function, and $\sigma$ …

Clausen’s identity for Legendre polynomials has the form (see, for example, A generating function of the squares of Legendre polynomials, by Wadim Zudilin: https://arxiv.org/abs/1210.2493) $$P_n(\cos{ …

Maybe the following reference will be useful https://arxiv.org/abs/1105.0060 (Signal Processing in Large Systems: a New Paradigm, by R. Couillet, M. Debbah). See also chapter 3 in the book "Random Mat …

According to 8.111 from Lewin's book "Polylogarithms and associated functions", it is expected that $$\int\limits_0^2\frac{\ln{(1-x)}\ln{(1+x)}}{x}\,dx=Li_3(-3)+\zeta(3)-2Li_3(3)+$$ $$\ln{3}\left[Li_2 …

It seems I figured it out. 8.111 in Lewin's book has the form $$\int\limits_0^x\frac{\ln{(1-y)}\ln{(1-cy)}}{y}\,dy=\mathrm{Li}_3\left(\frac{1-xc}{1-x}\right)+\mathrm{Li}_3\left(\frac{1}{c}\right)+\mat …

While studying some dark matter related stuff, I came across to the following interesting identities: $$\int\limits_0^\infty\sqrt{\frac{y}{xp}}\,e^{-y}\left(K(p)-E(p)\right)dy= \frac{\pi x}{4} \left[I …

The result $$\small R(e^{-2\pi\sqrt{5}})=\frac{\sqrt{5}}{1+\left[5^{3/4}\left(\frac{\sqrt{5}-1}{2}\right)^{5/2}-1\right]^{1/5}}-\frac{\sqrt{5}+1}{2}=\frac{\beta+2}{\beta+\sqrt[5]{\sqrt{1+\beta^{10}}-\ …

The following results are quoted in http://www.hindawi.com/journals/ijmms/2007/019381/abs/ (Integer Powers of Arcsin, by J.M. Borwein and M. Chamberland): $$\large I(4,1)=-\frac{3}{2}\mathrm{Li}_5(g^2 …