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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 …

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 …

indefinite integrals of the type $$\int x^pe^{qx}U(\alpha, \beta, x) d x$$ were considered in http://cdm16009.contentdm.oclc.org/cdm/ref/collection/p13011coll6/id/61450 (On some indefinite integrals o …

In the paper E. Mehlum, Appell and the apple (nonlinear splines in space), Technical Report No. 1676 (1981), Central institute for industrial research, Oslo (reproduced in the book Mathematical Method …

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 …

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 …

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}}-\ …

Physics oriented introduction is given in http://link.springer.com/book/10.1007/978-1-4757-5443-8 (Hypergeometric Functions and Their Applications, by James B. Seaborn). This review article http://io …

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 …

In http://iopscience.iop.org/0305-4470/38/8/005 (A differentiation formula for spherical Bessel functions) Boersma and Glasser proved the following interesting formula $$\left(1-\frac{\sqrt{z^2+a^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 …

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 …

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 …

The following inequality: $$\frac{\pi k}{\sinh{(\pi k)}}\;|J_{ik}(\tau)|^2\le 1,\;\;\;k,\tau\ge 0,$$ for Bessel function $J_{ik}(\tau)$, I found in http://link.springer.com/article/10.1134%2F1.558677 …

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}{ …