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Many special functions appear as solutions of differential equations or integrals of elementary functions. Most special functions have relationships with representation theory of Lie groups.

4 votes
0 answers
99 views

Boersma and Glasser formula

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}}{ …
Zurab Silagadze's user avatar
5 votes
Accepted

Is $\frac{\pi}{4}L_0(z) = \sum\limits_{n=1}^{+\infty} (-1)^{n+1} \frac{I_{2n-1}(z)}{2n-1}$ b...

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 …
Zurab Silagadze's user avatar
1 vote
1 answer
318 views

Value of the hypergeometric function

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 …
Zurab Silagadze's user avatar
4 votes

A good reference to grok hypergeometric functions?

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 …
Zurab Silagadze's user avatar
3 votes

Choice of branch cuts in logarithmic integral

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 …
Zurab Silagadze's user avatar
2 votes
Accepted

Asymptotic behaviour of function from integral representation

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}{ …
Zurab Silagadze's user avatar
1 vote
0 answers
78 views

Finite sum of spherical Bessel functions

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{\ …
Zurab Silagadze's user avatar
3 votes

Asymptotic forms of Legendre functions for large degree

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 …
Zurab Silagadze's user avatar
5 votes
1 answer
932 views

Identity involving Fresnel integrals

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 …
Zurab Silagadze's user avatar
6 votes
1 answer
708 views

Аrе thеsе integrals known?

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 …
Zurab Silagadze's user avatar
6 votes
2 answers
308 views

Choice of branch cuts in logarithmic integral

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 …
Zurab Silagadze's user avatar
3 votes
1 answer
461 views

An inequality involving Bessel functions of imaginary order

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 …
Zurab Silagadze's user avatar
6 votes

Integrals involving the Tricomi hypergeometric function

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 …
Zurab Silagadze's user avatar
4 votes

Rogers-Ramanujan continued fraction $R(e^{-2 \pi \sqrt 5})$

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}}-\ …
Zurab Silagadze's user avatar
0 votes

Reference request for Stieltjes Transform

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 …
Zurab Silagadze's user avatar

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