Search Results

A calculation of the dark matter density profile in a dissipative dark matter model leads to the integral $$f(x,\theta)=\int\limits_0^\infty\frac{y\,e^{-y}\,dy}{\sqrt{x^4+y^4+2x^2y^2\cos{2\theta}}}.$$ …

Are there any hyperelliptic generalizations of the following formula, first proved by Euler in 1782, $$\int\limits_0^1\frac{dx}{\sqrt{1-x^4}}\int\limits_0^1\frac{x^2\,dx}{\sqrt{1-x^4}}=\frac{\pi}{4}\, …

This is just a further detalization of the Carlo Beenakker's answer. Let us write $$K_{ik}=\int\left (\frac{\partial^2}{\partial r_i \partial r_k}\frac{1}{r}\right )\mu(\vec{y})d\vec{y},$$ where $r_i= …

The integral $$\int\limits_0^1 x^k P_m(x)P_n(x)dx$$ is evaluated in terms of the hypergeometric function $_3F_2$ in http://link.springer.com/article/10.1007/BF01650571 (Some integrals containing produ …

.$$ The integration domain in the last integral is determined by the conditions $0\le x,y\le 1$, which gives $$-\frac{\phi}{2}\le\theta\le\frac{\phi}{2},\;\;\;\phi-\frac{\pi}{2}\le\theta\le\frac{\pi}{ … \cos{\theta}},\;\;\;y=\frac{\sin{\theta}}{\cos{\phi}}.$$ Its Jacobian is $1-x^2y^2$ and it is applied to the integral $$\zeta(2)=\frac{4}{3}\int_0^1\int_0^1\frac{dx\,dy}{1-x^2y^2}.$$ In this case the integration …

I found the answer in http://retro.seals.ch/digbib/view?pid=elemat-001:2000:55::180 (A Property of Euler's Elastic Curve, by V.H. Moll, P.A. Neill, J.L. Nowalsky and L. Solanilla) where a Euler-type …

The fact that the integral is proportional to the difference of the arithmetic and geometric means can be established in the following way, without calculating any integral. Let consider $\alpha=\frac …

There is a numerical evidence that the following is approximatelly true $$\int\limits_0^1\frac{x^2(\pi-x)}{\pi\sin{x}}dx\approx\sin{\left(\frac{13\pi}{46}\right)}-\sin{\left(\frac{6\pi}{53}\right)},$$ …

See paragraph 3.8 in https://people.sc.fsu.edu/~%20jburkardt/presentations/truncated_normal.pdf and https://people.smp.uq.edu.au/YoniNazarathy/teaching_projects/studentWork/EricOrjebin_TruncatedNormal …

How this alleged multiple integral identity can be proved? $$\int\limits_{-\infty }^{\infty }ds_{1}\int\limits_{-\infty}^{s_{1}}ds_{2}\cdots \int\limits_{-\infty }^{s_{2n-1}}ds_{2n}\;\cos { (s_{1}^{2} …

Some time ago I stumbled on an alleged identity $$\int\limits_0^\infty \frac{dx}{x} \int\limits_0^x \frac{dy}{y} \int\limits_0^y \frac{dz}{z} [\sin{x}+\sin{(x-y)}-\sin{(x-z)}-\sin{(x-y+z)}]= -\frac{\p …

They write that they "have not attempted to perform the integration explicitly". Was this integral ever calculated explicitly? …

Numerical evidence shows the validity of the following identity $$\int\limits_0^z\frac{xdx}{\sin{x}\sqrt{\sin^2{z}-\sin^2{x}}}=\frac{\pi}{4\sin{z}}\ln{\frac{1+\sin{z}}{1-\sin{z}}},\tag{1}$$ if $0< z< …

Wolfram alpha calculates the integral $$\int\limits_0^\infty \frac{x^2\ln{x}}{e^x-1}dx=2\zeta^\prime(3)+3\zeta(3)-2\gamma\zeta(3).$$ However, I need to cite the source of this identity (the table of i …

The integral should be calculable by the Mellin-transform technique. See the calculation of the similar (but different) integral in http://www.opticsinfobase.org/josaa/abstract.cfm?uri=josaa-7-7-1218 …