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As for point (1), maybe the following references will be useful: The Euler–Maclaurin formula revisited, by D. Elliott The Euler–Maclaurin expansion and finite-part integrals, by G. Monegato, J.N. Ly …

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 …

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 …

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< …

.$$ The integration domain in the last integral is determined by the conditions $0\le x,y\le 1$, which gives $$-\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 …

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 $\sigma$ …

I think the following articles can give a clue: http://www.tandfonline.com/doi/abs/10.1080/10652460701688125?journalCode=gitr20 (Closed-form evaluation of some families of cotangent and cosecant integ …

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 …

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

\int\limits_0^1\cdots \int\limits_0^1\frac{\delta\left(1-\sum\limits_{i=1}^n x_i\right)dx_1\cdots dx_n}{(x_1A_1+x_2A_2+\cdots +x_nA_n)^n},$$ we get after the parametrization and subsequent trivial integration … \ln{(1+x+y)}- \ln{(1+x+z)}-\ln{(1+y+z)}\right]\theta(z-x)\theta(y-z).$$ $\theta(z-x)\theta(y-z)$ term is a consequence of integration with the help of the $\delta$-function and reflects $x_2=z-x>0$ and …

This is only a partial answer. The Beukers-Kolk-Calabi change of variables $$x_1=\frac{\sin{u_1}}{\cos{u_2}},\;\;x_2=\frac{\sin{u_2}}{\cos{u_3}},\ldots, \;x_{n-1}=\frac{\sin{u_{n-1}}}{\cos{u_n}},\;\;x …

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 …

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