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I am not sure this answer your question, but it is an approximation. It is not a reference, only that many years ago I wrote for my use an exposition of Euler-MacLaurin's formula. Defining first the …

This is not properly an answer, after the comments of Tao it is difficult to give an answer. Only an explanation of my comment above. I still think that my series and the integral are equal. We can w …

I have proved this equality by means of Cauchy’s Theorem applied to an adequate function. Since my solution is too long to post it here, I posted it in arXiv: Juan Arias de Reyna, Computation of a De …

We prove that $$I=-\frac{\pi}{4}(\gamma+\log 4).$$ $$I=\int_0^\infty\frac{t\arg\Gamma(\frac14+\frac{it}{2})}{(\frac14+t^2)^2}\,dt.$$ $I$ is the imaginary part of the complex integral $$\int_0^\infty …

This is a generous explanation of Lucia's comment above. The functions $$\mathcal H_n(x)=\frac{2^{1/4}}{(2^n n!)^{1/2}}H_n(\sqrt{2\pi}\; x) e^{-\pi x^2}$$ form an orthonormal system in $L^2(\textbf{R …

I consider the function $$f(t):=\int_0^t\frac{dx}{x}\int_0^x\frac{dy}{y}\int_0^y\frac{dz}{z}\bigl\{ \sin x+\sin(x-y)-\sin(x-z)-\sin(x-y+z)\bigr\}.$$ It has an asymptotic expansion with main terms $$f( …

Given a non real value of $z$, the line of integration, i.e. the real axis can be tilted a little, depending on $z$ without changing the value of the integral. … For this you have to use different positions of the line of integration. Each position give you the function on a half plane. …