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arXiv · 2407.05719

A naive integral

Abstract

In arXiv:2406.0243 two real functions $g(x,t)$ and $f(x,t)$ are defined, so that the Riemann-Siegel $Z$ function is given as \[Z(t)=\mathop{\mathrm{Re}}\Bigl\{\frac{u(t)e^{\frac{πi}{8}}}{\frac12+it}\int_0^\infty g(x,t)e^{i f(x,t)}\,dt\Bigr\},\] where $u(t)$ is a real function of order $t^{-1/4}$ when $t\to+\infty$. The function $g(x,t)$ is indefinitely differentiable and tends to $0$ as well as all its derivatives when $x\to0^+$ or $x\to+\infty$. Since, furthermore, for $t\to+\infty$ the function $f(x,t)$ tends to $+\infty$ we may expect that the integral depends essentially on the behavior of $g(x,t)$ at the extremes. As Polya in an analogous situation we consider the substitution of $ψ(x)$ by a simpler similar function. A simple function with this behavior is \[ψ_0(x):=2π(1+\tfrac{1}{4}x^{-5/2})e^{-πx-\fracπ{4x}}.\] Therefore, we define $J_0(t)$ replacing in the definition of $J(t)$ the function $ψ(x)$ by the simpler $ψ_0(x)$. \begin{equation} J_0(t)=2π\int_0^\infty (1+\tfrac{1}{4}x^{-\frac52})e^{-πx-\fracπ{4x}}(1-ix)^{\frac12(\frac12+it)}\,dx. \end{equation} The resulting $Z_0(t)$ disappoints us \[Z_0(t)\asymp \mathop{\mathrm{Re}}\Bigl\{\frac{2}{\sqrtπ}\exp\Bigl\{i\Bigl(\frac{t}{2}\log\frac{t}{2π}-\frac{t}{2}-\fracπ{8}\Bigr)\Bigr\}+\frac{2}{(2πt)^{1/4}}\exp\Bigl(πi\sqrt{\frac{t}{2π}}\;\Bigr)\Bigr\},\quad t\to+\infty.\] However, the integral $J_0(t)$ is interesting as a technical challenge. And still we have the possibility to get a better result improving $ψ_0(x)$. This is a preliminary version, and we set it as a challenge: to compute and study this integral.

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BibTeXRIS

Juan Arias de Reyna. 2024-07-08. A naive integral. https://arxiv.org/abs/2407.05719

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