For $x\in \mathbb R_+$, let us define $$ I_\lambda(x)=\frac2π\int_0^{+\infty}\frac{\sin t}{t\sqrt{1+t^2x^2}}\cos(\lambda \arctan (xt)) \,dt,\quad \lambda\in 1+2\mathbb N. $$ We see that $I_\lambda(0)=1$ and $\lim_{x\rightarrow+\infty}I_\lambda(x)=0.$ My hunch is that $$ \forall x>0,\quad I_\lambda(x)<1.\tag {$\ast$} $$ I do not know if the assumption on $\lambda$ is important and I would like to know if $(\ast)$ holds true.
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$\begingroup$ $I_1(x)=1-e^{-1/x}$ $\endgroup$– Carlo BeenakkerCommented Jul 7, 2016 at 12:28
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$\begingroup$ @Carlo Beenakker Thanks for this sharp calculation. With $$ F_{k}(x)=\int_{\mathbb R}\frac{\sin x t}{\pi t}\frac{(1+it)^{2k+1}}{(1+t^{2})^{k+1}} dt, $$ we get indeed $F_0(x)=1-e^{-x}$(your calculation), and also $F_1(x)=1-e^{-x}(1+2x)$, $F_2(x)=1-e^{-x}(1+2x^2)$, $F_3(x)=1-e^{-x}(1+2x-2x^2+4x^3/3),$ and a general formula involving the Laguerre polynomials. $\endgroup$– BazinCommented Oct 20, 2016 at 7:33
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