The question is to prove:
$$
\int\_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}ds < \arctan\left(\frac1x\right),\quad\forall x\ge1.
$$
Numerically it seems to hold true. So I have made some attempts to prove this analytically but have all failed.


I also wonder if there is a systematic approach to solve this kind of problem.


Thanks for your help.



Prove $\int_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}ds < \arctan\left(\frac1x\right),\quad\forall x\ge1$
Here is a proof of the inequality for $x\geq 2$. For the remaining range, see the Added section below.


Let $\lambda:=1-1/\sqrt{2}$, then by convexity we have
$$\frac{1}{\sqrt{1+s^2}}\leq 1-\lambda s^2,\qquad 0\leq s\leq 1.$$
Using this bound we can estimate
\begin{align\*}\int\_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}\,ds&<\int\_0^1\frac{1-\lambda s^2}{e^{sx}}\,ds+\int\_1^\infty\frac{1-\lambda}{e^{sx}}\,ds\\[6pt]
&=\frac{1}{x}-\lambda\left(\int\_0^1\frac{s^2}{e^{sx}}\,ds+\int\_1^\infty\frac{1}{e^{sx}}\,ds\right)\\[6pt]
&=\frac{1}{x}-2\lambda\frac{1-e^{-x}x-e^{-x}}{x^3}
.\end{align\*}
On the other hand, for $x>0$ we also have
$$\arctan\left(\frac{1}{x}\right)=\int\_0^{1/x}\frac{1}{1+s^2}\,ds>\int\_0^{1/x}(1-s^2)\,ds=\frac{1}{x}-\frac{1}{3x^3},$$
hence it suffices to verify that
$$\lambda(1-e^{-x}x-e^{-x})>\frac{1}{6},\qquad x\geq 2.$$
This is straightforward, so we proved the original inequality for $x\geq 2$.


**Added.** One can cover the remaining range $1\leq x<2$ as follows. Let $f(x)$ denote the LHS and $g(x)$ denote the RHS in the original inequality. These two functions are decreasing, hence it suffices to verify the following $20$ numeric inequalities:
$$f(1+(n-1)/20)<g(1+n/20),\qquad n=1,2,\dots,20.$$
These are likely to be true, e.g. I checked them with Mathematica's NIntegrate command.


