# Prove $\int_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}ds < \arctan\left(\frac1x\right),\quad\forall x\ge1$.

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.

• How is it related? Do you mean need proof $${\cal L}(\dfrac{1}{\sqrt{1+s^2}})<{\cal L}(\dfrac{\sin s}{s})$$ – Nosrati Aug 16 '18 at 12:59
• In the original post, Jack D'Aurizio deduced that $\int_0^x \frac{\sin t}{t}dt = \frac{\pi}2 - \int_0^{\infty} \frac{\cos x + s \sin x}{\left(1+s^2\right)e^{sx}}ds \ge \frac{\pi}2 - \int_0^{\infty} \frac{1}{e^{sx}\sqrt{1+s^2}}ds$. Using this and the identity $\arctan x = \frac{\pi}2 - \arctan \left( \frac{1}{x}\right)$ for $x>0$, we conclude: Once we prove $\int_0^{\infty} \frac{1}{e^{sx}\sqrt{1+s^2}}ds < \arctan\left(\frac1{x}\right)$ for $x\ge1$, it follows $\int_0^x \frac{\sin t}{t}dt > \arctan x$ for $x\ge 1$. – Ramanasa Aug 16 '18 at 13:39
• Posted also on MathOverflow: Prove $\int_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}ds < \arctan\left(\frac1x\right),\quad\forall x\ge1$. I think this answer offers very reasonable advice on cross-posting. – Martin Sleziak Aug 17 '18 at 13:03