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.

  • $\begingroup$ Actually this originates from math.stackexchange.com/questions/2879015/…, which has not been solved completely. $\endgroup$ – Ramanasa Aug 16 '18 at 12:45
  • $\begingroup$ How is it related? Do you mean need proof $${\cal L}(\dfrac{1}{\sqrt{1+s^2}})<{\cal L}(\dfrac{\sin s}{s})$$ $\endgroup$ – Nosrati Aug 16 '18 at 12:59
  • $\begingroup$ 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 $x\ge 1$. $\endgroup$ – Ramanasa Aug 16 '18 at 13:39
  • $\begingroup$ 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. $\endgroup$ – Martin Sleziak Aug 17 '18 at 13:03

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