# 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.

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.

• Why is 20 enough? Commented Aug 17, 2018 at 17:03
• @StevenGubkin: What I did in the Added section is that I decomposed the interval $[1,2]$ into $20$ equal subintervals such that on each of the subintervals, the maximum of $f$ is less than the minimum of $g$. So I proved a bit more than $f(x)<g(x)$ on $[1,2]$, and the advantage is that I only need to check finitely many numeric inequalities. The number $20$ has no significance except that it makes the idea work. For example, $15$ is not ok, because then on the first subinterval, the maximum of $f$ is greater than the minimum of $g$. Commented Aug 17, 2018 at 19:56
• Sorry for my late confirmation. Your method is surprisingly skillful, but at the same time clear. Actually, even if we prove only for x> 2, there is no big problem in the original problem. But as you added, the idea of proving the inequality by dividing the interval is also very impressive. Thanks for the great intuition and explanation. Commented Aug 20, 2018 at 17:35
• @Ramanasa: Thanks for your feedback and kind words. I am glad I could help! Commented Aug 20, 2018 at 17:40

The inequality holds by direct calculation for $x$ close to unity, when the two sides are well separated, but for large $x$ the two sides approach each other and we need to check that the inequality is not violated.

I denote $x\equiv 1/\epsilon$, and seek to prove that $$\int_0^{\infty}{\frac{\epsilon}{e^{s}\sqrt{1+\epsilon^2 s^2}}}ds < \arctan\epsilon,\;\;0<\epsilon\leq 1.$$ Series expansion of both sides gives $$\int_0^{\infty}{\frac{\epsilon}{e^{s}\sqrt{1+\epsilon^2 s^2}}}ds=\epsilon-\epsilon^3+{\cal O}(\epsilon^5),$$ $$\arctan\epsilon=\epsilon-\tfrac{1}{3}\epsilon^3+{\cal O}(\epsilon^5).$$ This proves the inequality for $\epsilon$ close to $0$, hence for large $x$.

• Well, one needs to be somewhat careful as the power series expansion of $1/\sqrt{1+\epsilon^2 s^2}$ only converges for $s<1/\epsilon$. Commented Aug 17, 2018 at 13:14
• I have developed your idea and made it concrete as follows. The inequality $$\frac1{\sqrt{1+t^2}} \le 1- \frac{1}{2}t^2 + \frac38 t^4$$ holds for all $t$. Using this the LHS is $$\epsilon\int_0^\infty \frac1{e^s\sqrt{1+s^2 x^2}}ds<\epsilon \int_0^\infty \frac1{e^s}\left(1-\frac12 s^2 \epsilon^2+\frac38 s^4 \epsilon^4\right)ds=\epsilon-\epsilon^3+9\epsilon^5$$ for all $\epsilon>0$, where the RHS is $$\arctan\epsilon > \epsilon-\frac13\epsilon^3$$ for all $\epsilon>0$. So it proves the inequality for $0<\epsilon<\frac2{3\sqrt3}$, or for $x>\frac{3\sqrt3}2\simeq2.598$. Commented Aug 20, 2018 at 17:36
• I could not find out that the left side has a series expansion, but your advice was a great help. Commented Aug 20, 2018 at 17:36
• I apologize for that I could choose only one answer on this system (Is there any other way?) while you provided also an excellent and clever solution. Commented Aug 20, 2018 at 17:44

Alternative solution.

We have the following bounds (easy) $$\frac{1}{\sqrt{1 + s^2}} \le 1 - \frac{20}{69}s^2, \quad \forall s\in [0, 1];$$ and $$\frac{1}{\sqrt{1 + s^2}} < \frac{1}{12}s^2 - \frac12 s + \frac98, \quad \forall s\in [1, \infty).$$

Thus, we have \begin{align*} \int_0^\infty \frac{1}{\mathrm{e}^{sx}\sqrt{1 + s^2}} \,\mathrm{d} s &< \int_0^1 \frac{1 - 20s^2/69}{\mathrm{e}^{sx}}\,\mathrm{d} s + \int_1^\infty \frac{\frac{1}{12}s^2 - \frac12 s + \frac98}{\mathrm{e}^{sx}}\,\mathrm{d} s\\ &= \frac{69x^2 - 40}{69x^3} + \frac{-x^2 + 136x + 412}{552x^3\mathrm{e}^x}. \end{align*}

It suffices to prove that $$\frac{69x^2 - 40}{69x^3} + \frac{-x^2 + 136x + 412}{552x^3\mathrm{e}^x} \le \arctan \frac{1}{x}.$$

We split into two cases.

Case 1. $$x\ge 1$$ and $$-x^2 + 136x + 412 < 0$$

It suffices to prove that $$\frac{69x^2 - 40}{69x^3} \le \arctan \frac{1}{x}.$$ Let $$g(x) := \arctan \frac{1}{x} - \frac{69x^2 - 40}{69x^3}.$$ We have $$g'(x) = - \frac{17x^2 + 40}{23x^4(x^2 + 1)} < 0.$$ Also, $$\lim_{x\to\infty} g(x) = 0$$. Thus, $$g(x) \ge 0$$ for all $$x \ge 1$$.

Case 2. $$x\ge 1$$ and $$-x^2 + 136x + 412\ge 0$$

Since $$\mathrm{e}^x = \mathrm{e}\cdot \mathrm{e}^{x-1} \ge \mathrm{e}(1 + x - 1)$$ and $$\frac{1}{\mathrm{e}^x} \le \frac{1}{\mathrm{e}x} \le \frac{215389350}{584654827x}$$, it suffices to prove that $$f(x) := \arctan \frac{1}{x} - \frac{69x^2 - 40}{69x^3} - \frac{-x^2 + 136x + 412}{552x^3}\cdot \frac{215389350}{584654827x} \ge 0.$$ We have $$f'(x) = \frac{(33 - 25x)(1435929x^3 + 504096475x^2 - 516362220x + 896367800)}{26894122042x^5(x^2 + 1)}.$$ Since $$1435929x^3 + 504096475x^2 - 516362220x + 896367800 > 0$$ on $$x\ge 0$$, we have $$f'(x) > 0$$ on $$(1, 33/25)$$ and $$f'(x) < 0$$ on $$(33/25, \infty)$$. Also, $$f(1) > 0$$ and $$\lim_{x\to \infty} f(x) = 0$$. Thus, $$f(x) \ge 0$$ for all $$x \ge 1$$.

We are done.