# Prove : $\int_0^1 \frac{ e^{\sin^2x}}{1+x^2}\mathrm dx > 1$

I want to prove the following inequality :

$$I = \int_0^1 \frac{ e^{\sin^2x}}{1+x^2}\mathrm dx > 1$$

Since I have been only introduced to elementary methods of integration, the indefinite integral appears non-solvable to me (I would be glad to see if there's a neat expression for indefinite integral).

To evaluate the integral, I used the property $$\int_a^b f(x) \mathrm dx = \int_a^b f(a+b-x) \mathrm dx$$ , to get $$I = \int_0^1 \frac{ e^{\sin^2(1-x)}}{x^2 - 2x + 2}\mathrm dx$$ which looks much uglier. Further, the limits of integral are not symmetrical with respect to $$x=0$$, hence the fact that the integrand is an even function is useless.

Please tell that how can I approach this inequality. Thanks !

• Perhaps you can try a Taylor approximation for $\mathrm{e}^{\sin^2 x}$.
– Gary
Commented Apr 30, 2023 at 1:17
• @Gary Thanks sir for hint. But unfortunately, I haven't learnt this technique yet :( Commented Apr 30, 2023 at 1:21

It suffices to prove that, for all $$x\in [0, 1]$$, $$\mathrm{e}^{\sin^2 x} \ge 1 + x^2$$ or $$\sin^2 x \ge \ln (1 + x^2).$$

Using the known inequality $$\sin x \ge x - x^3/6$$ for all $$x \ge 0$$, noting that $$x - x^3/6 \ge 0$$ on $$[0, 1]$$, it suffices to prove that $$(x - x^3/6)^2 \ge \ln(1 + x^2)$$ or $$\frac{1}{36}x^2(6 - x^2)^2 \ge \ln(1 + x^2).$$

Letting $$y = x^2$$, it suffices to prove that, for all $$y\in [0, 1]$$, $$f(y) := \frac{1}{36}y(6 - y)^2 - \ln(1 + y) \ge 0.$$

We have $$f'(y) = \frac{y(y^2 - 7y + 4)}{12(1+y)}.$$ Thus, $$f'(\frac{7-\sqrt{33}}{2}) = 0$$, and $$f'(y) > 0$$ on $$(0, \frac{7-\sqrt{33}}{2})$$, and $$f'(y) < 0$$ on $$(\frac{7-\sqrt{33}}{2},1]$$. Also, $$f(0) = 0$$ and $$f(1) > 0$$. Thus, $$f(y) \ge 0$$ on $$[0, 1]$$.

We are done.

Instead of Taylor series, use the simplest Padé approximant and $$e^{\sin ^2(x)} \leq \frac{5 x^2+6}{6-x^2} \qquad \text{for} \quad x \in (0,1)$$ So, for an upper bound, $$\int_0^1 \frac{ e^{\sin^2x}}{1+x^2}\,dx < \int_0^1 \frac{5 x^2+6}{(1+x) ^2(6-x^2)}\,dx$$ $$\frac{5 x^2+6}{(1+x) ^2(6-x^2)}=\frac{1}{7 \left(x^2+1\right)}-\frac{36}{7 \left(x^2-6\right)}$$ $$\int_0^1 \frac{5 x^2+6}{(1+x) ^2(6-x^2)}\,dx=\frac{\pi }{28}+\frac{6\sqrt{6}}{7} \tanh ^{-1}\left(\frac{1}{\sqrt{6}}\right)=1.02238$$

The lower bound is simple since $$e^{\sin ^2(x)} \gt 1+x^2$$

• "use the simplest Padé approximant and.." Simplest 🙂 ? Anyway thanks , I hope I will learn these things when I will learn advanced math. Commented Apr 30, 2023 at 12:37
• @An_Elephant. It is as simple as Taylor series but much more accurate. I am sure that you xill enjoy them as much as I do for 64+, years. Cheers :-° Commented Apr 30, 2023 at 14:58
• @ClaudeLeibovici Any good references for Padé approximations? Commented Apr 30, 2023 at 15:35
• @V.S.e.H.. I have an anwer on the site where I expain how to build them from Taylor series . math.stackexchange.com/questions/1809229/… . If you look at my answers, I already made 473 answers with them. I love !! Commented Apr 30, 2023 at 15:47
• @ClaudeLeibovici Your last statement is not obvious for me. For what values of $x$ does it hold? Do you have a rigorous proof?
– Gary
Commented May 1, 2023 at 0:33

$$e^{\sin^2x}=1+x^2+\frac{x^4}6-\frac{11}{90}x^6-\frac{71}{2520}x^8+\frac{1441}{113400}x^{10}+\ldots>1+x^2$$

So we have, $$I = \int_0^1 \frac{ e^{\sin^2x}}{1+x^2}\mathrm dx > \int_0^1 \frac{ 1+x^2}{1+x^2}\mathrm dx=1$$

• The first equality looks false. If it held for $0\le x\le 1$ then by analycity it would extend to the entire real line. For $x=\pi$ the equality fails. Commented Apr 30, 2023 at 2:42
• How $e^{\sin^2x}=1+x^2+\frac{x^4}6+....$ ? Is this called taylor series ? Commented Apr 30, 2023 at 2:57
• The Taylor series of $\mathrm{e}^{\sin^2 x}$ is $1 + x^2 + \frac16x^4 - \frac{11}{90}x^6 - \frac{71}{2520} x^8 + \cdots$. Commented Apr 30, 2023 at 6:23
• @Macavity The function is analytic so the radius of convergence is equal $+\infty.$ Commented Apr 30, 2023 at 10:30
• @MathFail From the Taylor expansion you can conclude that the inequality holds for sufficiently small $x$, but not necessarily for $0<x<1$. You have to use a remainder and estimate it.
– Gary
Commented Apr 30, 2023 at 22:06