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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 !

An_Elephant
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3 Answers3

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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.

River Li
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  • "Using the known inequality sinx≥x−x3/6" Can you give a link for this ? I have seen this first time. Thanks ! The solution is then perfect. – An_Elephant May 01 '23 at 08:13
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    @An_Elephant I have added the link for $\sin x \ge x - x^3/6$. More references: https://math.stackexchange.com/questions/1613919/proving-that-x-fracx36-sinx-for-0x-pi, and https://math.stackexchange.com/questions/2555669/estimate-from-below-of-the-sine-and-from-above-of-cosine, and https://math.stackexchange.com/questions/3905903/prove-sin-x-ge-x-fracx33 – River Li May 01 '23 at 08:56
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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 $$e^{\sin ^2(x)} \gt 1+x^2$$

Claude Leibovici
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  • "use the *simplest* Padé approximant and.." Simplest ? Anyway thanks , I hope I will learn these things when I will learn advanced math. – An_Elephant Apr 30 '23 at 12:37
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    @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 :-° – Claude Leibovici Apr 30 '23 at 14:58
  • @ClaudeLeibovici Any good references for Padé approximations? – V.S.e.H. Apr 30 '23 at 15:35
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    @V.S.e.H.. I have an anwer on the site where I expain how to build them from Taylor series . https://math.stackexchange.com/questions/1809229/pad%c3%a9-approximation/1809303#1809303 . If you look at my answers, I already made 473 answers with them. I love !! – Claude Leibovici Apr 30 '23 at 15:47
  • @ClaudeLeibovici Yes , I too hope :) Thanks sir for your kind blessings. – An_Elephant Apr 30 '23 at 16:41
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    @ClaudeLeibovici Your last statement is not obvious for me. For what values of $x$ does it hold? Do you have a rigorous proof? – Gary May 01 '23 at 00:33
  • @Gary. $0<x<1$ as required for the integral – Claude Leibovici May 01 '23 at 02:43
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$$e^{\sin^2x}=1+x^2+\frac{x^4}6-\frac{11}{90}x^6-\frac{71}{2520}x^8+\frac{1441}{113400}x^{10}+\ldots>1+x^2$$

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

MathFail
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    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. – Ryszard Szwarc Apr 30 '23 at 02:42
  • How $e^{\sin^2x}=1+x^2+\frac{x^4}6+....$ ? Is this called taylor series ? – An_Elephant Apr 30 '23 at 02:57
  • The Taylor series of $\mathrm{e}^{\sin^2 x}$ is $1 + x^2 + \frac16x^4 - \frac{11}{90}x^6 - \frac{71}{2520} x^8 + \cdots$. – River Li Apr 30 '23 at 06:23
  • @Macavity The function is analytic so the radius of convergence is equal $+\infty.$ – Ryszard Szwarc Apr 30 '23 at 10:30
  • @RyszardSzwarc Right. The problem seems to be that MathFail hasn't put down enough terms for you to make a conclusion that it doesn't converge at $x=\pi$. By your argument on the function being analytic, it should. For e.g. of the next 6 terms in the Taylor series, 4 are negative. – Macavity Apr 30 '23 at 10:48
  • Sorry for confusion, I didn't copy enough terms. @RyszardSzwarc – MathFail Apr 30 '23 at 15:22
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    @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 Apr 30 '23 at 22:06
  • @MathFail As Gary pointed out, your reasoning is not enough to prove the result. – River Li May 13 '23 at 02:32
  • @MathFail By the way, I did not downvote you although I criticize your answer. – River Li May 13 '23 at 05:54