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

Question

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 !

Answer 1

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.

Answer 2

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 $$e^{\sin ^2(x)} \gt 1+x^2$$

Answer 3

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

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