Find a big one between $$\int_0^{\frac{\pi}{2}}e^{-\sin^2 x}dx \quad \text{and}\quad \int_0^{\frac{\pi}{2}}e^{\frac{-2x}{\pi}}dx.$$

I tried to use jensen Inequality for integrals, but tt didnt work.

Also, some well-known inequalities (like $\frac{2x}{\pi}<\sin x<x$) also didnt work.

So I think I should approach with another way but I dont think how. Thanks for any help.

  • $\begingroup$ Very tight: The $\sin^2$ one is $\approx 1.01322$ and the other one is $\frac{(e-1) \pi }{2 e}\approx0.992933$ $\endgroup$ – User Mar 8 at 17:34

I found the following trick (maybe it's over-complicated ?). I'll let you fill the details.

First, note that the curves of equations $y=\sin^2 (x)$ and $y = 2x/\pi$ have a common central symmetry around $(\pi/4, 1/2)$. In order to take advantage of that, I change coordinates so as to center the integrals on $\pi/4$. Using a few trigonometric formulae (doubling of angle...) and a change of variables,

$$I_1 := \int_0^{\pi/2} e^{-\sin^2 (x)}\text{ d}x = \frac{1}{\sqrt{e}}\int_{-\pi/4}^{\pi/4} e^{-\frac{\sin(2x)}{2}}\text{ d}x,$$

$$I_2 := \int_0^{\pi/2} e^{-\frac{2x}{\pi}}\text{ d}x = \frac{1}{\sqrt{e}}\int_{-\pi/4}^{\pi/4} e^{-\frac{2x}{\pi}}\text{ d}x.$$

Now, $-\frac{\sin(2x)}{2}$ is inferior to $-\frac{2x}{\pi}$ on $(0,\pi/4)$, and superior on $(-\pi/4,0)$. We should expect $I_1$ to be larger than $I_2$ based on this and the fact that most of the mass of the integral comes from $(-\pi/4,0)$, but that needs to be proved, for instance by a convexity argument. Thus the second trick: symmetrize the integrals.

$$I_1 = \int_0^{\pi/2} e^{-\sin^2 (x)}\text{ d}x = \frac{1}{\sqrt{e}}\int_{-\pi/4}^{\pi/4} \cosh \left(\frac{\sin(2x)}{2}\right)\text{ d}x = \frac{2}{\sqrt{e}}\int_0^{\pi/4} \cosh \left(\frac{\sin(2x)}{2}\right)\text{ d}x,$$

$$I_2 = \int_0^{\pi/2} e^{-\frac{2x}{\pi}}\text{ d}x = \frac{2}{\sqrt{e}}\int_0^{\pi/4} \cosh \left(\frac{2x}{\pi}\right) \text{ d}x.$$

The convexity ox the exponential implies that $\cosh$ is convex with a minimum at $0$, and thus increasing on $[0,\pi/4]$. You should be able to conclude from there.

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  • $\begingroup$ Nice manipulation. It turns out we may just use Maclaurin series, but your approach is definitely more elegant than mine. (+1) $\endgroup$ – Jack D'Aurizio Mar 8 at 17:41

Curious exercise. We have $$ I_1 = \int_{0}^{\pi/2}e^{-\sin^2(x)}\,dx = \int_{0}^{1}\frac{e^{-u^2}}{\sqrt{1-u^2}}\,du = \int_{0}^{1}\frac{e^{-v}\,dv}{2\sqrt{v(1-v)}} $$ and $$ I_2 = \int_{0}^{\pi/2}e^{-2x/\pi}\,dx = \int_{0}^{1}\frac{\pi}{2}e^{-v}\,dv=\frac{\pi}{2}\left(1-\frac{1}{e}\right)=\frac{\pi}{2}\sum_{m\geq 0}\frac{(-1)^{m}}{(m+1)!} $$ so it is possible to exploit $$ I_1 = \sum_{n\geq 0}\frac{(-1)^n}{2n!}\int_{0}^{1}\frac{v^n}{\sqrt{v(1-v)}}\,dv =\frac{\pi}{2}\sum_{n\geq 0}\frac{(-1)^n \binom{2n}{n}}{4^n n!}$$ then couple consecutive terms in the series defining $\frac{2}{\pi}I_1$ and $\frac{2}{\pi}I_2$. We have $$ \frac{2}{\pi}I_2 = \sum_{m\geq 0}\left[\frac{1}{(2m+1)!}-\frac{1}{(2m+2)!}\right]=\sum_{m\geq 0}\frac{2m+1}{(2m+2)!}$$ and $$ \frac{2}{\pi}I_1 = \sum_{m\geq 0}\left[\frac{\binom{4m}{2m}}{16^m (2m)!}-\frac{\binom{4m+2}{2m+1}}{4^{2m+1}(2m+1)!}\right]=\sum_{m\geq 0}\frac{\left(2m+\frac{1}{4m+2}\right)\binom{4m}{2m}}{16^m(2m+1)!} $$ so $\color{red}{I_1 > I_2}$ by direct comparison of the main terms of the last series.

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COMMENT.-This is not an answer. We can see that $\int_0^{\frac{\pi}{2}}e^{-\sin^2 x}$ is greater in the two following ways:

$(1)$ Graphing the two functions to be integrated and looking at the two parts is that one is larger than the other are significantly different in magnitude so the integrals give the first one is greater than the second.

$(2)$ integrating the function $e^{-\sin^2 x}-e^{\frac{-2x}{\pi}}$ whose value is given by $\dfrac{\pi(1-e+\sqrt e I_0(\frac12))}{2e}\approx0.02028\gt0$ where $$I_0\left(\dfrac12\right)=\dfrac{1}{\pi}\int_0^{\pi}e^\dfrac{\cos(t)}{2}dt$$ is the modified Bessel function of the first kind with $\nu=0.$

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