Prove : $\int_{{π}/{4}}^{{π}/{2}} e^{\cos x + \cos^2{x}}\mathrm dx>\sqrt2$

Question

Exactly as the title says , we need to prove :

$$\int_\frac{π}{4} ^ \frac{π}{2} e^{\cos x + \cos^2{x}}\mathrm dx>\sqrt2$$

My unsuccessful approach:

I tried by searching for a function $f$ such that :

$$e^{\cos x + \cos^2{x}}>f$$ and

$$\int_\frac{π}{4} ^ \frac{π}{2}f\mathrm dx = \sqrt2$$

Since the limits are in terms of $π$ while the integral should be equal to $\sqrt 2$ , the indefinite integral should be of the form having denominator equal to $π$. Unfortunately, I have never encountered such type of functions or integrals.

For the same reasons above, I also cannot think of functions to apply sandwich theorem.

I am looking for some elementary methods but all level of answers are welcome. Thanks !

Answer 1

Let $\,g(x)=\cos x+\cos^2x\,.$

It results that$$e^{g(x)}>1+g(x)+\frac12g^2(x)+\frac16g^3(x)\,.$$ Moreover, $$\int_{\frac{\pi}4}^{\frac{\pi}2}\left[1+g(x)+\frac12g^2(x)+\frac16g^3(x)\right]\!\mathrm dx=\\=\dfrac{7956+3135\pi-6712\sqrt2}{5760}>\sqrt2\;.$$

Answer 2

Remarks: The trick is to use $\mathrm{e}^{u} = \mathrm{e}^a\mathrm{e}^{u - a} \ge \mathrm{e}^a (1 + u - a)$ by choosing appropriate constant $a$. More details are given at the end.

We have \begin{align*} I &= \int_{\pi/4}^{\pi/3} \mathrm{e}^{\cos x + \cos^2{x}}\,\mathrm{d} x + \int_{\pi/3}^{\pi/2} \mathrm{e}^{\cos x + \cos^2{x}}\,\mathrm{d} x\\[6pt] &= \int_{\pi/4}^{\pi/3} \mathrm{e}\cdot \mathrm{e}^{\cos x + \cos^2{x} - 1}\,\mathrm{d} x + \int_{\pi/3}^{\pi/2} \mathrm{e}^{1/2}\mathrm{e}^{\cos x + \cos^2{x} - 1/2}\,\mathrm{d} x\\[6pt] &\ge \int_{\pi/4}^{\pi/3} \mathrm{e}\cdot (1 + \cos x + \cos^2{x} - 1)\,\mathrm{d} x + \int_{\pi/3}^{\pi/2} \mathrm{e}^{1/2}(1 + \cos x + \cos^2{x} - 1/2)\,\mathrm{d} x \tag{1}\\[6pt] &= \mathrm{e}\left(- \frac14 + \frac{1}{24}\pi - \frac{\sqrt 2}{2} + \frac58 \sqrt 3\right) + \mathrm{e}^{1/2}\left(-\frac58\sqrt 3 + \frac16\pi + 1\right)\\[6pt] &> 2.7\left(- \frac14 + \frac{1}{24}\cdot 3.14 - \frac{\sqrt 2}{2} + \frac58 \sqrt 3\right) + 1.64\left(-\frac58\sqrt 3 + \frac16\cdot 3.14 + 1\right)\\[6pt] &> \sqrt 2 \end{align*} where we use $\mathrm{e}^u \ge 1 + u$ for all $u\in \mathbb{R}$ in (1), and $\mathrm{e} > 2.7$, and $\mathrm{e}^{1/2} > 1.64$, and $\pi > 3.14$.

More details for the trick:

First, we try \begin{align*} I &= \int_{\pi/4}^{\pi/2} \mathrm{e}^a\cdot\mathrm{e}^{\cos x + \cos^2{x} - a}\,\mathrm{d} x\\ &\ge \int_{\pi/4}^{\pi/2} \mathrm{e}^a\cdot (1 + \cos x + \cos^2{x} - a)\,\mathrm{d} x\\ &= \mathrm{e}^a \left(\frac34 + \frac38\pi - \frac{\sqrt 2}{2} - \frac{\pi}{4} a\right). \end{align*} Let $$f(a) := \mathrm{e}^a \left(\frac34 + \frac38\pi - \frac{\sqrt 2}{2} - \frac{\pi}{4} a\right).$$ We have $$f'(a) = \frac{\pi}{4}\mathrm{e}^a \left(\frac12 + \frac{3-2\sqrt 2}{\pi} -a \right).$$ Thus, $f(a)$ achieves its global maximum at $a_0 = \frac12 + \frac{3-2\sqrt 2}{\pi}$.

However $f(a_0) < \sqrt 2$. Thus, we split the integral into two parts i.e. $$I = \int_{\pi/4}^{\pi/3} \mathrm{e}^{\cos x + \cos^2{x}}\,\mathrm{d} x + \int_{\pi/3}^{\pi/2} \mathrm{e}^{\cos x + \cos^2{x}}\,\mathrm{d} x$$ and do something similar for each part.

Answer 3

Since, for the consiered range, the cosine is not negative, using $$e^{\cos(t)+\cos^2(t)}=\sum_{n=0}^\infty a_n\cos^n(t)$$ we shall just add a succession of $\color{red}{\text{positive}}$ terms

The $a_n$ are given by $$a_n=\frac{2 a_{n-2}+a_{n-1}}{n} \qquad \text{with} \qquad a_0=a_1=1$$

On the other side, we know the reduction formula for $$I_n=\int \cos^n(x)\,dx$$ and that $$J_n=\int_{\frac \pi 4}^{\frac \pi 2} \cos^n(x)\,dx ~~> ~0$$

We just need to look at the partial sum $$S_p=\sum_{n=0}^p a_n\, J_n$$

We do not need many terms since $$S_4=\frac{2632-2272 \sqrt{2}+1233 \pi }{2304}=1.42903 ~~>~\sqrt 2$$