# An inequality

• Jul 6th 2009, 04:28 AM
NonCommAlg
An inequality
The upper bound given in this problem is a good approximation for the integral:

Problem: Prove that $\int_0^1 e^{\cos x} \ dx \leq e \sqrt{2} \tan^{-1}(\sqrt{2}/2).$
• Jul 6th 2009, 11:53 AM
Sampras
By the Rectangle Rule, $\int_{0}^{1} e^{\cos x} \ dx \approx e^{\cos \frac{1}{2}}$. Or we know that $e^{\cos x}$ has a bounded derivative over $[0,1]$. Now by the MVT, $xf'(y_x) = f(x)-f(0)$ for some $y_x \in [a,x]$. Also, $x<1$. Then

$\left|\int_{0}^{1} e^{\cos x} \ dx -e^{\cos 1} \right| = \left|\int_{0}^{1} x f'(y_x) \ dx \right|$

$\left|\int_{0}^{1} e^{\cos x} \ dx -e^{\cos 1} \right| \leq \frac{1}{2} \sup_{0 \leq x \leq 1} |-e^{\cos x} \sin x|$
• Jul 6th 2009, 12:00 PM
halbard
OK, I have come up with an incredibly contrived solution suggested by the answer. Clearly it is equivalent to

$\int_0^1\mathrm e^{\cos x-1}\mathrm dx\leq\int_0^1\frac 2{2+x^2}\mathrm dx$, so this suggests the inequality $\mathrm e^{\cos x-1}\leq\frac 2{2+x^2}$. If there's a better way...

Here we go. Start with $\sin x\leq x$ and $\sin^2 x+\cos x=1+\cos x(1-\cos x)\geq 1$ on $[0,1]$. Therefore $\sin x\leq x(\sin^2 x+\cos x)$.

This rearranges to $\frac{\sin x-x\cos x}{\sin^2 x}\leq x$. The LHS is the derivative of $\frac x{\sin x}$, so integrate both sides from $0$ to $t$, where $t\in[0,1]$.

Thus $\frac t{\sin t}-1\leq \frac{t^2}2$ and so $\sin t\geq\frac t{1+t^2/2}$. Integrate this from $0$ to $x$ and $1-\cos x\geq \ln(1+x^2/2)$ pops up.

Guess what? $\cos x-1\leq\ln\left(\frac2{2+x^2}\right)$, and we arrive at our destination. Where's my hat? Rabbit, anyone?