Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I plan to prove the following integral inequality:

$$ \int_{0}^{1} \ln \sqrt{\frac{1+\cos x}{1-\sin x}}\le \ln 2$$

Since we have to deal with a convex function on this interval i thought of considering the area of the trapeze that can be formed if we unify the points $(0, f(0))$ and $(1, f(1))$, where the function $f(x) =\ln \sqrt{\frac{1+\cos x}{1-\sin x}}$, but things are ugly even if the method itself isn't complicated. So, I'm looking for something better if possible.

share|improve this question
1  
$$\int_0^1\log\frac{\cos\frac{x}{2}}{\cos\left(\frac{x}{2}+\frac\pi4\right)} dx\leq\log\,2$$ then? –  J. M. Jul 15 '12 at 12:01
1  
@J.M.: The inequality you noted is a magic one for me :) But I just know that for $0<x<y$, $\ln(\frac{y}{x})<\frac{y-x}{x}$ and this gives a big upper bound than $ln(2)$. –  B. S. Jul 15 '12 at 13:59
    
@Babak: huh? I just simplified that nasty square root within the logarithm, you know... –  J. M. Jul 15 '12 at 14:01
add comment

1 Answer 1

up vote 7 down vote accepted

Here's a (hopefully) corrected proof that uses convexity along with the trapezoid rule: You can rewrite what you're trying to prove as $$ \int_{0}^{1} \ln {\frac{1+\cos x}{1-\sin x}}\,dx\le 2\ln 2$$ Let $f(x) = \ln {\frac{1+\cos x}{1-\sin x}} = \ln(1 + \cos x) - \ln (1 - \sin x)$. Then $$f'(x) = -\frac{\sin x}{1 + \cos x} + \frac{\cos x}{1 - \sin x}$$ Using the tangent half-angle formula, this is the same as $$-\tan(x/2) + \tan(x/2 + \pi/4)$$ Therefore $$f''(x) = -(1/2)\sec^2(x/2) + 1/2\sec^2(x/2 + \pi/4)$$ Since $\sec$ is increasing on $(0,1/2 + \pi/4)$, we see that $f''(x) > 0$. So the integrand is convex. When applied to a convex function, the trapezoid rule always gives a result larger than the integral. But already with $2$ pieces, the trapezoid rule here gives $$1/4(\ln(1 + \cos(0)) - \ln(1 - \sin(0)) + 2(\ln(1 + \cos(1/2)) - \ln(1 - \sin(1/2)))$$ $$ +\ln(1 + \cos(1)) - \ln(1 - \sin(1)) )$$ $$= 1.3831395912690787...$$ This is slightly less than $2\ln2 = 1.3862943611198906...$, so the original integral is less than $\ln 2$ as needed.

share|improve this answer
1  
Well, you were using convexity, so Jensen's came to mind. $e^x$ is the most common convex function and your expression had $\ln$'s in it, so I tried to turn it into a Jensen problem using $e^x$. –  Zarrax Jul 15 '12 at 14:23
1  
however, i think you're wrong. It's 4 and not 8 on the right hand of the inequality and it doesn't hold. –  Chris's sis Jul 17 '12 at 9:15
    
yeah I did that one wrong, but I think my correction works, assuming I can do numerical calculations without making more mistakes –  Zarrax Jul 30 '12 at 22:12
    
By the way, using one trapezoid isn't enough as you'll get a result that is larger than $2\ln2$ (or $\ln2$ in the original question.) –  Zarrax Jul 30 '12 at 22:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.