Processing math: 14%

Pages

Bookmark and Share

Wednesday, September 1, 2010

Integral Inequality

Prove that:

\left( \int_\pi^\infty\frac{\cos x}{x}\ dx\right)^{2} < \frac{1}{{\pi}^{2}}

Solution

Integrate by parts:

\int \frac{\cos x}{x} dx = \frac{\sin x}{x} + \int \frac{\sin x}{x^2} dx

So
\int_\pi^\infty\frac{\cos x}{x}\ dx = \int_\pi^\infty \frac{\sin x}{x^2} dx

And
| \int_\pi^\infty\frac{\cos x}{x}\ dx| = |\int_\pi^\infty \frac{\sin x}{x^2} dx|
\leq \int_\pi^\infty \frac{| \sin x |}{x^2} dx
\leq \int_\pi^\infty \frac{1}{x^2} dx
= \frac{1}{\pi}

No comments:

Post a Comment