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

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