Thread: Integrals
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Old December 6th, 2007, 07:45 PM
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Quote:
Originally Posted by liyi View Post
Evaluate
\int_0^\infty\frac{1-\cos x}{x^2}\,dx
1 - \cos x = 2\sin^2 \frac{x}{2}
Thus,
2\int_0^{\infty} \frac{\sin^2 \frac{x}{2}}{x^2} dx
Let t=x/2,
2\int_0^{\infty} \frac{\sin^2 t}{4t^2} \cdot 2dt = \frac{\pi}{2}
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