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Krizalid · #4 December 7th, 2007, 07:04 AM

Quote:

Originally Posted by liyi

Evaluate

$\int_0^\infty\frac{1-\cos x}{x^2}\,dx$

Here's another proof.

We'll use the following fact $\int_0^\infty\frac{\sin x}x\,dx=\frac\pi2.$ (Proven here.)

Now $\frac{{1 - \cos x}}{x} = \int_0^1 {\sin (ux)\,du} .$

The integral becomes to $\int_0^\infty {\int_0^1 {\frac{{\sin (ux)}}{x}\,du} \,dx} = \int_0^1 {\int_0^\infty {\frac{{\sin (ux)}}{x}\,dx} \,du} .$

For the inner integral, substitute $\varphi=ux,$

$\int_0^\infty {\frac{{\sin (ux)}}{x}\,dx} = \frac{1}{u}\int_0^\infty {\frac{{u\sin \varphi }}{\varphi }\,d\varphi } = \frac{\pi }{2}.$

Finally

$\int_0^\infty {\frac{{1 - \cos x}}{{x^2 }}\,dx} = \frac{\pi }{2}\int_0^1 {du} = \frac{\pi }{2}\,\blacksquare$

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A related problem.

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As for $\int_0^\infty {\frac{{\sin ^2 x}}{{x^2 }}\,dx}$ can be found here.

The following users thank Krizalid for this useful post:
liyi
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