Thread: Sine squared integral
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Old November 15th, 2007, 01:11 PM
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Quote:
Originally Posted by liyi View Post
Evaluate \int_0^\infty\left(\frac{\sin x}x\right)^2\,dx
\int_0^\infty {\left( {\frac{{\sin x}}{x}} \right)^2 \,dx} = \int_0^\infty {\frac{{1 - \cos 2x}}{{2x^2 }}\,dx} .

Use the following parameter: \frac{1}{{x^2 }} = \int_0^\infty {ue^{ - ux} \,du} .

(The only reason that I'd ever create a double integral is that I can reverse the integration order.)

From there you can solve the rest, without problems.
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