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Krizalid · #5 November 30th, 2007, 09:00 PM

Quote:

Originally Posted by liyi

Evaluate $\int_{-\infty}^{+\infty}\frac{\sin x}x\,dx$

$\frac1x=\int_0^\infty e^{-ux}\,du.$ so,

$\int_{ - \infty }^{ + \infty } {\frac{{\sin x}}{x}\,dx} = 2\int_0^\infty {\int_0^\infty {e^{ - ux} \sin x\,du} \,dx} .$

Now the remaining challenge is to compute ${\int_0^\infty {e^{ - ux} \sin x\,dx} }.$ (Reverse integration order.)

This is nasty applyin' integration by parts, but it's quickly considering the following:

$\int_0^\infty {e^{ - ux} \sin x\,dx} = \text{Im} \int_0^\infty {e^{ - (u - i)x} \,dx} = \text{Im} \frac{1}{{u - i}} = \frac{1}{{u^2 + 1}}.$

And finally

$\int_{ - \infty }^{ + \infty } {\frac{{\sin x}}{x}\,dx} = 2\int_0^\infty {\frac{1}{{1 + u^2 }}\,du} = \pi .$

The following users thank Krizalid for this useful post:
liyi
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