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Possible Duplicate:
Proof for an integral involving sinc function

Prove that: $\displaystyle \int_{0}^{\infty }\frac{\sin t}{t}dt=\int_{0}^{\infty }\frac{\sin^{2}t}{t^{2}}dt$

Thank you in advance for any suggestion.

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    I don't know much about this subject, but I think you should take a look at this [reference](http://mathworld.wolfram.com/SineIntegral.html). (which defines the integral $\int_0^{\infty} \frac{sin t}{t} dt$)2012-01-27

1 Answers 1

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Integrating by parts:

$\eqalign{ \int_0^\infty {\sin^2 t\over t^2}\,dt&= {-\sin^2 t\over t}\biggl|_0^\infty + \int_0^\infty{ 2\sin t\cos t\over t}\,dt\cr &={-\sin^2 t\over t}\biggl|_0^\infty + \int_0^\infty{ 2 \sin 2t \over 2t}\,dt\cr &= 0 + \int_0^\infty {\sin u\over u}\,du } $

In the above, we computed $\lim\limits_{t\rightarrow0^+}{\sin^2 t\over t}=\lim\limits_{t\rightarrow0^+}{2\sin t \cos t\over 1}=0$; and in the last integral, we set $u=2t$.