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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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marked as duplicate by Aryabhata, Jonas Meyer, Nate Eldredge, t.b., Zev Chonoles Jan 28 '12 at 16:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Welcome to math.stackexchange Lilly. In the future, please avoid using imperative language- rather than saying "Prove X" phrase it as "Can someone help me prove X? Here is what I've tried, but I got stuck here...". Members here will give you more help if you do that. Anyway, for your problem, you should try integrating by parts. – Ragib Zaman Jan 27 '12 at 11:32
I don't know much about this subject, but I think you should take a look at this reference. (which defines the integral $\int_0^{\infty} \frac{sin t}{t} dt$) – joaopaulolf Jan 27 '12 at 12:26
The indefinite integral corresponding to the right side can be expressed in terms of the indefinite integral corresponding to the left side. – tzs Jan 27 '12 at 12:52

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 $u=2t$.

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