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.
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.
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 $u=2t$.