We know that ∫∞0(sinxx)2dx=∫∞0sinxxdx=π2.
How do I show that ∫∞0|sinxx|dx
converges?
Answer
It doesn't. Using the convexity of 1/x,
∫∞0|sinxx|dx=∞∑k=0∫(k+1)πkπ|sinxx|dx>∞∑k=0∫(k+1)πkπ|sinx|(k+1/2)πdx=2π∞∑k=01k+1/2,
which diverges since the harmonic series diverges.
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