show that
∫∞0sin3(x)x3dx=3π8
using different ways
thanks for all
Answer
Let f(y)=∫∞0sin3yxx3dx Then, f′(y)=3∫∞0sin2yxcosyxx2dx=34∫∞0cosyx−cos3yxx2dx f″(y)=34∫∞0−sinyx+3sin3yxxdx Therefore, f″(y)=94∫∞0sin3yxxdx−34∫∞0sinyxxdx
Now, it is quite easy to prove that ∫∞0sinaxxdx=π2signuma
Therefore, f″(y)=9π8signumy−3π8signumy=3π4signumy Then, f′(y)=3π4|y|+C Note that, f′(0)=0, therefore, C=0. f(y)=3π8y2signumy+D Again, f(0)=0, therefore, D=0.
Hence, f(1)=∫∞0sin3xx3=3π8
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