I am looking for a short proof that ∫0∞(sin⁡xx)2dx=π2 What do you think?

eiraszero11cu

eiraszero11cu

Answered

2021-12-31

I am looking for a short proof that
0(sinxx)2dx=π2
What do you think?

Answer & Explanation

twineg4

twineg4

Expert

2022-01-01Added 33 answers

Well, it's not hard to reduce this integral to f(x)=max{0,1|x|}. It is easy to calculate the Fourier transform
f^(ξ)=f(x)eixξ=(sin(ξ2)ξ2)2
Taking the inverse Fourier transform, we get
(sin(ξ2)ξ2)2eixξdξ=2πf(x),
and the result follows.
The second integral can be computed in a similar way. Just take f(x)=χ11(x) (the indicator function of the interval [1,1])
Karen Robbins

Karen Robbins

Expert

2022-01-02Added 49 answers

0sin2xx2dx=0121cos2xx2dx
=01cosxx2dx
=121cosxx2dx
=121cosxx2dx
=12R1e{ix}x2dx
=12R1eix+ix1+x2x2dx
=12R1eix+ix1+x2x2dx
Now close the contour in the upper half plane, enclosing the pole at x=i with residue 12i, yielding
0sin2xx2dx=122π12i=π2
Vasquez

Vasquez

Expert

2022-01-09Added 457 answers

From squaring the identity
sinnxsinx=einxeinxeixeix=k=0n1e(2kn+1)ix
and integrating we get
nπ=π/2π/2sin2nxsin2xdx
Let
In=π/2π/2sin2nxnx2dx=nπ/2nπ/2sin2yy2dy
Then
πIn=1nπ/2π/2sin2nx(csc2xx2x)dx
and so
|πIn|1nπ/2π/2|csc2xx2|dx=O(1/n)
as xcsc2xx2 extends to a continuous function on [π/2,π/2]. Hence Inπ as n and
π=sin2yy2dy

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