I would like to show that \(\displaystyle{\frac{{{1}}}{{{n}\pi}}}{\int_{{-\pi}}^{{\pi}}}{\frac{{{{\sin}^{{2}}{\left({n}\frac{{x}}{{2}}\right)}}}}{{{2}{{\sin}^{{2}}{\left(\frac{{x}}{{2}}\right)}}}}}{\left.{d}{x}\right.}={1}\)

poznateqojh

poznateqojh

Answered question

2022-04-08

I would like to show that
1nπππsin2(nx2)2sin2(x2)dx=1

Answer & Explanation

awalkbyfaithbzu6

awalkbyfaithbzu6

Beginner2022-04-09Added 21 answers

We have:
sin(nz)sinz=ezezeizeiz=eizeze2niz1e2iz1
=e(n1)izk=0n1e2kiz
so:
sin(nx2)sin(x2)=en12izk=0n1ekix
and:
(sin(nx2)sin(x2))2=e(n1)izj,k=0n1e(k+j)ix
so:
ππ(sin(nx2)sin(x2))2dx=2π[(j,k)[0,n1]2:j+k=n1]
and the claim:
ππ(sin(nx2)sin(x2))2dx=2πn
follows.
glajusv8iv

glajusv8iv

Beginner2022-04-10Added 10 answers

By some simple manipulations you can use what was shown in the related question:
ππsin2(nx)sin2(x)=2nπ (1)
So we want to use (1) to evaluate your integral. Note that by using ux2 one has
ππsin2(nx2)2sin2(x2)dx=π2π2sin2(ν)sin2(u)du (2)
It can be trivially shown that
0π2sinnxdx=120πsinnxdx (3)
Just evaluate both sides. Using (3), eaution (2) can be written as
ππsin2(nx2)2sin2(x2)dx=π2π2sin2(ν)sin2(u)du
=12ππsin2(ν)sin2(u)du=nπ
As wanted

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