Best solutions to - "I would like to show that1nπ∫−ππsin2(nx2)2sin2(x2)dx=1"

poznateqojh

Answered question

2022-04-08

I would like to show that

\frac{1}{n π} \int_{- π}^{π} \frac{\sin^{2} (n \frac{x}{2})}{2 \sin^{2} (\frac{x}{2})} d x = 1

awalkbyfaithbzu6

Beginner2022-04-09Added 21 answers

We have:

\frac{\sin (n z)}{\sin z} = \frac{e^{\in z} - e^{- \in z}}{e^{i z} - e^{- i z}} = \frac{e^{i z}}{e^{\in z}} \cdot \frac{e^{2 n i z} - 1}{e^{2 i z} - 1}

= e^{- (n - 1) i z} \sum_{k = 0}^{n - 1} e^{2 k i z}

so:

\frac{\sin (n \frac{x}{2})}{\sin (\frac{x}{2})} = e^{- \frac{n - 1}{2} i z} \sum_{k = 0}^{n - 1} e^{k i x}

and:

{(\frac{\sin (n \frac{x}{2})}{\sin (\frac{x}{2})})}^{2} = e^{- (n - 1) i z} \sum_{j, k = 0}^{n - 1} e^{(k + j) i x}

so:

\int_{- π}^{π} {(\frac{\sin (n \frac{x}{2})}{\sin (\frac{x}{2})})}^{2} d x = 2 π \cdot [(j, k) \in {[0, n - 1]}^{2} : j + k = n - 1]

and the claim:

\int_{- π}^{π} {(\frac{\sin (n \frac{x}{2})}{\sin (\frac{x}{2})})}^{2} d x = 2 π n

follows.

glajusv8iv

Beginner2022-04-10Added 10 answers

By some simple manipulations you can use what was shown in the related question:

\int_{- π}^{π} \frac{\sin^{2} (n x)}{\sin^{2} (x)} = 2 n π

(1)
So we want to use (1) to evaluate your integral. Note that by using

u \mapsto \frac{x}{2}

one has

\int_{- π}^{π} \frac{\sin^{2} (n \frac{x}{2})}{2 \sin^{2} (\frac{x}{2})} d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\sin^{2} (ν)}{\sin^{2} (u)} d u

(2)
It can be trivially shown that

\int_{0}^{\frac{π}{2}} \sin n x d x = \frac{1}{2} \int_{0}^{π} \sin n x d x

(3)
Just evaluate both sides. Using (3), eaution (2) can be written as

\int_{- π}^{π} \frac{\sin^{2} (n \frac{x}{2})}{2 \sin^{2} (\frac{x}{2})} d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\sin^{2} (ν)}{\sin^{2} (u)} d u

= \frac{1}{2} \int_{- π}^{π} \frac{\sin^{2} (ν)}{\sin^{2} (u)} d u = n π

As wanted

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