Evaluate the integral \int_{-\pi}^\pi|\sum_{n=1}^\infty\frac{1}{2^n}e^{inx}|^2dx

kuntungw3

kuntungw3

Answered question

2022-01-22

Evaluate the integral
ππ|n=112nex|2dx

Answer & Explanation

Flickkorbma

Flickkorbma

Beginner2022-01-23Added 17 answers

Define
g(x)=n=1ex2n
This is absolutely convergent everywhere, and defines a continuous function (by, for example, the Weierstrass M-test). The Fourier coefficients of g are easy to compute: Define
cn=12πππg(x)exdx
By orthogonality of the functions ex, we find that
cn=12πππk=1ei(kn)x2kdx=12π2π2n=12n
Then Parsevals
Devyn Figueroa

Devyn Figueroa

Beginner2022-01-24Added 10 answers

ππ|n=112nex|2=ππn=112nexm=112meimxdx=m,n=1ππ12m+nei(mn)xdx
=n=1=2π22n=2π4114=2π3
since
ππ12m+nei(mn)xdx=2π2m+nσm,n
RizerMix

RizerMix

Expert2022-01-27Added 656 answers

ππ|n=112neinx|2dx=ππ|eix/21eix/2|2dx =ππdx[2cos(x)]2+sin2(x) =20πdx54cos(x) =20154(1t2)/(1+t2)2dt1+t2 =4019t2+1dt =4301t2+1dt =43limtarctan(t) =43π2=23π

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?