Mathematica 2015

𝐿

1 𝐿

 𝐿



=

∫ ( )



−𝐿

∫ | |  𝜋 −𝜋 ∫   𝜋 0

1 2

=

=

beacause it is even

= 2 𝜋

( 𝑥𝑠𝑖𝑥

+ 
𝑠𝑥 2

) 0 𝜋

Using integration by part:

2  2

=

( − 1)

= − 4

So when n = 1,3,5…… 

𝜋 2

When n = 2,4,6 …… 

= 0

∫ | | 𝜋 −𝜋

= 1 𝜋

As to  0 ,  0

∫  𝜋 0

2

=

=

Then we consider the coefficient 

𝐿

1 𝐿

 𝐿



=

∫ ( )𝑖



−𝐿

∫ | |𝑖  𝜋 −𝜋

1

=

There is no need to use integration by part in order to get the answer. Because if we careful enough, we can easily see an interesting property of function | |𝑖 on the interval [−, ] that it is odd. Thus we can get  = 0 immediately. All considered, we express f(x) as the Fourier series:

∞

 0 2

 𝐿

 𝐿

( ) =

+∑(



+ 

𝑖

)

=1

∞

2

2  2

( − 1)

=

+∑

=1

2

4

1 1 2

1 3 2

1 5 2

=

−

(

 +

2 +

3 + ⋯ )

∞

2

4

1 (2 − 1) 2

=

−

∑

(2 − 1)

=1 Now, we have been familiar with writing a periodic function in form of Fourier series. Let’s again look at the final expression.

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