A half range Fourier series is a Fourier series defined on an interval 
  
    
      
        [
        0
        ,
        L
        ]
      
    
    
   instead of the more common 
  
    
      
        [
        −
        L
        ,
        L
        ]
      
    
    
  , with the implication that the analyzed function 
  
    
      
        f
        (
        x
        )
        ,
        x
        ∈
        [
        0
        ,
        L
        ]
      
    
    
   should be extended to 
  
    
      
        [
        −
        L
        ,
        0
        ]
      
    
    
   as either an even (f(-x)=f(x)) or odd function (f(-x)=-f(x)). This allows the expansion of the function in a series solely of sines (odd) or cosines (even). The choice between odd and even is typically motivated by boundary conditions associated with a differential equation satisfied by 
  
    
      
        f
        (
        x
        )
      
    
    
  .
Example
Calculate the half range Fourier sine series for the function 
  
    
      
        f
        (
        x
        )
        =
        cos
        
        (
        x
        )
      
    
    
   where 
  
    
      
        0
        <
        x
        <
        π
      
    
    
  .
Since we are calculating a sine series, 
  
    
      
        
          a
          
            n
          
        
        =
        0
         
        
        ∀
        n
      
    
    
   Now, 
  
    
      
        
          b
          
            n
          
        
        =
        
          
            2
            π
          
        
        
          ∫
          
            0
          
          
            π
          
        
        cos
        
        (
        x
        )
        sin
        
        (
        n
        x
        )
        
        
          d
        
        x
        =
        
          
            
              2
              n
              (
              (
              −
              1
              
                )
                
                  n
                
              
              +
              1
              )
            
            
              π
              (
              
                n
                
                  2
                
              
              −
              1
              )
            
          
        
        
        ∀
        n
        ≥
        2
      
    
    
  
When n is odd, 
  
    
      
        
          b
          
            n
          
        
        =
        0
         
      
    
    
   When n is even, 
  
    
      
        
          b
          
            n
          
        
        =
        
          
            
              4
              n
            
            
              π
              (
              
                n
                
                  2
                
              
              −
              1
              )
            
          
        
      
    
    
   thus 
  
    
      
        
          b
          
            2
            k
          
        
        =
        
          
            
              8
              k
            
            
              π
              (
              4
              
                k
                
                  2
                
              
              −
              1
              )
            
          
        
      
    
    
  
With the special case 
  
    
      
        
          b
          
            1
          
        
        =
        0
      
    
    
  , hence the required Fourier sine series is
  
    
      
        cos
        
        (
        x
        )
        =
        
          
            
              8
              π
            
          
          
            ∑
            
              n
              =
              1
            
            
              ∞
            
          
          
            
              n
              
                (
                4
                
                  n
                  
                    2
                  
                
                −
                1
                )
              
            
          
          sin
          
          (
          2
          n
          x
          )