In complex analysis, a branch of mathematics, the residue at infinity is a residue of a holomorphic function on an annulus having an infinite external radius.  The infinity 
  
    
      
        ∞
      
    
    {\displaystyle \infty }
  
 is a point added to the local space 
  
    
      
        
          C
        
      
    
    {\displaystyle \mathbb {C} }
  
 in order to render it compact (in this case it is a one-point compactification). This space denoted 
  
    
      
        
          
            
              
                C
              
              ^
            
          
        
      
    
    {\displaystyle {\hat {\mathbb {C} }}}
  
 is isomorphic to the Riemann sphere. One can use the residue at infinity to calculate some integrals.
 Given a holomorphic function f on an annulus 
  
    
      
        A
        (
        0
        ,
        R
        ,
        ∞
        )
      
    
    {\displaystyle A(0,R,\infty )}
  
 (centered at 0, with inner radius 
  
    
      
        R
      
    
    {\displaystyle R}
  
 and infinite outer radius), the residue at infinity of the function f can be defined in terms of the usual residue as follows:
 Thus, one can transfer the study of 
  
    
      
        f
        (
        z
        )
      
    
    {\displaystyle f(z)}
  
 at infinity to the study of 
  
    
      
        f
        (
        1
        
          /
        
        z
        )
      
    
    {\displaystyle f(1/z)}
  
 at the origin.
 Note that 
  
    
      
        ∀
        r
        >
        R
      
    
    {\displaystyle \forall r>R}
  
, we have
 Since, for holomorphic functions the sum of the residues at the isolated singularities plus the residue at infinity is zero, it can be expressed as:
 
  
    
      
        Res
        ⁡
        (
        f
        (
        z
        )
        ,
        ∞
        )
        =
        −
        
          ∑
          
            k
          
        
        Res
        ⁡
        
          (
          
            f
            
              (
              z
              )
            
            ,
            
              a
              
                k
              
            
          
          )
        
        .
      
    
    {\displaystyle \operatorname {Res} (f(z),\infty )=-\sum _{k}\operatorname {Res} \left(f\left(z\right),a_{k}\right).}
  

 One might first guess that the definition of the residue of 
  
    
      
        f
        (
        z
        )
      
    
    {\displaystyle f(z)}
  
 at infinity should just be the residue of 
  
    
      
        f
        (
        1
        
          /
        
        z
        )
      
    
    {\displaystyle f(1/z)}
  
 at 
  
    
      
        z
        =
        0
      
    
    {\displaystyle z=0}
  
. However, the reason that we consider instead 
  
    
      
        −
        
          
            1
            
              z
              
                2
              
            
          
        
        f
        
          (
          
            
              1
              z
            
          
          )
        
      
    
    {\displaystyle -{\frac {1}{z^{2}}}f\left({\frac {1}{z}}\right)}
  
 is that one does not take residues of functions, but of differential forms, i.e. the residue of 
  
    
      
        f
        (
        z
        )
        d
        z
      
    
    {\displaystyle f(z)dz}
  
 at infinity is the residue of 
  
    
      
        f
        
          (
          
            
              1
              z
            
          
          )
        
        d
        
          (
          
            
              1
              z
            
          
          )
        
        =
        −
        
          
            1
            
              z
              
                2
              
            
          
        
        f
        
          (
          
            
              1
              z
            
          
          )
        
        d
        z
      
    
    {\displaystyle f\left({\frac {1}{z}}\right)d\left({\frac {1}{z}}\right)=-{\frac {1}{z^{2}}}f\left({\frac {1}{z}}\right)dz}
  
 at 
  
    
      
        z
        =
        0
      
    
    {\displaystyle z=0}
  
.