Plot of the Dawson integral function F(z) in the complex plane from -2-2i to 2+2i with colors created with Mathematica 13.1 function ComplexPlot3D In mathematics , the Dawson function  or Dawson integral [ 1] H. G. Dawson [ 2] sine transform  of the Gaussian function.
Definition 
The Dawson function, 
  
    
      
        F 
        ( 
        x 
        ) 
        = 
        
          D 
          
            + 
           
         
        ( 
        x 
        ) 
        , 
       
     
    {\displaystyle F(x)=D_{+}(x),} 
   
  The Dawson function, 
  
    
      
        
          D 
          
            − 
           
         
        ( 
        x 
        ) 
        , 
       
     
    {\displaystyle D_{-}(x),} 
   
  The Dawson function is defined as either:
  
    
      
        
          D 
          
            + 
           
         
        ( 
        x 
        ) 
        = 
        
          e 
          
            − 
            
              x 
              
                2 
               
             
           
         
        
          ∫ 
          
            0 
           
          
            x 
           
         
        
          e 
          
            
              t 
              
                2 
               
             
           
         
        d 
        t 
        , 
       
     
    {\displaystyle D_{+}(x)=e^{-x^{2}}\int _{0}^{x}e^{t^{2}}\,dt,} 
   
 
  
    
      
        F 
        ( 
        x 
        ) 
       
     
    {\displaystyle F(x)} 
   
 
  
    
      
        D 
        ( 
        x 
        ) 
        , 
       
     
    {\displaystyle D(x),} 
   
 
  
    
      
        
          D 
          
            − 
           
         
        ( 
        x 
        ) 
        = 
        
          e 
          
            
              x 
              
                2 
               
             
           
         
        
          ∫ 
          
            0 
           
          
            x 
           
         
        
          e 
          
            − 
            
              t 
              
                2 
               
             
           
         
        d 
        t 
        . 
         
     
    {\displaystyle D_{-}(x)=e^{x^{2}}\int _{0}^{x}e^{-t^{2}}\,dt.\!} 
   
 
The Dawson function is the one-sided Fourier–Laplace sine transform  of the Gaussian function ,
  
    
      
        
          D 
          
            + 
           
         
        ( 
        x 
        ) 
        = 
        
          
            1 
            2 
           
         
        
          ∫ 
          
            0 
           
          
            ∞ 
           
         
        
          e 
          
            − 
            
              t 
              
                2 
               
             
            
              / 
             
            4 
           
         
        sin 
         
        ( 
        x 
        t 
        ) 
        d 
        t 
        . 
       
     
    {\displaystyle D_{+}(x)={\frac {1}{2}}\int _{0}^{\infty }e^{-t^{2}/4}\,\sin(xt)\,dt.} 
   
 
It is closely related to the error function  erf, as
  
    
      
        
          D 
          
            + 
           
         
        ( 
        x 
        ) 
        = 
        
          
            
              
                π 
               
             
            2 
           
         
        
          e 
          
            − 
            
              x 
              
                2 
               
             
           
         
        erfi 
         
        ( 
        x 
        ) 
        = 
        − 
        
          
            
              i 
              
                
                  π 
                 
               
             
            2 
           
         
        
          e 
          
            − 
            
              x 
              
                2 
               
             
           
         
        erf 
         
        ( 
        i 
        x 
        ) 
       
     
    {\displaystyle D_{+}(x)={{\sqrt {\pi }} \over 2}e^{-x^{2}}\operatorname {erfi} (x)=-{i{\sqrt {\pi }} \over 2}e^{-x^{2}}\operatorname {erf} (ix)} 
   
 where erfi is the imaginary error function, erfi(x ) = −i  erf(ix ). 
  
    
      
        
          D 
          
            − 
           
         
        ( 
        x 
        ) 
        = 
        
          
            
              π 
             
            2 
           
         
        
          e 
          
            
              x 
              
                2 
               
             
           
         
        erf 
         
        ( 
        x 
        ) 
       
     
    {\displaystyle D_{-}(x)={\frac {\sqrt {\pi }}{2}}e^{x^{2}}\operatorname {erf} (x)} 
   
 
In terms of either erfi or the Faddeeva function  
  
    
      
        w 
        ( 
        z 
        ) 
        , 
       
     
    {\displaystyle w(z),} 
   
 complex plane :[ 3] 
  
    
      
        F 
        ( 
        z 
        ) 
        = 
        
          
            
              
                π 
               
             
            2 
           
         
        
          e 
          
            − 
            
              z 
              
                2 
               
             
           
         
        erfi 
         
        ( 
        z 
        ) 
        = 
        
          
            
              i 
              
                
                  π 
                 
               
             
            2 
           
         
        
          [ 
          
            
              e 
              
                − 
                
                  z 
                  
                    2 
                   
                 
               
             
            − 
            w 
            ( 
            z 
            ) 
           
          ] 
         
        , 
       
     
    {\displaystyle F(z)={{\sqrt {\pi }} \over 2}e^{-z^{2}}\operatorname {erfi} (z)={\frac {i{\sqrt {\pi }}}{2}}\left[e^{-z^{2}}-w(z)\right],} 
   
 
  
    
      
        
          D 
          
            + 
           
         
        ( 
        x 
        ) 
        = 
        F 
        ( 
        x 
        ) 
        = 
        
          
            
              π 
             
            2 
           
         
        Im 
         
        [ 
        w 
        ( 
        x 
        ) 
        ] 
       
     
    {\displaystyle D_{+}(x)=F(x)={\frac {\sqrt {\pi }}{2}}\operatorname {Im} [w(x)]} 
   
 
  
    
      
        
          D 
          
            − 
           
         
        ( 
        x 
        ) 
        = 
        i 
        F 
        ( 
        − 
        i 
        x 
        ) 
        = 
        − 
        
          
            
              π 
             
            2 
           
         
        
          [ 
          
            
              e 
              
                
                  x 
                  
                    2 
                   
                 
               
             
            − 
            w 
            ( 
            − 
            i 
            x 
            ) 
           
          ] 
         
       
     
    {\displaystyle D_{-}(x)=iF(-ix)=-{\frac {\sqrt {\pi }}{2}}\left[e^{x^{2}}-w(-ix)\right]} 
   
 
  
    
      
        x 
        . 
       
     
    {\displaystyle x.} 
   
 
For 
  
    
      
        
          | 
         
        x 
        
          | 
         
       
     
    {\displaystyle |x|} 
   
 F (x ) ≈ x .
  
    
      
        
          | 
         
        x 
        
          | 
         
       
     
    {\displaystyle |x|} 
   
 F (x ) ≈ 1/(2x ).
  
    
      
        F 
        ( 
        x 
        ) 
        = 
        
          ∑ 
          
            k 
            = 
            0 
           
          
            ∞ 
           
         
        
          
            
              ( 
              − 
              1 
              
                ) 
                
                  k 
                 
               
              
                2 
                
                  k 
                 
               
             
            
              ( 
              2 
              k 
              + 
              1 
              ) 
              ! 
              ! 
             
           
         
        
          x 
          
            2 
            k 
            + 
            1 
           
         
        = 
        x 
        − 
        
          
            2 
            3 
           
         
        
          x 
          
            3 
           
         
        + 
        
          
            4 
            15 
           
         
        
          x 
          
            5 
           
         
        − 
        ⋯ 
        , 
       
     
    {\displaystyle F(x)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}\,2^{k}}{(2k+1)!!}}\,x^{2k+1}=x-{\frac {2}{3}}x^{3}+{\frac {4}{15}}x^{5}-\cdots ,} 
   
 
  
    
      
        x 
       
     
    {\displaystyle x} 
   
 
  
    
      
        F 
        ( 
        x 
        ) 
        = 
        
          
            1 
            
              2 
              x 
             
           
         
        + 
        
          
            1 
            
              4 
              
                x 
                
                  3 
                 
               
             
           
         
        + 
        
          
            3 
            
              8 
              
                x 
                
                  5 
                 
               
             
           
         
        + 
        ⋯ 
        . 
       
     
    {\displaystyle F(x)={\frac {1}{2x}}+{\frac {1}{4x^{3}}}+{\frac {3}{8x^{5}}}+\cdots .} 
   
 
More precisely 
  
    
      
        
          | 
          
            F 
            ( 
            x 
            ) 
            − 
            
              ∑ 
              
                k 
                = 
                0 
               
              
                N 
               
             
            
              
                
                  ( 
                  2 
                  k 
                  − 
                  1 
                  ) 
                  ! 
                  ! 
                 
                
                  
                    2 
                    
                      k 
                      + 
                      1 
                     
                   
                  
                    x 
                    
                      2 
                      k 
                      + 
                      1 
                     
                   
                 
               
             
           
          | 
         
        ≤ 
        
          
            
              C 
              
                N 
               
             
            
              x 
              
                2 
                N 
                + 
                3 
               
             
           
         
        . 
       
     
    {\displaystyle \left|F(x)-\sum _{k=0}^{N}{\frac {(2k-1)!!}{2^{k+1}x^{2k+1}}}\right|\leq {\frac {C_{N}}{x^{2N+3}}}.} 
   
 
  
    
      
        n 
        ! 
        ! 
       
     
    {\displaystyle n!!} 
   
 double factorial . 
  
    
      
        F 
        ( 
        x 
        ) 
       
     
    {\displaystyle F(x)} 
   
 
  
    
      
        
          
            
              d 
              F 
             
            
              d 
              x 
             
           
         
        + 
        2 
        x 
        F 
        = 
        1 
         
     
    {\displaystyle {\frac {dF}{dx}}+2xF=1\,\!} 
   
 
  
    
      
        F 
        ( 
        0 
        ) 
        = 
        0. 
       
     
    {\displaystyle F(0)=0.} 
   
 
  
    
      
        F 
        ( 
        x 
        ) 
        = 
        
          
            1 
            
              2 
              x 
             
           
         
        , 
       
     
    {\displaystyle F(x)={\frac {1}{2x}},} 
   
 x  = ±0.92413887... (OEIS : A133841 F (x ) = ±0.54104422... (OEIS : A133842 
Inflection points follow for
  
    
      
        F 
        ( 
        x 
        ) 
        = 
        
          
            x 
            
              2 
              
                x 
                
                  2 
                 
               
              − 
              1 
             
           
         
        , 
       
     
    {\displaystyle F(x)={\frac {x}{2x^{2}-1}},} 
   
 x  = ±1.50197526... (OEIS : A133843 F (x ) = ±0.42768661... (OEIS : A245262 inflection point  at 
  
    
      
        x 
        = 
        0 
        , 
       
     
    {\displaystyle x=0,} 
   
 
  
    
      
        F 
        ( 
        x 
        ) 
        = 
        0. 
       
     
    {\displaystyle F(x)=0.} 
   
 
The Hilbert transform  of the Gaussian is defined as
  
    
      
        H 
        ( 
        y 
        ) 
        = 
        
          π 
          
            − 
            1 
           
         
        
          P 
          . 
          V 
          . 
         
         
        
          ∫ 
          
            − 
            ∞ 
           
          
            ∞ 
           
         
        
          
            
              e 
              
                − 
                
                  x 
                  
                    2 
                   
                 
               
             
            
              y 
              − 
              x 
             
           
         
        d 
        x 
       
     
    {\displaystyle H(y)=\pi ^{-1}\operatorname {P.V.} \int _{-\infty }^{\infty }{\frac {e^{-x^{2}}}{y-x}}\,dx} 
   
 
P.V. denotes the Cauchy principal value , and we restrict ourselves to real 
  
    
      
        y 
        . 
       
     
    {\displaystyle y.} 
   
 
  
    
      
        H 
        ( 
        y 
        ) 
       
     
    {\displaystyle H(y)} 
   
 
  
    
      
        1 
        
          / 
         
        u 
       
     
    {\displaystyle 1/u} 
   
 generalized function  or distribution, and use the Fourier representation
  
    
      
        
          
            1 
            u 
           
         
        = 
        
          ∫ 
          
            0 
           
          
            ∞ 
           
         
        d 
        k 
        sin 
         
        k 
        u 
        = 
        
          ∫ 
          
            0 
           
          
            ∞ 
           
         
        d 
        k 
        Im 
         
        
          e 
          
            i 
            k 
            u 
           
         
        . 
       
     
    {\displaystyle {1 \over u}=\int _{0}^{\infty }dk\,\sin ku=\int _{0}^{\infty }dk\,\operatorname {Im} e^{iku}.} 
   
 
With 
  
    
      
        1 
        
          / 
         
        u 
        = 
        1 
        
          / 
         
        ( 
        y 
        − 
        x 
        ) 
        , 
       
     
    {\displaystyle 1/u=1/(y-x),} 
   
 
  
    
      
        sin 
         
        ( 
        k 
        u 
        ) 
       
     
    {\displaystyle \sin(ku)} 
   
 
  
    
      
        x 
       
     
    {\displaystyle x} 
   
 
  
    
      
        π 
        H 
        ( 
        y 
        ) 
        = 
        Im 
         
        
          ∫ 
          
            0 
           
          
            ∞ 
           
         
        d 
        k 
        exp 
         
        [ 
        − 
        
          k 
          
            2 
           
         
        
          / 
         
        4 
        + 
        i 
        k 
        y 
        ] 
        
          ∫ 
          
            − 
            ∞ 
           
          
            ∞ 
           
         
        d 
        x 
        exp 
         
        [ 
        − 
        ( 
        x 
        + 
        i 
        k 
        
          / 
         
        2 
        
          ) 
          
            2 
           
         
        ] 
        . 
       
     
    {\displaystyle \pi H(y)=\operatorname {Im} \int _{0}^{\infty }dk\,\exp[-k^{2}/4+iky]\int _{-\infty }^{\infty }dx\,\exp[-(x+ik/2)^{2}].} 
   
 
We can shift the integral over 
  
    
      
        x 
       
     
    {\displaystyle x} 
   
 
  
    
      
        
          π 
          
            1 
            
              / 
             
            2 
           
         
        . 
       
     
    {\displaystyle \pi ^{1/2}.} 
   
 
  
    
      
        
          π 
          
            1 
            
              / 
             
            2 
           
         
        H 
        ( 
        y 
        ) 
        = 
        Im 
         
        
          ∫ 
          
            0 
           
          
            ∞ 
           
         
        d 
        k 
        exp 
         
        [ 
        − 
        
          k 
          
            2 
           
         
        
          / 
         
        4 
        + 
        i 
        k 
        y 
        ] 
        . 
       
     
    {\displaystyle \pi ^{1/2}H(y)=\operatorname {Im} \int _{0}^{\infty }dk\,\exp[-k^{2}/4+iky].} 
   
 
We complete the square with respect to 
  
    
      
        k 
       
     
    {\displaystyle k} 
   
 
  
    
      
        
          π 
          
            1 
            
              / 
             
            2 
           
         
        H 
        ( 
        y 
        ) 
        = 
        
          e 
          
            − 
            
              y 
              
                2 
               
             
           
         
        Im 
         
        
          ∫ 
          
            0 
           
          
            ∞ 
           
         
        d 
        k 
        exp 
         
        [ 
        − 
        ( 
        k 
        
          / 
         
        2 
        − 
        i 
        y 
        
          ) 
          
            2 
           
         
        ] 
        . 
       
     
    {\displaystyle \pi ^{1/2}H(y)=e^{-y^{2}}\operatorname {Im} \int _{0}^{\infty }dk\,\exp[-(k/2-iy)^{2}].} 
   
 
We change variables to 
  
    
      
        u 
        = 
        i 
        k 
        
          / 
         
        2 
        + 
        y 
        : 
       
     
    {\displaystyle u=ik/2+y:} 
   
 
  
    
      
        
          π 
          
            1 
            
              / 
             
            2 
           
         
        H 
        ( 
        y 
        ) 
        = 
        − 
        2 
        
          e 
          
            − 
            
              y 
              
                2 
               
             
           
         
        Im 
         
        i 
        
          ∫ 
          
            y 
           
          
            i 
            ∞ 
            + 
            y 
           
         
        d 
        u 
          
        
          e 
          
            
              u 
              
                2 
               
             
           
         
        . 
       
     
    {\displaystyle \pi ^{1/2}H(y)=-2e^{-y^{2}}\operatorname {Im} i\int _{y}^{i\infty +y}du\ e^{u^{2}}.} 
   
 
The integral can be performed as a contour integral around a rectangle in the complex plane. Taking the imaginary part of the result gives
  
    
      
        H 
        ( 
        y 
        ) 
        = 
        2 
        
          π 
          
            − 
            1 
            
              / 
             
            2 
           
         
        F 
        ( 
        y 
        ) 
       
     
    {\displaystyle H(y)=2\pi ^{-1/2}F(y)} 
   
 
  
    
      
        F 
        ( 
        y 
        ) 
       
     
    {\displaystyle F(y)} 
   
 
The Hilbert transform of 
  
    
      
        
          x 
          
            2 
            n 
           
         
        
          e 
          
            − 
            
              x 
              
                2 
               
             
           
         
       
     
    {\displaystyle x^{2n}e^{-x^{2}}} 
   
 
  
    
      
        
          H 
          
            n 
           
         
        = 
        
          π 
          
            − 
            1 
           
         
        
          P 
          . 
          V 
          . 
         
         
        
          ∫ 
          
            − 
            ∞ 
           
          
            ∞ 
           
         
        
          
            
              
                x 
                
                  2 
                  n 
                 
               
              
                e 
                
                  − 
                  
                    x 
                    
                      2 
                     
                   
                 
               
             
            
              y 
              − 
              x 
             
           
         
        d 
        x 
        . 
       
     
    {\displaystyle H_{n}=\pi ^{-1}\operatorname {P.V.} \int _{-\infty }^{\infty }{\frac {x^{2n}e^{-x^{2}}}{y-x}}\,dx.} 
   
 
Introduce
  
    
      
        
          H 
          
            a 
           
         
        = 
        
          π 
          
            − 
            1 
           
         
        
          P 
          . 
          V 
          . 
         
         
        
          ∫ 
          
            − 
            ∞ 
           
          
            ∞ 
           
         
        
          
            
              e 
              
                − 
                a 
                
                  x 
                  
                    2 
                   
                 
               
             
            
              y 
              − 
              x 
             
           
         
        d 
        x 
        . 
       
     
    {\displaystyle H_{a}=\pi ^{-1}\operatorname {P.V.} \int _{-\infty }^{\infty }{e^{-ax^{2}} \over y-x}\,dx.} 
   
 
The 
  
    
      
        n 
       
     
    {\displaystyle n} 
   
 
  
    
      
        
          
            
              
                ∂ 
                
                  n 
                 
               
              
                H 
                
                  a 
                 
               
             
            
              ∂ 
              
                a 
                
                  n 
                 
               
             
           
         
        = 
        ( 
        − 
        1 
        
          ) 
          
            n 
           
         
        
          π 
          
            − 
            1 
           
         
        
          P 
          . 
          V 
          . 
         
         
        
          ∫ 
          
            − 
            ∞ 
           
          
            ∞ 
           
         
        
          
            
              
                x 
                
                  2 
                  n 
                 
               
              
                e 
                
                  − 
                  a 
                  
                    x 
                    
                      2 
                     
                   
                 
               
             
            
              y 
              − 
              x 
             
           
         
        d 
        x 
        . 
       
     
    {\displaystyle {\partial ^{n}H_{a} \over \partial a^{n}}=(-1)^{n}\pi ^{-1}\operatorname {P.V.} \int _{-\infty }^{\infty }{\frac {x^{2n}e^{-ax^{2}}}{y-x}}\,dx.} 
   
 
We thus find
  
    
      
        
          
            
              
                H 
                
                  n 
                 
               
              = 
              ( 
              − 
              1 
              
                ) 
                
                  n 
                 
               
              
                
                  
                    
                      ∂ 
                      
                        n 
                       
                     
                    
                      H 
                      
                        a 
                       
                     
                   
                  
                    ∂ 
                    
                      a 
                      
                        n 
                       
                     
                   
                 
               
             
            | 
           
          
            a 
            = 
            1 
           
         
        . 
       
     
    {\displaystyle \left.H_{n}=(-1)^{n}{\frac {\partial ^{n}H_{a}}{\partial a^{n}}}\right|_{a=1}.} 
   
 
The derivatives are performed first, then the result evaluated at 
  
    
      
        a 
        = 
        1. 
       
     
    {\displaystyle a=1.} 
   
 
  
    
      
        
          H 
          
            a 
           
         
        = 
        2 
        
          π 
          
            − 
            1 
            
              / 
             
            2 
           
         
        F 
        ( 
        y 
        
          
            a 
           
         
        ) 
        . 
       
     
    {\displaystyle H_{a}=2\pi ^{-1/2}F(y{\sqrt {a}}).} 
   
 
  
    
      
        
          F 
          ′ 
         
        ( 
        y 
        ) 
        = 
        1 
        − 
        2 
        y 
        F 
        ( 
        y 
        ) 
        , 
       
     
    {\displaystyle F'(y)=1-2yF(y),} 
   
 
  
    
      
        
          H 
          
            n 
           
         
        = 
        
          P 
          
            1 
           
         
        ( 
        y 
        ) 
        + 
        
          P 
          
            2 
           
         
        ( 
        y 
        ) 
        F 
        ( 
        y 
        ) 
       
     
    {\displaystyle H_{n}=P_{1}(y)+P_{2}(y)F(y)} 
   
 
  
    
      
        
          P 
          
            1 
           
         
       
     
    {\displaystyle P_{1}} 
   
 
  
    
      
        
          P 
          
            2 
           
         
       
     
    {\displaystyle P_{2}} 
   
 
  
    
      
        
          H 
          
            1 
           
         
        = 
        − 
        
          π 
          
            − 
            1 
            
              / 
             
            2 
           
         
        y 
        + 
        2 
        
          π 
          
            − 
            1 
            
              / 
             
            2 
           
         
        
          y 
          
            2 
           
         
        F 
        ( 
        y 
        ) 
        . 
       
     
    {\displaystyle H_{1}=-\pi ^{-1/2}y+2\pi ^{-1/2}y^{2}F(y).} 
   
 
  
    
      
        
          H 
          
            n 
           
         
       
     
    {\displaystyle H_{n}} 
   
 recurrence relation  (for 
  
    
      
        n 
        ≥ 
        0 
       
     
    {\displaystyle n\geq 0} 
   
 
  
    
      
        
          H 
          
            n 
            + 
            1 
           
         
        ( 
        y 
        ) 
        = 
        
          y 
          
            2 
           
         
        
          H 
          
            n 
           
         
        ( 
        y 
        ) 
        − 
        
          
            
              ( 
              2 
              n 
              − 
              1 
              ) 
              ! 
              ! 
             
            
              
                
                  π 
                 
               
              
                2 
                
                  n 
                 
               
             
           
         
        y 
        . 
       
     
    {\displaystyle H_{n+1}(y)=y^{2}H_{n}(y)-{\frac {(2n-1)!!}{{\sqrt {\pi }}2^{n}}}y.} 
   
 
See also 
References 
^ Temme, N. M. (2010), "Error Functions, Dawson's and Fresnel Integrals" , in Olver, Frank W. J. ; Lozier, Daniel M.; Boisvert, Ronald F.; Clark, Charles W. (eds.), NIST Handbook of Mathematical Functions ISBN  978-0-521-19225-5 , MR  2723248  ^ Dawson, H. G. (1897). "On the Numerical Value of 
  
    
      
        
          
            ∫ 
            
              0 
             
            
              h 
             
           
          exp 
           
          ( 
          
            x 
            
              2 
             
           
          ) 
          d 
          x 
         
       
     
    {\displaystyle \textstyle \int _{0}^{h}\exp(x^{2})\,dx} 
   
  . Proceedings of the London Mathematical Society . s1-29 (1): 519– 522. doi :10.1112/plms/s1-29.1.519 . ^ Mofreh R. Zaghloul and Ahmed N. Ali, "Algorithm 916: Computing the Faddeyeva and Voigt Functions ," ACM Trans. Math. Soft.  38  (2), 15 (2011). Preprint available at arXiv:1106.0151 . 
  
External links