PWGLF/FORWARD/doc/landgaus.tex

   1 \documentclass[11pt]{article}
   2 \usepackage{geometry}
   3 \usepackage{amsmath}
   4 \usepackage{amstext}
   5 \begin{document}
   6
   7
   8 The Landau function $L$ is a function of $x$ with the parameters
   9 $\Delta_p$ and $\xi$.  It can be written as
  10 \begin{align}
  11   \phi(u) &= \frac1\pi\int_0^\infty dt\,\sin(\pi t) e^{-t\log{t}-ut}
  12   \label{eq:phi}\\
  13   L(x;\Delta_p,\xi) &= \frac1\xi
  14   \phi\left(\frac{x-\Delta_p}{\xi}\right)\label{eq:land}
  15 \end{align}
  16 %%
  17 A Landau convolved with a Gaussian is given by
  18 \begin{align}
  19   L_{G}(x;\Delta_p,\xi,\sigma) &= \int_{-\infty}^{+\infty}
  20   ds\, L(s;\Delta_p,\xi) \frac{1}{\sqrt{2\pi}\sigma}
  21   e^{-\frac{(x-s)^2}{2\sigma^2}}\nonumber\\
  22   &=
  23   \frac{1}{\sqrt{2\pi}\sigma\pi\xi}\int_{-\infty}^{+\infty}ds\,\int_0^\infty
  24   dt\,\sin(\pi t)e^{-t\log{t}-\frac{s-\Delta_p}{\xi}t}
  25   e^{-\frac{(x-s)^2}{2\sigma^2}}\nonumber\\
  26   &= \frac{1}{\sqrt{2\pi}\sigma\pi\xi}\int_0^\infty
  27   dt\,\sin(\pi t) e^{-t\log{t}+\frac{\Delta_p}{\xi}t}\int_{-\infty}^{+\infty}
  28   ds\,e^{-\frac{(x-s)^2}{2\sigma^2}}e^{-\frac{st}{\xi}}\label{eq:lg}
  29 \end{align}
  30 %%
  31 Carrying out the inner integral of \eqref{eq:lg},
  32 %%
  33 \begin{align}
  34   \int_{-\infty}^{+\infty}
  35   ds\,e^{-\frac{(x-s)^2}{2\sigma^2}}e^{-\frac{st}{\xi}} &=
  36   \int_{-\infty}^{+\infty}
  37   ds\,e^{\frac{-s^2-x^2+2xs}{2\sigma^s}-\frac{st}{\xi}}\nonumber\\
  38   &= \int_{-\infty}^{+\infty}
  39   ds\,e^{-\frac{s^2}{2\sigma^2}+\left(\frac{2x}{2\sigma^2}-\frac{t}{\xi}\right)s-\frac{x^2}{2\sigma^2}}\nonumber\\
  40   &= \int_{-\infty}^{+\infty}
  41   ds\,e^{-\frac{1}{2\sigma^2}\left[s^2+\left(\frac{2\sigma^2t}{\xi}-2x\right)s\right]-\frac{x^2}{2\sigma^2}}\nonumber\\
  42   \intertext{Completing the square}
  43   %%
  44   &= \int_{-\infty}^{+\infty}
  45   ds\,e^{-\frac{1}{2\sigma^2}\left[s^2+2\left(\frac{\sigma^2t}{\xi}-x\right)s
  46       +\left(\frac{\sigma^2t}{\xi}-x\right)^2
  47       -\left(\frac{\sigma^2t}{\xi}-x\right)^2\right]-\frac{x^2}{2\sigma^2}}\nonumber\\
  48   &= \int_{-\infty}^{+\infty}
  49   ds\,e^{-\frac{1}{2\sigma^2}\left[s + \left(\frac{\sigma^2t}{\xi}-x\right)\right]^2
  50     +\frac{1}{2\sigma^2}\left(\frac{\sigma^2t}{\xi}-x\right)^2
  51     -\frac{x^2}{2\sigma^2}}\nonumber\\
  52   \intertext{Recognizing this as a Gaussian integral, we get}
  53   %% Maxima:
  54   %% integrate(%e^(-1/(2*sigma^2)*(s+sigma^2*t/xi-x)^2 +
  55   %%               1/(2*sigma^2)*(sigma^2*t/xi-x)^2-x^2/(2*sigma^2)),s,-inf,inf);
  56   &= \sqrt{2\pi}\sigma
  57    e^{-\frac{1}{2\xi^2}\left(2tx\xi-\sigma^2t^2\right)}\nonumber\\
  58    &= \sqrt{2\pi}\sigma e^{\frac{t}{2\xi}\left(\frac{\sigma^2t}{\xi}-2x\right)}
  59   \label{eq:inner}
  60 \end{align}
  61 %%
  62 Substituting \eqref{eq:inner} back into \eqref{eq:lg} yields
  63 %%
  64 \begin{align}
  65   L_{G}(x;\Delta_p,\xi,\sigma) &= \frac{1}{\sqrt{2\pi}\sigma\pi\xi}
  66    \int_0^{\infty}dt\,e^{-t\log{t} + \frac{\Delta_p t}{\xi}}\sin{\pi t}
  67    \sqrt{2\pi}\sigma e^{\frac{t}{2\xi}\left(\frac{\sigma^2t}{\xi}-2x\right)}\nonumber\\
  68    &= \frac{1}{\pi\xi}\int_o^{\infty}dt\,e^{-t\log{t} + \frac{\Delta_pt}{\xi}+
  69      \frac{t}{2\xi}\left(\frac{\sigma^2t}{\xi}-2x\right)}\sin{\pi
  70      t}\nonumber\\
  71    &= \frac{1}{\pi\xi}\int_o^{\infty}dt\,e^{-t\log{t} - \frac{(x
  72        -\Delta_p)t}{\xi} + \frac12\left(\frac{\sigma}{\xi}t\right)^2}\sin{\pi
  73        t}\quad.
  74 \end{align}
  75 %%
  76 Defining $v=\frac{\sigma}{\xi}$, and
  77 %%
  78 \begin{align}
  79   \Phi(u;v) &= \frac{1}{\pi}\int_0^{\infty}dt\,e^{-t\log{t} - ut +
  80     \frac12v^2t^2}\sin{\pi t}\label{eq:Phi}\quad,
  81 \end{align}
  82 we can write \eqref{eq:lg} as
  83 \begin{align}
  84   L_{G}(x;\Delta_p,\xi,\sigma) &=
  85   \frac1\xi\Phi\left(\frac{x-\Delta_p}{\xi},\frac\sigma\xi\right)\label{eq:lg2}
  86 \end{align}
  87 The normalization follows from the fact that
  88 $u=\frac{x-\Delta_p}{\xi}$ and $\frac{dv}{du}=f(u)$ so that
  89 \begin{align*}
  90   du &= \frac{dx}{\xi}\quad\text{and}\\
  91   \frac{dv}{dx/\xi} &= f\left(\frac{x-\Delta_p}{\xi}\right)\\
  92   \intertext{so that}
  93   \frac{dv}{dx} &= \frac1\xi f\left(\frac{x-\Delta_p}{\xi}\right)
  94 \end{align*}
  95
  96 %%
  97 %% Maxima input:
  98 %%  diff(integrate(1/%pi*%e^(-t * log(t)-u*t+1/2*v^2*t^2)*sin(%pi *
  99 %%  t), t, 0, inf),u);
 100 %% v is non-zero
 101 %% u is positive
 102 Differentiating \eqref{eq:Phi} with respect to $u$ gives
 103 \begin{align}
 104   \frac{d\Phi}{du} &=
 105   \frac{1}{\pi}
 106   \int_0^{\infty}dt\,e^{-t\log{t}-tu+\frac12t^2v^2}t\sin{\pi t}
 107 \end{align}
 108
 109 Now turning to the maximum of \eqref{eq:phi} which is known to be at
 110 $u=u_p\neq0$ we get that the most probable value $x=x_p$ of
 111 \eqref{eq:land} is given by
 112 \begin{align}
 113   \frac{x_p-\Delta_p}{\xi} &= u_p\nonumber\\
 114   \intertext{so that}
 115   x_p &= u_p\xi + \Delta_p\quad\text{and}\nonumber\\
 116   \Delta_p &= x_p - \xi u_p
 117 \end{align}
 118
 119 The maximum of \eqref{eq:Phi} must be a function solely of $v$, and we
 120 have the most probable value $x=x'_p$ of \eqref{eq:lg2}
 121 \begin{align}
 122   f(v) &= \frac{x'_p-\Delta_p}{\xi} \nonumber\\
 123   &= \frac{x'_p - x_p + \xi u_p}{\xi} \nonumber\\
 124   f(v) - u_p &=  \frac{x'_p - x_p}{\xi}\nonumber\\
 125   x_p'-x_p &= \xi\left(f(v)-u_p\right)\nonumber\\
 126   x_p' &= \xi\left(f(v)-u_p\right)+x_p
 127   \intertext{and it follows that}
 128   x_p' &= \xi g(v)
 129 \end{align}
 130
 131
 132
 133 \end{document}
 134
 135 %  LocalWords:  convolved