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ca4a424f | 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 |