Integration,

• Probability and distribution function (2) (distribution function and density function), a liberal translation
  http://maldoror-ducasse.cocolog-nifty.com/blog/2010/02/2-7481.html
  This time, p j ≡p (x=x j) Σ j=1 XXINF p j =1 the condition for becoming is satisfied

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  这时， p j ≡p (x=x j) Σ j=1 XXINF p j =1成为的条件是满意的
  
  
• The perturbation theory in the quantum theory of fields (2)
  http://maldoror-ducasse.cocolog-nifty.com/blog/2010/11/2-0753.html
  From these shapes, the unitary characteristic of s (conservation of unitality= probability): s + s=ss + it is clear for =1 to be formed,

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  从这些形状， s (unitality=可能性的保护的单一的特征) ： s + s=ss +被形成=1是确切的，
  
  
• Japanese weblog
  http://maldoror-ducasse.cocolog-nifty.com/blog/2010/04/5-f9e1.html
  Because e= ∫ t1 t2 {s (r and t) n (t 0 ')} (dt/dt 0') dt 0 'you can write this, the dispatch point z (t 0') with formula of the energy emissivity regarding time t 0 ': d e/d t 0 '= {s (r and t) n (t 0')} (dt/dt 0') = {s (r and t) n (t 0 ')} Α (t 0') was obtained

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  由于e= ∫ t1 T2 {s (r和t) n (t 0 ‘)} (dt/dt 0’) dt 0 ‘您能写此，与能量发射性的惯例的急件点z (t 0’)关于时间t 0的’ ： d e/d t 0 ‘= {s (r和t) n (t 0’)} (dt/dt 0’) = {s (r和t) n (t 0’)} Î ‘ (t 0 ')获得了
  
  
• Be covariant quantization of electromagnetic field (3) (Nakanishi - Lautrap theory)
  http://maldoror-ducasse.cocolog-nifty.com/blog/2009/06/3-lautrap-2191.html
  As for this, b (x) =∫d 3 zd (x-z) ∂ z 0 ⇔ b (z), a Μ (x) =∫d 3 z [d (x-z) ∂ z 0 ⇔ a Μ (z) + (1-α) e (x-z) ∂ z 0 ⇔ ∂ z Μ b (z)]It is something which is based on the expression which becomes

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  关于此， b (x) =∫d 3 zd (x-z) ∂ z 0 ⇔ b (z)， Î œ (x) =∫d 3 z [d (x-z) ∂ z 0 ⇔ Î œ (z) + (1-α) e (x-z) ∂ z 0 ⇔ ∂ z Î œ b (z)]它是根据表示成为的事
  
  
• Japanese Letter
  http://maldoror-ducasse.cocolog-nifty.com/blog/2010/04/2-b969.html
  As for the effective method of solving this equation eigen value equation: Σ a k ba Δ (e a - e) it is to form the eigenfunction of k which fills up f aa = k a f ba

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  关于解决这个等式eigen价值等式有效的方法： Σ k ba Î ” (e a - e)它是形成填满f aa = k f ba k的特征函数
  
  
• weblog title
  http://maldoror-ducasse.cocolog-nifty.com/blog/2009/06/2-3e4f.html
  As for this, vacuum expectation of t product: . Δ f (x-y) ≡<0|t (Φ (x) Φ (y)))|0>=θ (x 0 - y 0) <0|Φ (x) Φ (y)|0>+θ (y 0 - x 0) <0|Φ (y) Φ (x)|In this which is the function which is given with the 0> Φ (x) = (2π) -3/2 ∫d 3 k (2ω k) -1/2 {a^ (k) exp (- ikx) +a^ + (k) exp (ikx)}When it substitutes,

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  关于此， t产品的真空期望： . Î ” f (X - Y的) ≡=θ (x 0 - y0) +θ (y0 - x 0) Φ (x) = (2π) -3/2 ∫d 3 k (2ω k) -1/2 {a^ (k) exp (- ikx) +a^ + (k) exp (ikx)}当它替代，