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Integral w/ DiracDelta gives different results on different versions

Chris Sims

Posted 7 years ago

POSTED BY: Chris Sims

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Alexander Trounev

Alexander Trounev, A&E Trounev IT Consulting

Posted 7 years ago

In[1]:= $Version Out[1]= "10.2.0 for Microsoft Windows (64-bit) (July 28, 2015)" In[2]:= Assuming[ Element[a, Reals] && Element[y, Reals] && Element[z, Reals], Integrate[ DiracDelta[y - (x + a)/z] PDF[NormalDistribution[0, s], x], {x, -Infinity, Infinity}]] Out[2]= (E^(-((a - y z)^2/(2 s^2))) Abs[z])/(Sqrt[2 \[Pi]] s) In[1]:= $Version Out[1]= "11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)" In[2]:= Assuming[ Element[a, Reals] && Element[y, Reals] && Element[z, Reals], Integrate[ DiracDelta[y - (x + a)/z] PDF[NormalDistribution[0, s], x], {x, -Infinity, Infinity}]] Out[2]= (E^(-((a - y z)^2/(2 s^2))) Abs[z])/(Sqrt[2 \[Pi]] s) $Version "8.0 for Microsoft Windows (64-bit) (October 7, 2011)" Assuming[Element[a, Reals] && Element[y, Reals] && Element[z, Reals], Integrate[ DiracDelta[y - (x + a)/z] PDF[NormalDistribution[0, s], x], {x, -Infinity, Infinity}]] (E^(-((a - y z)^2/(2 s^2))) Abs[z])/(Sqrt[2 \[Pi]] s)

In[1]:= $Version

Out[1]= "10.2.0 for Microsoft Windows (64-bit) (July 28, 2015)"

In[2]:= Assuming[
 Element[a, Reals] && Element[y, Reals] && Element[z, Reals], 
 Integrate[
  DiracDelta[y - (x + a)/z] PDF[NormalDistribution[0, s], 
    x], {x, -Infinity, Infinity}]]

Out[2]= (E^(-((a - y z)^2/(2 s^2))) Abs[z])/(Sqrt[2 \[Pi]] s)
In[1]:= $Version

Out[1]= "11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)"

In[2]:= Assuming[
 Element[a, Reals] && Element[y, Reals] && Element[z, Reals], 
 Integrate[
  DiracDelta[y - (x + a)/z] PDF[NormalDistribution[0, s], 
    x], {x, -Infinity, Infinity}]]

Out[2]= (E^(-((a - y z)^2/(2 s^2))) Abs[z])/(Sqrt[2 \[Pi]] s)
$Version

"8.0 for Microsoft Windows (64-bit) (October 7, 2011)"

Assuming[Element[a, Reals] && Element[y, Reals] && Element[z, Reals], 
 Integrate[
  DiracDelta[y - (x + a)/z] PDF[NormalDistribution[0, s], 
    x], {x, -Infinity, Infinity}]]

(E^(-((a - y z)^2/(2 s^2))) Abs[z])/(Sqrt[2 \[Pi]] s)

POSTED BY: Alexander Trounev