Integrating over Dirac delta












2












$begingroup$


TL;DR: How does one show that $(clubsuit)$ holds.




Some context and how I arrived at my problem. In a Thermodynamics problem set I was asked to calculate the partition function of the ideal gas in the microcanonic ensemble using the Hamiltionian
$$mathcal{H} = frac{1}{2m}sum_{i=1}^N p_i^2.$$



The problem seems quite straight forward, but I fail at a certain point when trying to solve the integral:



$$begin{align}Z &= int_{mathbb{R}^{6N}} frac{d^{3N}pd^{3N}q}{h^{3N}N!}deltaleft(E-mathcal{H}right)\
&= frac{1}{h^{3N}N!}int_{mathbb{R}^{3N}}d^{3N}q int_{mathbb{R}^{3N}} d^{3N}p, deltaleft(E-sum_{i=1}^N frac{p_i^2}{2m}right)\
&= frac{V^N(2m)^{3N/2}}{h^{3N}N!}int_{mathbb{R}^{3N}}d^{3N}u, delta left(E-sum_{i=1}^Nu_i^2right)\
&= frac{V^N(2m)^{3N/2}}{h^{3N}N!} int_0^infty delta (E-r^2)r^{3N-1}dr int d^{3N-1}Omega\
&{=} frac{V^N(2m)^{3N/2}}{h^{3N}N!}frac { 2 pi ^ { 3 N / 2 } } { Gamma left( frac { 3 N } { 2 } right) }int_0^infty delta (E-r^2)r^{3N-1}dr.end{align}$$



And here is where the trouble starts. I'm not really sure on how to proceed from here on. It seems the right thing to do would be a coordinate transformation to get something of the form $delta(x-r)$, but I just can't figure out how to do that. Comparing my current result with the solution I expect something like
$$int_0^infty delta (E-r^2)r^{3N-1}dr=frac { 1 } { 2 } E ^ { 3 N / 2 - 1 }.tag{$clubsuit$}$$






integration definite-integrals physics distribution-theory dirac-delta







share|cite|improve this question











$endgroup$












  • $begingroup$
    Isn't it just $dr$? Why would it be $d^{3N}r$?
    $endgroup$
    – John Doe
    Jan 12 at 18:56










  • $begingroup$
    @JohnDoe You‘re right! I will correct it in a couple of minutes..
    $endgroup$
    – Sito
    Jan 12 at 18:58
















2












$begingroup$


TL;DR: How does one show that $(clubsuit)$ holds.




Some context and how I arrived at my problem. In a Thermodynamics problem set I was asked to calculate the partition function of the ideal gas in the microcanonic ensemble using the Hamiltionian
$$mathcal{H} = frac{1}{2m}sum_{i=1}^N p_i^2.$$



The problem seems quite straight forward, but I fail at a certain point when trying to solve the integral:



$$begin{align}Z &= int_{mathbb{R}^{6N}} frac{d^{3N}pd^{3N}q}{h^{3N}N!}deltaleft(E-mathcal{H}right)\
&= frac{1}{h^{3N}N!}int_{mathbb{R}^{3N}}d^{3N}q int_{mathbb{R}^{3N}} d^{3N}p, deltaleft(E-sum_{i=1}^N frac{p_i^2}{2m}right)\
&= frac{V^N(2m)^{3N/2}}{h^{3N}N!}int_{mathbb{R}^{3N}}d^{3N}u, delta left(E-sum_{i=1}^Nu_i^2right)\
&= frac{V^N(2m)^{3N/2}}{h^{3N}N!} int_0^infty delta (E-r^2)r^{3N-1}dr int d^{3N-1}Omega\
&{=} frac{V^N(2m)^{3N/2}}{h^{3N}N!}frac { 2 pi ^ { 3 N / 2 } } { Gamma left( frac { 3 N } { 2 } right) }int_0^infty delta (E-r^2)r^{3N-1}dr.end{align}$$



And here is where the trouble starts. I'm not really sure on how to proceed from here on. It seems the right thing to do would be a coordinate transformation to get something of the form $delta(x-r)$, but I just can't figure out how to do that. Comparing my current result with the solution I expect something like
$$int_0^infty delta (E-r^2)r^{3N-1}dr=frac { 1 } { 2 } E ^ { 3 N / 2 - 1 }.tag{$clubsuit$}$$






integration definite-integrals physics distribution-theory dirac-delta







share|cite|improve this question











$endgroup$












  • $begingroup$
    Isn't it just $dr$? Why would it be $d^{3N}r$?
    $endgroup$
    – John Doe
    Jan 12 at 18:56










  • $begingroup$
    @JohnDoe You‘re right! I will correct it in a couple of minutes..
    $endgroup$
    – Sito
    Jan 12 at 18:58














2












2








2





$begingroup$


TL;DR: How does one show that $(clubsuit)$ holds.




Some context and how I arrived at my problem. In a Thermodynamics problem set I was asked to calculate the partition function of the ideal gas in the microcanonic ensemble using the Hamiltionian
$$mathcal{H} = frac{1}{2m}sum_{i=1}^N p_i^2.$$



The problem seems quite straight forward, but I fail at a certain point when trying to solve the integral:



$$begin{align}Z &= int_{mathbb{R}^{6N}} frac{d^{3N}pd^{3N}q}{h^{3N}N!}deltaleft(E-mathcal{H}right)\
&= frac{1}{h^{3N}N!}int_{mathbb{R}^{3N}}d^{3N}q int_{mathbb{R}^{3N}} d^{3N}p, deltaleft(E-sum_{i=1}^N frac{p_i^2}{2m}right)\
&= frac{V^N(2m)^{3N/2}}{h^{3N}N!}int_{mathbb{R}^{3N}}d^{3N}u, delta left(E-sum_{i=1}^Nu_i^2right)\
&= frac{V^N(2m)^{3N/2}}{h^{3N}N!} int_0^infty delta (E-r^2)r^{3N-1}dr int d^{3N-1}Omega\
&{=} frac{V^N(2m)^{3N/2}}{h^{3N}N!}frac { 2 pi ^ { 3 N / 2 } } { Gamma left( frac { 3 N } { 2 } right) }int_0^infty delta (E-r^2)r^{3N-1}dr.end{align}$$



And here is where the trouble starts. I'm not really sure on how to proceed from here on. It seems the right thing to do would be a coordinate transformation to get something of the form $delta(x-r)$, but I just can't figure out how to do that. Comparing my current result with the solution I expect something like
$$int_0^infty delta (E-r^2)r^{3N-1}dr=frac { 1 } { 2 } E ^ { 3 N / 2 - 1 }.tag{$clubsuit$}$$






integration definite-integrals physics distribution-theory dirac-delta







share|cite|improve this question











$endgroup$




TL;DR: How does one show that $(clubsuit)$ holds.




Some context and how I arrived at my problem. In a Thermodynamics problem set I was asked to calculate the partition function of the ideal gas in the microcanonic ensemble using the Hamiltionian
$$mathcal{H} = frac{1}{2m}sum_{i=1}^N p_i^2.$$



The problem seems quite straight forward, but I fail at a certain point when trying to solve the integral:



$$begin{align}Z &= int_{mathbb{R}^{6N}} frac{d^{3N}pd^{3N}q}{h^{3N}N!}deltaleft(E-mathcal{H}right)\
&= frac{1}{h^{3N}N!}int_{mathbb{R}^{3N}}d^{3N}q int_{mathbb{R}^{3N}} d^{3N}p, deltaleft(E-sum_{i=1}^N frac{p_i^2}{2m}right)\
&= frac{V^N(2m)^{3N/2}}{h^{3N}N!}int_{mathbb{R}^{3N}}d^{3N}u, delta left(E-sum_{i=1}^Nu_i^2right)\
&= frac{V^N(2m)^{3N/2}}{h^{3N}N!} int_0^infty delta (E-r^2)r^{3N-1}dr int d^{3N-1}Omega\
&{=} frac{V^N(2m)^{3N/2}}{h^{3N}N!}frac { 2 pi ^ { 3 N / 2 } } { Gamma left( frac { 3 N } { 2 } right) }int_0^infty delta (E-r^2)r^{3N-1}dr.end{align}$$



And here is where the trouble starts. I'm not really sure on how to proceed from here on. It seems the right thing to do would be a coordinate transformation to get something of the form $delta(x-r)$, but I just can't figure out how to do that. Comparing my current result with the solution I expect something like
$$int_0^infty delta (E-r^2)r^{3N-1}dr=frac { 1 } { 2 } E ^ { 3 N / 2 - 1 }.tag{$clubsuit$}$$






integration definite-integrals physics distribution-theory dirac-delta




integration definite-integrals physics distribution-theory dirac-delta






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 22:19









Qmechanic

4,98711855




4,98711855










asked Jan 12 at 18:40









SitoSito

1706




1706












  • $begingroup$
    Isn't it just $dr$? Why would it be $d^{3N}r$?
    $endgroup$
    – John Doe
    Jan 12 at 18:56










  • $begingroup$
    @JohnDoe You‘re right! I will correct it in a couple of minutes..
    $endgroup$
    – Sito
    Jan 12 at 18:58


















  • $begingroup$
    Isn't it just $dr$? Why would it be $d^{3N}r$?
    $endgroup$
    – John Doe
    Jan 12 at 18:56










  • $begingroup$
    @JohnDoe You‘re right! I will correct it in a couple of minutes..
    $endgroup$
    – Sito
    Jan 12 at 18:58
















$begingroup$
Isn't it just $dr$? Why would it be $d^{3N}r$?
$endgroup$
– John Doe
Jan 12 at 18:56




$begingroup$
Isn't it just $dr$? Why would it be $d^{3N}r$?
$endgroup$
– John Doe
Jan 12 at 18:56












$begingroup$
@JohnDoe You‘re right! I will correct it in a couple of minutes..
$endgroup$
– Sito
Jan 12 at 18:58




$begingroup$
@JohnDoe You‘re right! I will correct it in a couple of minutes..
$endgroup$
– Sito
Jan 12 at 18:58










2 Answers
2






active

oldest

votes


















1












$begingroup$

You have the integral$$int_0^infty delta (E-r^2)r^{3N-1}drtag1$$
Then let $g(r)=-r^2+E$. Of course, $g(r)$ is only zero when $r=sqrt E$ ($E>0$ since it is an energy, and we don't care about $r=-sqrt E$ since this is not within out integration bounds), so we use the definition for the composition of the Dirac delta function with a different function given on Wikipedia to get the delta function into a more managable form.



$$delta(g(x))=sum_ifrac{delta(x-x_i)}{|g'(x_i)|}$$ where the sum is over all the roots of $g(x)$.



For our function $g(r)$, this gives $$g(-r^2+E)=frac{delta(r-sqrt E)}{2sqrt E}$$



Plugging this into $(1)$, $$int_0^infty delta (E-r^2)r^{3N-1}dr=frac1{2sqrt E}int_0^inftydelta(r-sqrt E)r^{3N-1}dr=frac12 E^{-1/2}E^{3N/2} E^{-1/2}=frac12E^{{3N}/2-1}$$ as required.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Beautiful! Didn't know about the composition, so thank you for that!
    $endgroup$
    – Sito
    Jan 12 at 19:14



















1












$begingroup$

User John Doe has already given a good answer. Here is a slightly different calculation:



$$ int_{mathbb{R}_+}! mathrm{d}r~r^{3N-1}delta(r^2-E)
~stackrel{u=r^2}{=}~
frac{1}{2}int_{mathbb{R}_+}! mathrm{d}u~u^{3N/2-1}delta(u-E)
~=~frac{1}{2}theta(E)E^{3N/2-1},tag{$clubsuit$}$$
where $theta$ denotes the Heaviside step function.






share|cite|improve this answer









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    2 Answers
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    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    1












    $begingroup$

    You have the integral$$int_0^infty delta (E-r^2)r^{3N-1}drtag1$$
    Then let $g(r)=-r^2+E$. Of course, $g(r)$ is only zero when $r=sqrt E$ ($E>0$ since it is an energy, and we don't care about $r=-sqrt E$ since this is not within out integration bounds), so we use the definition for the composition of the Dirac delta function with a different function given on Wikipedia to get the delta function into a more managable form.



    $$delta(g(x))=sum_ifrac{delta(x-x_i)}{|g'(x_i)|}$$ where the sum is over all the roots of $g(x)$.



    For our function $g(r)$, this gives $$g(-r^2+E)=frac{delta(r-sqrt E)}{2sqrt E}$$



    Plugging this into $(1)$, $$int_0^infty delta (E-r^2)r^{3N-1}dr=frac1{2sqrt E}int_0^inftydelta(r-sqrt E)r^{3N-1}dr=frac12 E^{-1/2}E^{3N/2} E^{-1/2}=frac12E^{{3N}/2-1}$$ as required.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Beautiful! Didn't know about the composition, so thank you for that!
      $endgroup$
      – Sito
      Jan 12 at 19:14
















    1












    $begingroup$

    You have the integral$$int_0^infty delta (E-r^2)r^{3N-1}drtag1$$
    Then let $g(r)=-r^2+E$. Of course, $g(r)$ is only zero when $r=sqrt E$ ($E>0$ since it is an energy, and we don't care about $r=-sqrt E$ since this is not within out integration bounds), so we use the definition for the composition of the Dirac delta function with a different function given on Wikipedia to get the delta function into a more managable form.



    $$delta(g(x))=sum_ifrac{delta(x-x_i)}{|g'(x_i)|}$$ where the sum is over all the roots of $g(x)$.



    For our function $g(r)$, this gives $$g(-r^2+E)=frac{delta(r-sqrt E)}{2sqrt E}$$



    Plugging this into $(1)$, $$int_0^infty delta (E-r^2)r^{3N-1}dr=frac1{2sqrt E}int_0^inftydelta(r-sqrt E)r^{3N-1}dr=frac12 E^{-1/2}E^{3N/2} E^{-1/2}=frac12E^{{3N}/2-1}$$ as required.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Beautiful! Didn't know about the composition, so thank you for that!
      $endgroup$
      – Sito
      Jan 12 at 19:14














    1












    1








    1





    $begingroup$

    You have the integral$$int_0^infty delta (E-r^2)r^{3N-1}drtag1$$
    Then let $g(r)=-r^2+E$. Of course, $g(r)$ is only zero when $r=sqrt E$ ($E>0$ since it is an energy, and we don't care about $r=-sqrt E$ since this is not within out integration bounds), so we use the definition for the composition of the Dirac delta function with a different function given on Wikipedia to get the delta function into a more managable form.



    $$delta(g(x))=sum_ifrac{delta(x-x_i)}{|g'(x_i)|}$$ where the sum is over all the roots of $g(x)$.



    For our function $g(r)$, this gives $$g(-r^2+E)=frac{delta(r-sqrt E)}{2sqrt E}$$



    Plugging this into $(1)$, $$int_0^infty delta (E-r^2)r^{3N-1}dr=frac1{2sqrt E}int_0^inftydelta(r-sqrt E)r^{3N-1}dr=frac12 E^{-1/2}E^{3N/2} E^{-1/2}=frac12E^{{3N}/2-1}$$ as required.






    share|cite|improve this answer











    $endgroup$



    You have the integral$$int_0^infty delta (E-r^2)r^{3N-1}drtag1$$
    Then let $g(r)=-r^2+E$. Of course, $g(r)$ is only zero when $r=sqrt E$ ($E>0$ since it is an energy, and we don't care about $r=-sqrt E$ since this is not within out integration bounds), so we use the definition for the composition of the Dirac delta function with a different function given on Wikipedia to get the delta function into a more managable form.



    $$delta(g(x))=sum_ifrac{delta(x-x_i)}{|g'(x_i)|}$$ where the sum is over all the roots of $g(x)$.



    For our function $g(r)$, this gives $$g(-r^2+E)=frac{delta(r-sqrt E)}{2sqrt E}$$



    Plugging this into $(1)$, $$int_0^infty delta (E-r^2)r^{3N-1}dr=frac1{2sqrt E}int_0^inftydelta(r-sqrt E)r^{3N-1}dr=frac12 E^{-1/2}E^{3N/2} E^{-1/2}=frac12E^{{3N}/2-1}$$ as required.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 12 at 19:34

























    answered Jan 12 at 18:57









    John DoeJohn Doe

    11.1k11238




    11.1k11238












    • $begingroup$
      Beautiful! Didn't know about the composition, so thank you for that!
      $endgroup$
      – Sito
      Jan 12 at 19:14


















    • $begingroup$
      Beautiful! Didn't know about the composition, so thank you for that!
      $endgroup$
      – Sito
      Jan 12 at 19:14
















    $begingroup$
    Beautiful! Didn't know about the composition, so thank you for that!
    $endgroup$
    – Sito
    Jan 12 at 19:14




    $begingroup$
    Beautiful! Didn't know about the composition, so thank you for that!
    $endgroup$
    – Sito
    Jan 12 at 19:14











    1












    $begingroup$

    User John Doe has already given a good answer. Here is a slightly different calculation:



    $$ int_{mathbb{R}_+}! mathrm{d}r~r^{3N-1}delta(r^2-E)
    ~stackrel{u=r^2}{=}~
    frac{1}{2}int_{mathbb{R}_+}! mathrm{d}u~u^{3N/2-1}delta(u-E)
    ~=~frac{1}{2}theta(E)E^{3N/2-1},tag{$clubsuit$}$$
    where $theta$ denotes the Heaviside step function.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      User John Doe has already given a good answer. Here is a slightly different calculation:



      $$ int_{mathbb{R}_+}! mathrm{d}r~r^{3N-1}delta(r^2-E)
      ~stackrel{u=r^2}{=}~
      frac{1}{2}int_{mathbb{R}_+}! mathrm{d}u~u^{3N/2-1}delta(u-E)
      ~=~frac{1}{2}theta(E)E^{3N/2-1},tag{$clubsuit$}$$
      where $theta$ denotes the Heaviside step function.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        User John Doe has already given a good answer. Here is a slightly different calculation:



        $$ int_{mathbb{R}_+}! mathrm{d}r~r^{3N-1}delta(r^2-E)
        ~stackrel{u=r^2}{=}~
        frac{1}{2}int_{mathbb{R}_+}! mathrm{d}u~u^{3N/2-1}delta(u-E)
        ~=~frac{1}{2}theta(E)E^{3N/2-1},tag{$clubsuit$}$$
        where $theta$ denotes the Heaviside step function.






        share|cite|improve this answer









        $endgroup$



        User John Doe has already given a good answer. Here is a slightly different calculation:



        $$ int_{mathbb{R}_+}! mathrm{d}r~r^{3N-1}delta(r^2-E)
        ~stackrel{u=r^2}{=}~
        frac{1}{2}int_{mathbb{R}_+}! mathrm{d}u~u^{3N/2-1}delta(u-E)
        ~=~frac{1}{2}theta(E)E^{3N/2-1},tag{$clubsuit$}$$
        where $theta$ denotes the Heaviside step function.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 12 at 22:17









        QmechanicQmechanic

        4,98711855




        4,98711855






























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