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I am reading the article of Deriving probability distributions using the Principle of Maximum Entropy

and I am trying to derive some of the equations in it automatically using Mathematica.

1. Derivation of maximum entropy probability distribution with no other constraints (uniform distribution)

First, we solve for the case where the only constraint is that the distribution is a pdf, which we will see is the uniform distribution. To maximize entropy, we want to minimize the following function: $$ J(p)=\int_a^b p(x) \ln p(x) d x-\lambda_0\left(\int_a^b p(x) d x-1\right) $$ . Taking the derivative with respect ot $p(x)$ and setting to zero, $$ \frac{\delta J}{\delta p(x)}=1+\ln p(x)-\lambda_0=0 $$

Can I derive the second equation automatically using Mathematica?

if just $ J(p)=\int_a^b p(x) \ln p(x) dx$, I can do

Needs["VariationalMethods`"]
VariationalD[p[x] Log[p[x]], p[x], x]

the output is 1 + Log[p[x]] as expected.

But here $ J(p)=\int_a^b p(x) \ln p(x) d x-\lambda_0\left(\int_a^b p(x) d x-1\right) $, how can I do that?

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    $\begingroup$ Why not VariationalD[p[x] Log[p[x]] - λ0 ( p[x] - 1), p[x], x]? What's the issue? $\endgroup$
    – bmf
    Jan 12 at 11:40
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    $\begingroup$ @user64494 that's weird. on my laptop it returns 1 - λ0 + Log[p[x]] which is the expected answer from the OP $\endgroup$
    – bmf
    Jan 12 at 11:55
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    $\begingroup$ @bmf's code works as expected also in 12.3.1 (Windows) and 13.2.0 (Wolfram Cloud). $\endgroup$
    – Domen
    Jan 12 at 11:56
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    $\begingroup$ @benjaminchanming I mean that you wrote that if $J$ has the simple form you know how to do the variational derivative in Mathematica. For the complicated form I am unsure why you got stuck, but I showed you that you can write the same command effectively and get the result that you want for $\tfrac{\delta J}{\delta \rho}$ and now you say you are unsure about the result, which is the one you wanted. Unless, I am missing something $\endgroup$
    – bmf
    Jan 12 at 12:18
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    $\begingroup$ @benjaminchanming, linearity is one of the basic properties of integral, and the equality holds (if both of the integrals exist). However, for ill-behaved (!!) functions, the equality might not hold, and that is why Mathematica returns False. $\endgroup$
    – Domen
    Jan 12 at 12:42

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