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This question is inspired by a recent course I did on random matrix theory and also from common mistakes high-schoolers make in algebra :).

In random matrix theory, one often encounters somewhat intractable integrals involving a logarithmic term that are often made more amenable to analysis using the replica trick, that is, say, in the case of trying to find the expectation of logarithm of a partition function, $Z(x)$, one can use the identity:

$$\mathbb{E}\big[\log Z(x)\big] = \lim_{n\to 0} \frac{1}{n}\log \mathbb{E}\big[Z(x)\big]^n.$$

This helps to evaluate integral more easily because the partition function usually takes the form of an exponential multiplied by another function.

My question is much simpler than this. Under what conditions would it just justifiable, or through the use of what technique would it be possible that either of the following relations

$$ \int \exp(f(x)) dx = \exp\left(\int f(x) dx\right) $$

or

$$ \int \log(f(x)) dx = \log\left(\int f(x) dx\right) $$

holds. Here $f\in \mathbb{C}^{\infty}$ and the integral denotes the Riemann integral.

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    $\begingroup$ you're asking when $e^a+e^b=e^{a+b}$? $\endgroup$
    – Carlo Beenakker
    Commented May 3, 2017 at 12:20
  • $\begingroup$ Could you specify better what is $f$ and the meaning of the integral? $\endgroup$
    – Pietro Majer
    Commented May 3, 2017 at 12:21
  • $\begingroup$ Not exactly. I'm asking for which non-trivial functions or which non-trivial conditions does the relation $int\exp(f(x)) = exp(\int f(x))$ hold. $\endgroup$
    – user119264
    Commented May 3, 2017 at 12:23
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    $\begingroup$ I guess that you are interested in integrals against probability measures - if so, you are basically asking "For what functions is Jensen's inequality with $\exp$ an equality?". For the domain $[0,1]$ a simple calculation shows that among the linear function $f(x) = ax+b$ only the constants satisfy $\exp(\int_0^1 f(x)dx) = \int_0^1\exp(f(x))dx$. $\endgroup$
    – Dirk
    Commented May 3, 2017 at 13:11
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    $\begingroup$ For any strictly convex function $g$ and any probability measure $\mu$, $g\left(\int f(x) d\mu(x)\right) \le \int g(f(x)) d\mu(x)$, with equality only when $f$ is a.s. constant. Similarly for strictly concave functions, with $\le$ replaced by $\ge$. $\endgroup$
    – Robert Israel
    Commented May 3, 2017 at 15:53

2 Answers 2

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Here's one way to get lots of examples.

Start with a finite interval $[a,b]$ and a function $g$ on this interval such that $\int_a^b g(x)\; dx > 0$, and for $c > 0$ and $d > 0$ consider $f$ defined on $[ca, cb]$ by $f(x) = d g(x/c)$. Then $$ A(c,d) := \exp\int_{ca}^{cb} f(x)\; dx = \exp\left(c d \int_a^b g(t)\; dt\right) $$ $$ B(c,d) := \int_{ca}^{cb} \exp(f(x))\; dx = c \int_a^b \exp(d g(t))\; dt $$ We have $A(0,d) = 1$ and $B(0,d) = 0$, while for any fixed $d > 0$, $A(c,d) > B(c,d)$ if $c$ is sufficiently large. Thus by the Intermediate Value Theorem, there will be some $c$ that makes $A(c,d) = B(c,d)$.

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Given $f(x)>0$ for $x>0$, we could extend the function to negative $x$ by demanding that: $$f(-x)=f(x)-\log(e^{f(x)}-1)$$ then $$e^{f(x)}+e^{f(-x)}=\exp[f(x)+f(-x)],$$ hence $\int e^f\,dx = \exp\left(\int f\,dx\right)$. A discontinuity at $x=0$ can avoided if $f(0)=\log 2$.

As pointed out by Pietro Majer, this construction needs a finite domain $-a<x<a$ for a convergent integral. [Alternatively, the domain could be displaced to $(a,b)$ by relating $f(x)$ to $f(a+b-x)$.]

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  • $\begingroup$ Is the integral finite? Since $u(x):=e^{f(x)}$ and $v(x):=e^{f(-x)}$ satisfy $u+v=uv$, hence $(u-1)(v-1)=1$, we have $\mathbb{R}=\{|u-1|\ge1\}\cup\{|v-1|\ge1\}$, so either $\int_{\mathbb{R}} u$ or $\int_{\mathbb{R}} v$ should be infinite... $\endgroup$
    – Pietro Majer
    Commented May 3, 2017 at 14:22
  • $\begingroup$ true, I need a finite domain --- thanks for correcting me. $\endgroup$
    – Carlo Beenakker
    Commented May 3, 2017 at 14:55

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