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December 7[edit]

$\int _{I}\exp(-f(x))dx\geq \int _{I}\exp(-g(x))dx$ if $\int _{I}(f(x)-g(x))\exp(-g(x))dx=0$ [edit]

If for two functions $f(x)$ and $g(x)$ we have that:

\int _{I}(f(x)-g(x))\exp(-g(x))dx=0

(1)

where $I$ is an arbitrary interval on the real line, then

\int _{I}\exp(-f(x))dx\geq \int _{I}\exp(-g(x))dx

(2)

This follows in a straightforward way from the Bogoliubov inequality. It seems to be a reasonable powerful tool to get to sharp bounds for certain integrals or summations, but it doesn't seem to be used a lot for this purpose in mathematics. I was wondering if a more straightforward proof can be given for the special case of integrals and summations.

This can be used to obtain sharp lower bounds for integrals of the form $\int _{I}\exp(-f(x))dx$ by writing down a function $g(x)$ containing parameters, and then imposing the constraint (1) to eliminate one parameter and then maximizing $\int _{I}\exp(-g(x))dx$ w.r.t. the remaining parameters. A simple example is to take $f(x)=x^{2}$ and $g(x)=a+bx$ for $b>0$ and $I$ the positive real line, which yields the inequality $\pi \geq e$ .

Count Iblis (talk) 00:11, 7 December 2020 (UTC)[reply]

December 7[edit]

∫ I exp ⁡ ( − f ( x ) ) d x ≥ ∫ I exp ⁡ ( − g ( x ) ) d x {\displaystyle \int _{I}\exp(-f(x))dx\geq \int _{I}\exp(-g(x))dx} if ∫ I ( f ( x ) − g ( x ) ) exp ⁡ ( − g ( x ) ) d x = 0 {\displaystyle \int _{I}(f(x)-g(x))\exp(-g(x))dx=0} [edit]

$\int _{I}\exp(-f(x))dx\geq \int _{I}\exp(-g(x))dx$ if $\int _{I}(f(x)-g(x))\exp(-g(x))dx=0$ [edit]