I want to simplify the integral $$I=\int_y \log \left( \int_x f(y) \delta(x-y) dx \right)dy,$$ where $x$, $y$ are real numbers, $f$ is a "nice" real fuction of real argument (eg. exp) and $\delta$ stands for Dirac Delta function.

My idea is to somehow rescale the measure $dy$ by a function $g(x,y)$ such that $$I=\log \left(\int_y \int_x f(y) \delta(x-y) g(x,y) dx dy\right).$$ The question is how to find such a function $g(x,y)$ that would enable to use the sampling property of Dirac Delta function to simplify the integral $I$.

  • $\begingroup$ Is it the Kronecker delta or the Dirac delta? $\endgroup$ – G. Gare Oct 7 '19 at 14:17
  • $\begingroup$ Dirac delta, thanks for the comment, I fixed the question $\endgroup$ – Pavel Prochazka Oct 7 '19 at 14:19
  • $\begingroup$ Since the logarithm is concave, you can define $\mu(dx) = \delta(x - y) dx$ and bring the logarithm inside with Jensen's inequality (this gives you a lower bound). Then, Fubini what you get and in the end you get a "nice" lower bound $\endgroup$ – G. Gare Oct 7 '19 at 14:24
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    $\begingroup$ Note that $\displaystyle\int_x f(y)\,\delta(x-y)\,dx=f(y)\int_x\delta(x-y)\,dx=f(y).$ $\endgroup$ – Adrian Keister Oct 7 '19 at 14:26

We have that \begin{align*} I &=\int_y \ln \left( \int_x f(y)\, \delta(x-y)\, dx \right)dy\\ &=\int_y \ln \left(f(y) \int_x \, \delta(x-y)\, dx \right)dy\\ &=\int_y \ln \left(f(y) \right)\,dy\\ &=y\ln(f(y))-\int_y y\,\frac{f'(y)}{f(y)}\,dy, \end{align*} using by-parts once. Without knowing what $f(y)$ is, this is likely as far as you can go, if you want exact simplifications and not approximations.


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