Working over the real line, I am interested in the integral of the form $$ I := \int_a^b dx \int_a^x dy f(y) \delta'(y), $$ where $a < 0 < b < \infty$, $f$ is some smooth test function with compact support such that $(a,b) \subset supp(f)$, and $\delta$ is the usual Dirac delta "function".
At first, when attempting to evaluate this integral, I naively used that $\delta'(f) = -\delta(f')=-f'(0)$ so that the above integral would evaluate to $(a-b) f'(0)$. However, looking at it more closely, the result $\delta'(f) = -\delta(f')$ which can be attributed to "integration by parts", is really only working if the integration is actually over the support of $f$. This is not the case for the first integral.
Instead, I modified the relation such that $$ I = \int_a^b dx \left( [f(y) \delta(y)]_a^x - \int_a^x dyf'(y) \delta(y) \right)= \int_a^b dx \left(f(x) \delta(x) - f'(0) \right) = f(0) + (a-b) f'(0). $$ Is this the correct way to do this? I simply generalised the "integration by parts" rule. Is there a rigourous explanation why this is the correct answer?
The global strategy is good, but there are some caveat. The integration by parts is correct, but the next steps should be : $$ \begin{array}{rcl} I &=&\displaystyle \int_a^b\mathrm{d}x \left( [f(y) \delta(y)]_a^x - \int_a^x\mathrm{d}y\,f'(y) \delta(y) \right) \\ &=&\displaystyle \int_a^b\mathrm{d}x \left( f(x)\delta(x) - f'(0)H(x) \right) \\ &=&\displaystyle f(0) - f'(0) \int_a^b H(x) \,\mathrm{d}x \\ &=&\displaystyle f(0) - f'(0)[xH(x)]_a^b \\ &=&\displaystyle f(0) - bf'(0) \color{white}{\frac{1}{1}} \end{array} $$ where $H$ is the Heaviside function, which plays the role of Dirac delta's antiderivative. Indeed, since $x\in(a,b)$ can be negative, the condition $0\in(a,x)$ is not guaranteed, that is why $$ \int_a^x f'(y)\delta(y) \,\mathrm{d}y \equiv f'(0) \int_a^x \delta(y) \,\mathrm{d}y = f'(0) [H(y)]_a^x = f'(0)H(x) \neq f'(0). $$