I know that you can define the derivative of the delta function as:-
$$\delta'(x)=-\frac{1}{x}\delta(x)$$
If i use this to calculate the integral with $f(x)$, I get 2 different results.
Method 1:-
$$\int_{-\infty}^{+\infty}\delta'(x)f(x)dx = -f'(0) $$ by using integration by parts.
Method 2:-
$$\int_{-\infty}^{+\infty}-\frac{1}{x}\delta(x)f(x)dx = \lim_{x\to0} -\frac{f(x)}{x}$$
I think I am doing something wrong in method 2 as the 2 results do not match.
I do not know much about distribution theory and would appreciate any help.
Best Answer
You could use the symmetry of the integrand, assuming that these integrals make sense in some weak sense. $$ \int_{\Bbb R}\frac{f(x)}xδ(x)\,dx =\frac12\int_{\Bbb R}\left(\frac{f(x)}x+\frac{f(-x)}{-x}\right)δ(x)\,dx =\int_{\Bbb R}\left(\frac{f(x)-f(-x)}{2x}\right)δ(x)\,dx $$ If $f$ is continuously differentiable, the difference quotient has a continuous continuation to $x=0$ with value $f'(0)$. Then the original definition of the Dirac delta applies to give the integral exactly this value at $x=0$.