# Using identity for the derivative of Dirac Delta function

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.

• Looks like you're close. You need to be careful with plus and minus signs, and $( f(x) - f(0) )/x$ is indeed the derivative of $f$ (in the limit). So you're on the right track. – Jake Mirra Nov 3 at 15:23
• How do you justify the step from the symbolic integral expression to the limit? – Dr. Lutz Lehmann Nov 3 at 16:31
• @Dr.LutzLehmann As far as I know, the integral of the dirac delta function is the value of the function at 0. To define it for an arbitrary function, I took the limit. – Rishabh Jain Nov 3 at 20:02
• But why this specific limit? Why not some kind of average like $\lim_{x\to 0}\frac12\left(\frac{f(x)}x+\frac{f(-x)}{-x}\right)$? And how do you make sure that all such limites give the same result (or some result to begin with)? – Dr. Lutz Lehmann Nov 3 at 21:29

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 $$x=0$$.
• I think your original equality is wrong, it should be $$xδ'(x)=-δ(x),$$ as for test functions $f$ you have $$\int_{-\infty}^\infty xδ'(x)f(x)dx=-\int_{-\infty}^\infty δ(x)(xf(x))'dx=-f(0).$$ – Dr. Lutz Lehmann Nov 4 at 9:00
If one wants to go strictly by definition, for every multi-index $$\alpha$$, test function $$f:\mathbb R^n\to\mathbb C$$ and tempered distribution $$u$$ one defines $$(D^\alpha u)(f)=(-1)^{|\alpha|} u(D^\alpha f)$$ (cf. Theorem 7.13 in Rudin's "Functional Analysis"). This definition ensures that "the usual formal rules of calculus hold" (Rudin, Ch.6.1).
Taking $$u=\delta$$ with $$\delta(f):=f(0)$$ being the delta distribution one immediatly gets $$\int_{-\infty}^\infty \delta'(x) f(x)\,dx\overset{(\leftarrow\text{ formally)}}=\delta'(f)= -\delta(f')\Big(\overset{\text{(formally }\rightarrow)}=- \int_{-\infty}^\infty \delta(x)f'(x)\,dx \Big)=-f'(0)\,.$$ This is how one would obtain the derivative of the delta distribution rigorously without using the purely formal integral expression with $$\delta(x)$$ which, of course, is not even a function.