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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.

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  • $\begingroup$ 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. $\endgroup$ – Jake Mirra Nov 3 at 15:23
  • $\begingroup$ How do you justify the step from the symbolic integral expression to the limit? $\endgroup$ – Dr. Lutz Lehmann Nov 3 at 16:31
  • $\begingroup$ @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. $\endgroup$ – Rishabh Jain Nov 3 at 20:02
  • $\begingroup$ 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)? $\endgroup$ – Dr. Lutz Lehmann Nov 3 at 21:29
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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 $x=0$.

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  • $\begingroup$ The test function used with dirac delta is always supposed to be differentiable. However, even if I put the limit in your expression, the answer is still missing a negative sign. $\endgroup$ – Rishabh Jain Nov 4 at 8:51
  • $\begingroup$ I think your original equality is wrong, it should be $$xδ'(x)=-δ(x),$$ as for test functions $f$ you have $$\int_{-\infty}^\infty xδ'(x)f(x)dx=-\int_{-\infty}^\infty δ(x)(xf(x))'dx=-f(0).$$ $\endgroup$ – Dr. Lutz Lehmann Nov 4 at 9:00
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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.

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