how to solve integrals involving derivative of delta function

Question

How to solve this kind of integrals? $$ I=\int dx\int dx_1 f(x)g(x_1) \bigg[\frac{d \delta (x-x_1)}{d(x-x_1)}\bigg] $$ here $f(x)$ and $g(x)$ are two functions. The derivative is w.r.t. difference of $x$ and $x_1$

Answer 1

First put $y = x-x_1$, we get: $$ I=\int d(y+x_1) \bigg(\int dx_1 f(y+x_1)g(x_1) \bigg[\frac{d \delta (y)}{dy}\bigg]\bigg) $$ For the inner integraion, use integration by parts $\int u dv = uv-\int vdu $. Let $u = f(y+x_1)g(x_1) $, and $dv = dx_1 \bigg[\frac{d \delta (y)}{dy}\bigg]$. And we get $du = dx_1 \frac{d}{dx_1}[f(y+x_1)g(x_1)]$, and $v=\delta(y)$. So, the inner integral become: $$ f(y+x_1)g(x_1)\delta(y) - \int dx_1 \frac{d}{dx_1}[f(y+x_1)g(x_1)] \delta (y) $$ put the value of $y$: $$ f(x)g(x_1)\delta(x-x_1) - \int dx_1 \frac{d}{dx_1}[f(x)g(x_1)] \delta (x-x_1) $$ So, the total integral is: $$ I=\int dx \bigg(f(x)g(x_1)\delta(x-x_1) - \int dx_1 \frac{d}{dx_1}[f(x)g(x_1)] \delta (x-x_1)\bigg) \\ I=\int dx f(x)g(x_1)\delta(x-x_1) - \int dx \int dx_1 \frac{d}{dx_1}[f(x)g(x_1)] \delta (x-x_1) \\ I=f(x_1)g(x_1)- \int dx_1 \frac{d}{dx_1}[f(x_1)g(x_1)]\\ I=f(x_1)g(x_1)- f(x_1)g(x_1)=0 $$

Answer 2

$$ \int dx\int dx_1 \, f(x) g(x_1) \delta'(x-x_1) = \{ y := x-x_1 \} = \int dy \int dx_1 \, f(y+x_1) g(x_1) \delta'(y) \\ = \int dx_1 \left( \int dy \, f(y+x_1) \delta'(y) \right) g(x_1) = \int dx_1 \left( -\int dy \, f'(y+x_1) \delta(y) \right) g(x_1) \\ = - \int dx_1 \, f'(x_1) g(x_1) $$