Is the integration of Dirac delta function and another function correct?

Question

I have the first derivative of a function, and I want to solve the original function through its indefinite integral. $$ \frac{d\phi}{dx}=\left[ \frac{1}{\left( EI \right) _0}+\sum_{i=1}^N{\frac{1}{k_i}\delta \left( x-x_i \right)} \right] g\left( x \right) $$ But the answer I saw was different from the result I solved. I'm not sure if this is because of the series.

This is the answer I saw. $$ \phi \left( x \right) =\frac{g^1\left( x \right)}{\left( EI \right) _0}+\sum_{i=1}^N{\frac{g\left( x_i \right)}{k_i}H\left( x-x_i \right)}+C_1 $$ where, $$ g^1\left( x \right) =\int{g\left( x \right) dx} $$ The integral rule of Dirac delta function I found is this： $$ \int{f\left( x \right) \delta \left( x-x_0 \right) dx\ =}f\left( x_0 \right) $$ Therefore, I think there is no step function in the solution： $$ \phi \left( x \right) =\frac{g^1\left( x \right)}{\left( EI \right) _0}+\sum_{i=1}^N{\frac{g\left( x_i \right)}{k_i}}+C_1 $$ This confuses me. Please tell me that the result is correct, thank you.

Answer 1

Let $g\in C_C^\infty$ and $\phi$ be a distribution such that $\phi'$ is given by

$$\phi'(x)=g(x)\left(\frac1{(EI)_0}+\sum_{i=1}^N\frac1{k_i}\delta(x-x_i)\right)$$

Then, we have for any $\psi\in C^\infty_C$

$$\begin{align} \langle \phi',\psi\rangle&=\frac1{(EI)_0}\int_{-\infty}^\infty g(x)\psi(x)\,dx+\sum_{i=1}^N \frac1{k_i}g(x_i)\psi(x_i)\\\\ &=-\langle \phi,\psi'\rangle \end{align}$$

Then, $\phi$ is the distribution given by

$$\phi(x)=\frac1{(EI)_0}\int_{-\infty}^x g(t)\,dt+\sum_{k=1}^N\frac1{k_i}g(x_i)H(x-x_i)+C$$

for a constant $C$.

Answer 2

If $x>0$ and $x_i>0$ , the result of definite integral form is that: $$ \int_0^x{f\left( x \right) \delta \left( x-x_i \right) =f\left( x_i \right) H\left( x-x_i \right)} $$