I have an integral I need to evaluate as far as possible that contains the delta function, and I haven't been able to solve it the way I usually do. Here it is:
$$\int_{-\infty}^\infty dx\ \delta(x-\pi)sin(x)f(x)$$
Assuming that $f(x)$ is a general smooth (infinitely differentiable) function.
The way I have been solving these is using substitution, and here is my attempt at solving this problem:
$$\int_{-\infty}^\infty dx\ \delta(g(x))sin(x)f(x)$$ $$g(x)=x-\pi=0 \\ x=\pi$$ $$y(x)=sin(x)f(x)$$ $$\int_{-\infty}^\infty dx\ \delta(g(x))sin(x)f(x)=\sum_{a<x_i<b,\ \ \ g(x_i)=0} {y(x_i)\over |g'(x_i)|}$$ $$g'(x)=1, \ \ g'(\pi)=1$$ $$y(\pi)=sin(\pi)f(\pi)=0$$ $$\int_{-\infty}^\infty dx\ \delta(g(x))sin(x)f(x)=0$$
Normally the formula I have been given is this: $$\sum_{a<x_i<b,\ \ \ g(x_i)=0} {f(x_i)\over |g'(x_i)|}$$ but $f(x)$ was already in the integral and I thought it would be a good idea to combine the two of them together to make it into a form that I know how to solve.
This answer doesn't seem right to me. I would think that it wouldn't be $0$ because we are working with the interval $(-\infty,\infty)$ and not including $\pi$, which is the only place $sin(x)$ would be $0$. Is this the correct approach?
Theorem: Let $\delta$ denote the Dirac Delta function. Then for all $a$ and $f$, $\int_\mathbb R dx \delta(x-a)f(x)=f(a)$
From this, we immediately have that the answer is zero since $\sin(\pi)=0$.
I think you’re confused with when we integrate across a subset of the domain, call it $S$. In this case we have that $\int_S dx\delta(x-a)f(x)$ gives $f(a)$ if $a\in S$ and $0$ otherwise. However we aren’t considering an integral of this form, and even if we were $a$ lies inside the domain of the integral (since $\pi\in\mathbb R$) and even if it didn’t both options tell you the answer is $0$.
You’re way over complicating this.