Let's say that $y=\sin{x}$. Then the first order taylor series approximation about $c$ is $g(x)=\sin{(c)}+\cos{(c)}(x-c)$. Note that this is also equivalent to the line tangent to the curve $\sin{x}$ with point $(c,\sin{c})$ and slope $\cos c$.
Now, imagine you take two lines: $g(k)$, and $g(k+\delta)$, letting $\delta$ go to $0$ from the right. It seems obvious to me that this limit implies that the two lines will in fact be the same. And in that sense, $\lim_{\delta\rightarrow0^+}\int_a^b(g(x)-g(x+\delta))\ \text{dx}=0$ for finite $a$ and $b$.
My question is then, what can we say about $$\lim_{\delta\rightarrow0^+}\int_{-\infty}^\infty(g(k)-g(k+\delta))\ \text{dk}$$ My intuition thinks it may end up being $f(c)=\sin c$.
The answer may be a little disappointing. Observe that $$g(x)-g(x+\delta)=-\delta\cos(c), $$ which is constant in $x$. So the integral $$\int_{-\infty}^\infty g(x)-g(x+\delta)\ dx=\lim_{A\to\infty}\int_{-A}^A g(x)-g(x+\delta)\ dx=\lim_{A\to\infty}-2\delta A\cos(c)$$ converges only if $\cos(c)=0$. Thus, it either makes no sense to talk about the limit as $\delta\to0^+$, or the limit is trivially $0$.