Derivative of $F(x):= \int_{-\infty } ^x f(t) \, dt $

Question

It is one of the main results in Calculus that if $f $ is Riemann integrable and continous at a point $x_0 $ then for every $a < x$

$$F(x):= \int_{a} ^x f(t) \, dt $$

has derivative

$$F'(x_0)=f(x_0)$$

I wonder what can be said of the derivative of of

$$G(x):= \int_{-\infty } ^x f(t) \, dt $$

at a point $x_0 $ if we further assume that $$\lim_{t \to - \infty } f(t) = 0 $$ (if necessary one could also assume that $f $ approaches zero as $t \to - \infty $ in some particular order $O(g(x)) $, or some other condition).

Is it possible to do some argument that $G(x) \overset{\sim } {= } F(x) + \epsilon$ and then apply the result for $F $?

Thanks in advance!

Answer 1

We can split up the integral for $G(x)$ into two parts as follows: $$G(x)=\int_a^x f(t) \, dt+\int_{-\infty \ }^a f(t) \, dt=F(x)+\int_{-\infty }^a f(t) \, dt=F(x)+C$$ However, the last integral does not necessarily converge, even if $\underset{t\to -\infty }{\text{lim}}f(t)=0$. A basic example is $f(t)=\frac{1}{t}$. That said, $C$ is independent of $x$, so the derivative of $G(x)$ w.r.t. $x$ is unaffected even if the constant is undefined or infinity. It is possible to apply the result for $F(x)$ to $G(x)$.