Suppose that $f(x)$ is Riemann integrable on $[0, \infty)$ and the improper integral $\int_0^{+\infty} f(x) \mathrm{d} x$ converges. Prove that:
$$ \lim _{a \rightarrow 0} \int_0^{+\infty} \mathrm{e}^{-a x} f(x) \mathrm{d} x=\int_0^{+\infty} f(x) \mathrm{d} x $$ MY TRY:Since $\int_0^{+\infty} f(x) \mathrm{d} x$ converges, when $a \geqslant 0$ and $x \geqslant 0$, we have $\left|e^{-a x}\right| \leq 1$.
By Abel's Test, we know that:
$$ \begin{aligned} & I(a)=\int_0^{+\infty} e^{-a x} f(x) \mathrm{d} x, \text { converges uniformly } \\ & \lim _{a \rightarrow 0^{+}} \int_0^{+\infty} e^{-a x} f(x) \mathrm{d} x=\int_0^{+\infty} \lim _{a \rightarrow 0^{+}} e^{-a x} f(x) \mathrm{d} x=\int_0^{+\infty} f(x) \mathrm{d} x \end{aligned} $$
How should we handle the case when $a<0$, that is, $\lim _{a \rightarrow 0^{-}} \int_0^{+\infty} e^{-a x} f(x) \mathrm{d} x$ ?
