How to prove $\int^{\infty}_{0} e^{-\alpha x} f(x)dx \rightarrow \int ^{\infty}_{0} f(x) dx$ when $\alpha \rightarrow 0$? I think I could show that if $$\int ^{\infty}_{0} f(x) dx$$ is convergent, then $$\int^{\infty}_{0} e^{-\alpha x} f(x)dx$$ is also convergent for $\alpha>0$. I'm wondering is my proof true?

Here's my proof outline:
Because of The Archimedean Property of real numbers:
$$\forall \alpha >0 \; \exists x_{0}\in \mathbb{R} \; (\alpha x>1)$$ 
We also have:
$$\int^{\infty}_{0} e^{-\alpha x} f(x)dx = \int^{x_{0}}_{0} e^{-\alpha x} f(x)dx + \int^{\infty}_{x_{0}} e^{-\alpha x} f(x)dx$$ 
Trivially $\int^{x_{0}}_{0} e^{-\alpha x} f(x)dx$ is finite, and $$\int^{\infty}_{x_{0}} e^{-\alpha x} f(x)dx < \frac{1}{e}\int^{\infty}_{x_{0}} f(x)dx < \infty$$
So $\int^{\infty}_{0} e^{-\alpha x} f(x)dx$ converges.

And I know it is true to say $$\int^{\infty}_{0} e^{-\alpha x} f(x)dx \rightarrow \int ^{\infty}_{0} f(x) dx$$ when $\alpha \rightarrow 0$, but how can I prove it?
Thanks for help.
 A: If $f$ is absolutely integrable, that is $\int_0^\infty |f(x)| \, dx < \infty$, then the proof is easy. 
In this case -- because $|e^{-\alpha x}f(x)| \leqslant |f(x)|$ for all $\alpha \geqslant 0$ -- the improper integral converges uniformly by the Weierstrass M-test.  This implies that $I(\alpha) = \int_0^\infty e^{-\alpha x} f(x) \, dx$ is continuous and $\lim_{\alpha\to 0+}I(\alpha) = I(0)$.
More generally -- for example if the improper integral $\int_0^\infty f(x) \, dx$ is only conditionally convergent --  the result is true as a consequence of Abel's test for the uniform convergence of an improper integral.  Note that  $x \mapsto e^{-\alpha x}$ is monotone with respect to $x \in [0,\infty)$ and is uniformly bounded for all $x \geqslant 0$ and $\alpha \geqslant 0$ since $|e^{-\alpha x}| \leqslant 1$. Additionally, $\int_0^\infty f(x) \, dx$ is convergent by hypothesis and, therefore, uniformly convergent for $\alpha \geqslant 0$ (trivially since there is no dependence on $\alpha$). Both condition of Abel's test are met and the improper integral is uniformly convergent for $\alpha \geqslant 0$. 
To see why $\lim_{\alpha\to 0+}I(\alpha) = I(0)$ is implied by uniform convergence note that for any $ \epsilon > 0$ there exists $C_1>0$ such that for all $c > C_1$ we have (by integrability of $f$)
$$\left|\int_c^\infty f(x) \, dx\right| < \frac{\epsilon}{3},$$
and there exists $C_2>0$ such that for all $c > C_2$ and all $\alpha \geqslant 0$ we have (by uniform convergence of $I(\alpha)$)
$$\left|\int_c^\infty e^{-\alpha x} f(x) \, dx\right| < \frac{\epsilon}{3}$$
Thus, for any $c > \max(C_1,C_2)$ we have
$$\left|\int_0^\infty e^{-\alpha x} f(x) \, dx - \int_0^\infty f(x) \, dx \right| \\\leqslant \left|\int_0^c e^{-\alpha x} f(x) \, dx - \int_0^c f(x) \, dx \right|+ \left|\int_c^\infty e^{-\alpha x} f(x) \, dx  \right|+ \left|\int_c^\infty f(x) \, dx  \right| \\ \leqslant \int_0^c |e^{-\alpha x} - 1||f(x)| \, dx + \frac{2\epsilon}{3} \\ \leqslant (1- e^{-\alpha c}) \int_0^c |f(x)| \, dx + \frac{2\epsilon}{3}$$
The first term on the RHS can be made smaller than $\epsilon /3$ by choosing $\alpha$ sufficiently close to $0$ and the result $\lim_{\alpha \to 0+} I(\alpha) = I(0)$ follows.
A: There are two ways to pass a limit through an integral; the first is by proving that the limit converges uniformly. The second way is by a significantly more powerful theorem called Lebesgue Dominated Convergence theorem. Either of these should do the trick - 
$$ \lim_{a \to 0} \int_{0}^{\infty} e^{-ax}f(x) dx = \int_0^{\infty} \lim_{a \to 0} e^{-ax} f(x) dx = \int_0^{\infty} f(x) dx < + \infty. $$
Can you show that either the convergence is uniform or that the Lebesgue Dominated Convergence Theorem applies?
A: We can show that $\int ^{\infty}_{0} e^{-\alpha x}f(x)dx$ is convergent by using Second Mean Value Theorem.

Here is the outline:
There exists a $c \in (a, b]$ such that:
$$ \int ^{b}_{a}e^{-\alpha x}f(x)dx = (lim _{x \rightarrow a^+} e^{-\alpha x})\int ^c_{a}f(x)dx = e^{-a x}\int ^c_{a}f(x)dx$$
We know that $\mid \int ^c_{a}f(x)dx \mid$ is less than arbitrary $\epsilon >0$ for large enough $a$, so $\mid \int ^{b}_{a}e^{-\alpha x}f(x)dx \mid$ have this property, hence $\int ^{\infty}_{0}e^{-\alpha x}f(x)dx$ is convergent.

Now I want to show that $\int^{\infty}_{0} e^{-\alpha x} f(x)dx \rightarrow \int ^{\infty}_{0} f(x) dx$ when $\alpha \rightarrow 0$.
We know:
$$\int^{\infty}_{0} f(x)dx - \int ^{\infty}_{0} e^{-\alpha x}f(x) dx = \int ^{A}_{0} (1-e^{-\alpha x})f(x) dx +\int ^{\infty}_{A}f(x) dx -\int ^{\infty}_{A} e^{-\alpha x}f(x) dx$$
Because of convergence the integrals, for large enough $A$ and arbitrary $\epsilon >0$ we have: $$\mid \int ^{\infty}_{A}f(x) dx \mid + \mid \int ^{\infty}_{A} e^{-\alpha x}f(x) dx \mid < \frac{2\epsilon}{3}$$
We can also make $\int ^{A}_{0} (1-e^{-\alpha x})f(x) dx$ smaller than $\frac{\epsilon}{3}$ by taking $\alpha$ close enough to $0$.
So $\int^{\infty}_{0} e^{-\alpha x} f(x)dx \rightarrow \int ^{\infty}_{0} f(x) dx$ when $\alpha \rightarrow 0$.
