Some gamma function questions... I have shown that $\Gamma(a+1)=a\Gamma(a)$ for all $a>0$. 
But I'd also like to show the following 2 things:
1) Using the previous fact, I'd like to show that $\lim_{a \to 0^{+}}a\Gamma(a) = \Gamma(1)=1$
The second equality is obvious, but how about the limit equality?
2) Using 1) then, I'd like to calculate $\int_{0}^{\infty} {{e^{-x}-e^{-3x}}\over{x}}$. The suggestion is to, using the first two facts above, calculate
$$\lim_{a \to 0^{+}}\int_{0}^{\infty} x^{a}{{e^{-x}-e^{-3x}}\over{x}}$$.
I can sort of prove 1), provided I am allowed to switch the $a$ limit and the $M$ limit outside the integral (we replace the upper limit $\infty$ with $M$ and take the limit as $M \to \infty$ using the MCT.) However, though I get the correct answer of $1$, it seems fishy that I am allowed to switch the limits like this.
Number 2) I have no idea.
 A: *

*Since $\Gamma(x)$ is continuous at $x=1$, we get $$\lim_{a\to0^+}a\Gamma(a)=\lim_{a\to0^+}\Gamma(a+1)=\Gamma(1)=1$$

*Here is how I usually approach this type of integral, but it does not use the Gamma function, just the fact that for $x\gt0$, $-1\le\frac{e^{-x\vphantom{\Large1}}-1}{x}\le0$:
$$
\begin{align}
\int_0^\infty\frac{e^{-x}-e^{-3x}}{x}\,\mathrm{d}x
&=\lim_{a\to0}\int_a^\infty\frac{e^{-x}-e^{-3x}}{x}\,\mathrm{d}x\\
&=\lim_{a\to0}\left(\int_a^\infty\frac{e^{-x}}{x}\,\mathrm{d}x-\int_a^\infty\frac{e^{-3x}}{x}\,\mathrm{d}x\right)\\
&=\lim_{a\to0}\left(\int_a^\infty\frac{e^{-x}}{x}\,\mathrm{d}x-\int_{3a}^\infty\frac{e^{-x}}{x}\,\mathrm{d}x\right)\\
&=\lim_{a\to0}\int_a^{3a}\frac{e^{-x}}{x}\,\mathrm{d}x\\
&=\lim_{a\to0}\int_a^{3a}\frac1{x}\,\mathrm{d}x+\lim_{a\to0}\int_a^{3a}\frac{e^{-x}-1}{x}\,\mathrm{d}x\\[8pt]
&=\log(3)
\end{align}
$$
A: Here is the solution to part (2) using the Gamma function and the result of part (1):
$$\int_0^\infty x^a \frac{e^{-x}-e^{-3x}}{x}dx  = \int_0^\infty x^{a-1}e^{-x}dx-\int_0^\infty x^{a-1}e^{-3x}dx.$$
For the second integral on the RHS, make the change of variables $u = 3x$ to obtain
$$\int_0^\infty x^a \frac{e^{-x}-e^{-3x}}{x}dx  = \int_0^\infty x^{a-1}e^{-x}dx-3^{-a}\int_0^\infty u^{a-1}e^{-u}dx.$$
Using the definition of the Gamma function,
$$\int_0^\infty x^a \frac{e^{-x}-e^{-3x}}{x}dx  = (1-3^{-a})\Gamma(a) = \frac{1-3^{-a}}{a}a\Gamma(a).$$
From part (1) we have $\lim_{a \rightarrow 0+} a\Gamma(a) = 1$, and using L'Hospital's rule we know that
$$\lim_{a \rightarrow 0+}\frac{1-3^{-a}}{a} = \log(3)$$
Therefore 
$$\lim_{a \rightarrow 0+}\int_0^\infty x^a \frac{e^{-x}-e^{-3x}}{x}dx  =  \lim_{a \rightarrow 0+}\frac{1-3^{-a}}{a} \lim_{a \rightarrow 0+} a\Gamma(a) = \log(3).$$
A: The limit follows from the functional property of the Gamma function:
$$
       a\Gamma(a)=\Gamma(a+1).
$$
