Are these funcitons linearly independent? Let $a,b,c,\ldots$ be a finite set of distinct positive real numbers.
Are the functions
$(a+x)^{-r}$, where $r$ is a positive real number, linearly independent function on $[0,\infty)$? Are there any references for this? Would the answer depend on r?

Let $0 < a_1 < \dots < a_n$ and assume that $\left( (a_i + x)^{-r} \right)_{i=1}^n$ are linearly dependent over $[0,\infty)$ and so we can find $b_1,\dots,b_n \in \mathbb{R}$ with
$$ \sum_{i=1}^n \frac{b_i}{(a_i + x)^r} = \sum_{i=1}^n \frac{b_i}{e^{r \ln(a_i + x)}}=  0 $$
for all $x \in [0,\infty)$. Note that the function $\sum_{i=1}^n \frac{b_i}{(a_i + x)^r}$ is in fact defined on the interval $(-a_1,\infty)$. In fact, it can be extended to a complex analytic function $z \mapsto \sum_{i=1}^n \frac{b_i}{e^{r \ln(a_i + z)}}$ on the set $\{ z \in \mathbb{C} \, | \, \Re(z) > -a_1 \}$ and by assumption, it vanishes on $[0,\infty)$ and hence it must actually vanish on $\{ z \in \mathbb{C} \, | \, \Re(z) > -a_1 \}$. In particular, we must have
$$ 0 = \lim_{x \to -a_1} \sum_{i=1}^n \frac{b_i}{(a_i + x)^r} = \sum_{i=2}^n \frac{b_i}{(a_i - a_1)^r} + \lim_{x \to -a_1} \frac{b_1}{(x + a_1)^r}$$
which shows that $b_1 = 0$. But then $\sum_{i=1}^n \frac{b_i}{(a_i + x)^r} = \sum_{i=2}^n \frac{b_i}{(a_i + x)^r}$ is actually well-defined on $(-a_2,\infty)$ and repeating the argument we see that $b_2 = 0$, etc.
