Prove $\lim_{\alpha\to -1} \biggl(\frac{(-1)^{\alpha}}{(x)^{\alpha + 1}}-(-1)^\alpha\biggr)\operatorname{\Gamma}(\alpha+1)=\operatorname{ln}(x)$

Question

How do I prove:$$\lim_{\alpha\to -1} \biggl(\frac{(-1)^{\alpha}}{(x)^{\alpha + 1}}-(-1)^\alpha\biggr)\operatorname{\Gamma}(\alpha+1)=\operatorname{ln}(x)$$ I'm completely lost on how to solve this limit because of the Gamma/Factorial function.

Answer 1

1) If $f(a)=g(a)$ for some $a \in R$ and $f'(x)=g'(x)$ for all $x \in \mathbb D$, we can say $f(x)=g(x)$. $\mathbb D$ and $R$ is respective domain and range.

Taking $z=\alpha +1$ we change the limit as below.

$L(x)=\lim \limits_{z\to 0} (e^{iπz}-\frac{e^{iπz}}{x^z}){\Gamma(z)}$ and this limit equals to $0$ when $x=1$. That means at $x=1$, $L(1)=0=ln(1)$.

Now, $L'(x)=\lim \limits_{z \to 0} (\frac{e^{iπz}z}{x^{z+1}}){\Gamma(z)}$

or, $L'(x)={\frac{1}{x}}{\lim \limits_{z \to 0} e^{iπz}z{\Gamma(z)}}$

or, $L'(x)={\frac{1}{x}}{\lim \limits_{z \to 0} z{\Gamma(z)}}$

And, ${\lim \limits_{z \to 0} z{\Gamma(z)}}=1 \Rightarrow L'(x)={\frac{1}{x}}{\lim \limits_{z \to 0} z{\Gamma(z)}}=\frac{1}{x}=\frac{\text{d ln}(x)}{\text{d}x}$

for all $x>0$

So 1 implies that $L(x)=\text{ln}(x)$ for all $x>0$