I have this integral, and the solution gives an indeterminate form for the value $\alpha = 1$, can you explain to me how to solve the indeterminate form?
$$\int_{\beta}^{+\infty} x^{-\alpha} dx = \frac{1}{1-\alpha} \left[ x^{-\alpha + 1}\right]_{\beta}^{+\infty} = \frac{1}{0} \left[ +\infty^{0} - 1\right] =?$$
Little side note, I know the solution of this integral I am really just curious od knowing how to tackle the indeterminate form.
It's not really indeterminate. What you are trying to do is $$ \lim_{\alpha\to 1^+}\int_\beta^\infty x^{-\alpha}\,\, dx = \lim_{\alpha\to 1^+}\frac{x^{1-\alpha}}{1-\alpha}. $$ But this is of the form $\displaystyle \frac{1}{0}$ which is not indeterminate.
Your issue is that to reach that last formula in your post you have to assume the integral exists.