Determine the value of $α$ to prove the convergence of the following generalised integral : $ α∈ℝ,α>0, \int^{1}_{0} {\frac{1}{t^{α}} dt}$

Question

I'm stuck on this one... $$ α∈ℝ,α>0, \int^{1}_{0} {\frac{1}{t^{α}} dt}$$

I need to find the values of $α$ when the integral is convergent. The answer is that: "The integral converges if α < 1 and diverges otherwise."

I've calculated the integral, and found: $$ \frac{t^{1-α}}{1-α}$$ with $t$ evaluated between $0$ and $1$, and that gives me: $$ \frac{1^{1-α}}{1-α}- \frac{0^{1-α}}{1-α } = \frac{1^{1-α}}{1-α}$$ But then! In my mind, for every $α\neq1$, the expression is convergent!

Not only for $α<1$, cause the numerator will be equal to $1$ for every $α$, so we can rewrite it as: $$\frac{1}{1-α}$$ And that is convergent for every $α\neq1$...

I still don't understand why I'm wrong, or where I've made a mistake, if anyone can help?

Thank you in advance.

Answer 1

Note that $0^{1-\alpha} =0$ only for $1-\alpha>0 \Rightarrow \alpha <1$.

Otherwise, if $\alpha >1$, then we will get a form of $\frac {1}{0} $ which is not defined.