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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.

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    The fundamental theorem of calculus does not hold if $\alpha<1$, for in that case the function $\frac{1}{t^\alpha}$ is not defined at the endpoint $0$. In this case, we have to prove that the integral diverges, which is easy to see. – 2017-01-11

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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.

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    Oh yea... thanks! – 2017-01-11
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    And so if it's undefined you can say that it diverges? – 2017-01-11
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    @John Adams Yes you are right. – 2017-01-11
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    Ok! Thank you for your answer – 2017-01-11