Convergence of $\int_1^2 \frac{\mathrm{d}x}{(x-1)^a}$

Question

Why does the integral $$\int_1^2 \frac{1}{(x-1)^a}$$ converge only for $a<1$ and diverge otherwise?

I would compute the integral: $\int_1^2 \frac{1}{(x-1)^a}=\frac{1}{(a-1)}$ and this gives a finite result for any $a$ different from $1$. So I don't understand why the integral would diverge for example for $a=3$ since plugging in this value of a into the integral gives $\int_1^2 \frac{1}{(x-1)^3}=\frac{1}{(3-1)}$ which is finite.

Answer 1

This improper integral can be explicitly computed. The integral is convergent if the following limit exists and is finite. $$ \lim_{k \to 1^+} \int_k^2 \frac{1}{(x-1)^a} dx $$

And the limit can be easily computed. If $a \ne 1$, \begin{align*} \lim_{k \to 1^+} \int_k^2 \frac{1}{(x-1)^a} dx =& \lim_{k \to 1^+} \left[\frac{(x-1)^{-a+1}}{-a+1}\right]_k^2 = \frac{1}{1-a}\lim_{k \to 1^+}(1-(k-1)^{1-a})\\ =&\begin{cases} \frac{1}{1-a}, & a <1\\ \infty, & a >1\end{cases} \end{align*}

If $a=1$, the primitive is a logarithm and you also conclude that the integral is divergent.

Answer 2

Let $1<t<2$ and compute the integral $F_a(t):=\int_t^2 \frac{\mathrm{d}x}{(x-1)^a}$.

Distinguish two cases: $a=1$ and $a \ne 1.$

We have $\int_1^2 \frac{\mathrm{d}x}{(x-1)^a}$ is convergent $ \iff \lim_{t \to 1+1}F_a(t)$ exists in $\mathbb R$.

Hence compute $ \lim_{t \to 1+1}F_a(t)$ and look what happens !

Answer 3

What you've found is analogous to extending the formula

$$1+x+x^2+x^3+x^4+\cdots={1\over 1-x}$$

which holds when $|x|\lt1$, to all $x\not=1$, including amusements such as

$$1+2+4+8+16+\cdots={1\over1-2}=-1$$

In each case, the left hand side specifies a computation that requires the taking of a limit. (In particular, the integral in the OP is improper when $a\lt0$, which means it is computed as the limit of the integral from $h$ to $2$, as $h$ tends to $1$ from the right.) Sometimes the limit exists, sometimes it doesn't. The fact that the answer you get when the limit exists can be extended into an "answer" that seems to make sense when the limit doesn't exist is well worth noting -- it's the basis for what's known as analytic continuation -- but it doesn't change the fact that the limit doesn't exist in the cases where it doesn't exist.