Prove that $\int_0^1 \frac{1}{x^\delta} \,dx$ convergent $\Leftrightarrow \ \delta <1$

Question

I want to prove the following equation: $$ \int_0^1 \frac{1}{x^\delta} \,dx \text{ convergent }\Leftrightarrow \ \delta <1 $$

$"\Rightarrow":$

I know that $\frac{1}{x^\delta}$ is not defined for $x=0$. Thus I need to take $a \in (0,1] $ and calculate $$ \int_a^1 \frac{1}{x^\delta} \,dx $$

But how do I continue from this point on? Should I calculate the antiderivative and if yes, how? I can't figure it out because $\delta$ can be any number.

$"\Leftarrow":$

I don't have any idea how to prove this direction.

Answer 1

by definition of improper integral you get \begin{align*} \int_0^1\frac1{x^{\delta}}\,dx &=\lim_{a\to0^+}\int_a^1\frac1{x^{\delta}}\,dx\\ &=\lim_{a\to0^+}\left(\frac{x^{-\delta+1}}{-\delta+1}\right|_a^1\\ &=\frac1{1-\delta}\lim_{a\to0^+}\left(1-\frac1{a^{\delta-1}}\right) \end{align*} which is finite iff $\delta-1<0$

Answer 2

Hint

• If $\delta \neq 1$ an antiderivative of $\frac{1}{x^\delta}=x^{-\delta}$ is $\frac{1}{{1-\delta}} x^{1-\delta}$ so: $$\int_a^1 \frac{1}{x^\delta} dx=\left[ \frac{1}{{1-\delta}} x^{1-\delta}\right]_a^1=\frac{1}{1-\delta} \left(1-a^{1-\delta}\right)$$ so it goes to $-\infty$ if $1-\delta <0$ and converges to $\frac{1}{1-\delta}$ if $1-\delta >0$.
• If $\delta=1$ an antiderivative of $1/x$ is $\ln(x)$ so: $$\int_a^1 \frac{1}{x^1} dx=\left[ \ln(x)\right]_a^1=-\ln(a)$$ which goes to $-\infty$ as $a \to 0$.

Answer 3

Yes you should compute the antiderivative. You should already be familiar with it. Remember that $$\frac{1}{x^\delta}=x^{-\delta}$$ so you can call $-\delta = n$ and compute the antiderivative of $x^n$. Be careful because there is a special case ($\delta=1$), and you should know what the antiderivative is in that case.

Answer 4

For $\delta \neq 1$, $$\int_a^1 \frac 1{x^\delta} \mathrm d x = \left[\frac {x^{1-\delta}}{1-\delta}\right]_a^1 = \frac 1{1-\delta} \left(1-a^{1-\delta}\right)$$ This quantity has a finite limit when $a\to 0$ if and only if $1-\delta > 0$ so $\delta < 1$.

Now if $\delta = 1$, $$\int_a^1 \frac 1{x^\delta} \mathrm d x = \left[\ln x\right]_a^1 = -\ln a \to_{a \to 0} + \infty$$

Answer 5

The crux of showing that $\int^1_0 \frac{1}{x^{\delta}} dx$ converges for $\delta > 0$, note that $d\left(\frac{x^{1-\delta}}{1-\delta}\right)/dx$ is $\frac{1}{x^{\delta}}$, AND that for all $x \in [0,1]$, the value of $\frac{x^{1-\delta}}{1-\delta}$ is no more than $\frac{1}{1-\alpha}$.

For the other direction, here is a hint: What is $\int^1_{\epsilon} \frac{1}{x} dx$. And at each point in $(0,1)$ what happens to the value of the function $\frac{1}{x^{\delta}}$ as $\delta$ increases.