For which values of p would this integral converge?

Question

I have:

$$\int_{0}^{1} \frac1{x^p}dx$$

My question is what values of $p$ would this converge. I know that

$$\int_{1}^{\infty} \frac1{x^p}dx$$ converges when $p>1$, but I am unsure how that would relate to the question I have.

Answer 1

The two integrals are related by the substitution $y=\frac1x$ as shown by Mark Viola.

We can also prove directly that $\int_{0}^{1} \frac1{x^p}dx$ converges for $p<1$, indeed in this case we have that $$\int_{0}^{1} \frac1{x^p}dx=\left[\frac{x^{-p+1}}{-p+1}\right]_0^1=\frac{1}{p-1}\in\mathbb{R}$$

and in the limit case for $p=1$

$$\int_{0}^{1} \frac1{x}dx=\left[\log x\right]_0^1=\infty$$

and for $p>1$ since for $0<x<1$

$$\frac1{x^2}\ge\frac1x\implies \int_{0}^{1} \frac1{x^p}\,dx>\int_{0}^{1} \frac1{x}dx=\infty$$

Answer 2

Hint : Break into two cases. For $p=1$ and for $p \neq 1$.

For $p=1$, you very well know the value of indefinite integral $(\ln x)$, and can therefore comment on definite integral too.

For $p \neq 1$, integrate to get $\dfrac{x^{p+1}}{p+1}$ and put the limits then to check when does it converge.

Answer 3

You have $$ \int_a^b \frac1{x^p}d x =\bigg[\frac{x^{1-p}}{1-p}\bigg]_a^b$$ so (when $p\neq 1$), it is easy to see when this integral converges or not as $a$ tends to $0$ or $b$ tends to $\infty$.

Answer 4

HINT:

Note that by enforcing the substitution $x\mapsto 1/x$, we see that

$$ \int_1^\infty \frac1{x^p} \,dx=\int_0^1 \frac1{x^{2-p}}\,dx$$

Now, using the fact that the integral on the left-hand side converges for $p>1$, can you answer the question you have?

Answer 5

That's simple since we know the antiderivative: $$\int \frac{\mathrm d x}{x^p}=\begin{cases}-\dfrac1{(p-1)x^{p-1}}&\text{if }p\ne 1,\\ \ln x&\text{if }p=1.\end{cases}$$ This shows that $\;\displaystyle\int_0^1 \frac{\mathrm d x}{x^p}$:

• diverges if $p> 1$, since $\;\displaystyle\lim_{\varepsilon\to 0}\int_\varepsilon^1 \frac{\mathrm d x}{x^p}=-\frac1{p-1}+\lim_{\varepsilon\searrow\,0}\dfrac1{(p-1)\varepsilon^{p-1}}=+\infty$,
• clearly diverges if $p=1$,
• converges is $p<1$, since $\;\displaystyle\lim_{\varepsilon\searrow\, 0}\dfrac1{(p-1)\varepsilon^{p-1}}=\lim_{\varepsilon\searrow\,0}\dfrac{\varepsilon^{1-p}}{p-1}=0.$

Answer 6

If this is a Riemann integral, it will never converge for any $p>0$.

For Riemann integrals, both the upper and lower Riemann limits must exist and converge to the same value as the partition mesh size goes to zero, irregardless of how the function is sampled within each partition. In this case, you can always pick the left-most sample to be as large as you want, and thus ensure that the upper Riemann sum is always larger than any finite value. Thus, the upper Riemann limit doesn't exist, and the integral is said to diverge.

If this is a Lebesgue integral, it will converge for $p<1$, as other answers illustrate.

If this is a improper Riemann integral, it will also converge, because now we're interpreting it as:

$$\lim_{a\to0} \int_a^1 \frac{1}{x^p} dx$$

So, the answer really depends on how you define the term integral.