# Show that the function $\dfrac{1}{\sqrt{x}}$ is Riemann integrable on $[2^{-2n} , 2^{-2n+2}]$ for any positive integer $n$.

Show that the function $\dfrac{1}{\sqrt{x}}$ is Riemann integrable on $[2^{-2n} , 2^{-2n+2}]$ for any positive integer $n$.

So that readers may be able to give precise feedback as to where my understanding is erroneous, I will lay out my calculations and reasoning for each step.

My Work

I know that the formula for the Riemann sum is $lim_{n \to \infty} \sum_{j = 1}^n f(c_j)\Delta x_j$.

$c_j$ is some element in the interval of integration $[x_{j-1}, x_j]$. Since we're taking the Riemann sum ($n \to \infty$ and the norm/magnitude of each partition goes to $0$), the value of $x$ in the interval of integration for which we take the function value at does not matter.

$\Delta x_j$ represents the magnitude/norm of each partition. It is calculated using $\dfrac{b - a}{n} = \Delta x_j$ where $b$ and $a$ are the bounds of integration.

For this problem, the interval/domain of integration is $[2^{-2n}, 2^{-2n+2}]$ for all positive integers $n$.

Therefore, we can calculate the norm/magnitude of this partition (for any positive integer $n$) as $2^{-2n+2} - 2^{-2n} = 3 \cdot 2^{-2n}.$ This is our $\Delta x_j$.

For $c_j$, we can select any point in the interval $[2^{-2n}, 2^{-2n+2}]$. I will select $c_j = c_n = 2^{-2n}$.

Therefore, the Riemann sum is $\sum_{n = 1}^\infty f(2^{-2n}) \cdot 3 \cdot 2^{-2n} = 3 \cdot \sum_{n = 1}^\infty \dfrac{1}{\sqrt{\dfrac{1}{2^{2n}}}} \cdot 2^{-2n}$

I have no idea how to continue from here.

I would greatly appreciate it if people could please take the time to review my work and provide feedback.

• Maybe do you mean function $\dfrac{1}{\sqrt{x}}$? Otherwise, the function is constant and any constant is Riemann integrable. – user261263 Mar 27 '17 at 19:12
• @EugenCovaci You are correct. Thank you for the correction. – The Pointer Mar 27 '17 at 19:12
• It is mandatory to use the definition? The function $\dfrac{1}{\sqrt{x}}$ is continuous on a compact interval therefore Riemann integrable . – user261263 Mar 27 '17 at 19:15
• @EugenCovaci I don't think it is mandatory, but I thought that this was the best way to "prove" that it is Riemann integrable? – The Pointer Mar 27 '17 at 19:16
• See this question math.stackexchange.com/questions/56393/… – user261263 Mar 27 '17 at 19:19