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

This is question involves a problem from one of my previous questions:

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 $n$.

My Work

$\int^{2^{-2n+2}}_{x = 2^{-2n}} \dfrac{1}{\sqrt{x}} = 2(2^{-n})$

Therefore, for any $n \ge 1$, the function is bounded and continuous on the interval $[2^{-2n} , 2^{-2n+2}]$. Therefore, the function is Riemann integrable on this interval.

When finding the area under the curve, we get $\sum^{\infty}_{n=1} 2(2^{-n}) = 2$. Therefore, the area $= 2$ is defined.

However, although the area is defined, apparently the function is not Riemann integrable.

I am really struggling to understand this. We found that the function $\dfrac{1}{\sqrt{x}}$ is Riemann integrable on the interval $[2^{-2n} , 2^{-2n+2}]$ for any $n \ge 1$. We also found that the area is defined for any $n \ge 1$. But apparently the function is not Riemann integrable? This seems contradictory.

I would greatly appreciate it if people could please take the time to help me understand what is going on here.

• The function is not Riemann-integrable on $[0,1]$. – Wojowu Apr 18 '17 at 12:09
• To add on what have been said, Just because a function is integrable on $[\epsilon,1]$ for all $\epsilon > 0$ does not mean it is on integrable on $[0,1]$ – Rab Apr 18 '17 at 12:14

## 1 Answer

You have to distinguish between different things:

1) For any $n$, the function $x\mapsto 1/\sqrt{x}$ is Riemann integrable on $[2^{-2n},2^{-2n+2}]$ since the function is continuous on that compact interval. (That is a standard theorem in the theory of Riemann integrals.

2) On $(0,1)$ the function $x\mapsto 1/\sqrt{x}$ is not bounded, and thus, according to a theorem not Riemann integrable on that interval.

3) If $\lim_{\epsilon \to 0+}\int_\epsilon^1 f(x)\,dx$ exists then the generalized Riemann integral $\int_0^1 f(x)\,dx$ exists (and is defined to be that limit). Your function satisfies this property.

• Thanks for the response. I don't understand where the interval $(0, 1)$ is coming from? I didn't even mention such an interval in my question. What am I missing? – The Pointer Apr 18 '17 at 12:20
• When you were discussing the area under the curve and summing, it seemed to me that you were looking at that interval. However, the important thing is that $1/\sqrt{x}$ is not integrable at $0$. The upper limit $1$ is not too important. – mickep Apr 18 '17 at 12:22
• Oh, I see where $(0,1)$ comes from: It's what we get for $[2^{-2n} , 2^{-2n+2}]$ for $n \ge 1$. But shouldn't it be $(0, 1]$ since we can have $n = 1$? – The Pointer Apr 18 '17 at 12:31
• As you wish, but that does not make any difference (it is just one point). – mickep Apr 18 '17 at 12:37
• Your last comment (now deleted) shows that your level of confusion is on another level than I thought. I will let my answer stand as it is, but I don't see how I could continue this in a comment. I hope that your teacher can point you in the right direction. Good luck! – mickep Apr 18 '17 at 14:04