Riemann Integration

xyz · #1 $Old$ November 17th, 2009, 09:08 PM

This is an integrable function. However, how is this integral solved? I can perfectly visualize this function. I don't see how the value of integral can be anything besides 0. But I cannot put in words or show work as to how the integral is solved. Please help

RockHard · #2 $Old$ November 17th, 2009, 09:19 PM

I believe this what they call a improper integral, and because it is obviously undefined at 0, for something like this I believe you will have to do

$\lim_{x\to0+}{\int_B}^1\frac{1}{x}$

This should get you started

xyz · #3 $Old$ November 17th, 2009, 09:23 PM

What is B in the integral you entered? I don't understand that.

Drexel28 · #4 $Old$ November 17th, 2009, 09:48 PM

Quote:

Originally Posted by xyz $View Post$

This is an integrable function. However, how is this integral solved? I can perfectly visualize this function. I don't see how the value of integral can be anything besides 0. But I cannot put in words or show work as to how the integral is solved. Please help

Note that over any subinterval $[a,b]\subset [0,1]$ that $\inf_{x\in[a,b]}f(x)=0$ ....so

chisigma · #5 $Old$ November 18th, 2009, 12:20 AM

Is $f(x)=0$ everywhere with the exception of $x=\frac{1}{n}$ , $n \in \mathbb{N}$ so that you can devide the integration interval in subintervals of the type...

$\frac{3}{4}<x\le 1$ , $\frac{5}{12} < x \le \frac{3}{4}$ , $\dots$ , $\frac{n+\frac{3}{2}}{n+2}< x \le \frac{n+\frac{1}{2}}{n+1}$ , $\dots$

... so that in each interval is $f(x)=0$ with the only exception of $x=\frac{1}{n}$ , i.e. in only one point. In each interval the function is Riemann integrable and is...

$\int_{x_{n+1}}^{x_{n}} f(x)\cdot dx=0$ (1)

From (1)...

$\int_{0}^{1} f(x)\cdot dx = \sum_{n} \int_{x_{n+1}}^{x_{n}} f(x)\cdot dx=0$ (2)

Kind regards

$\chi$ $\sigma$

xyz · #6 $Old$ November 18th, 2009, 12:29 AM

Thanks a lot for all the help. I already got the idea from Drexel's hint. It was easy after that. But your solution is also very interesting!