# Interesting integral

1. Feb 19, 2005

### msmith12

An interesting integral appeared on a test that I took today. I had no idea how to solve it, and was wondering if any of you all could possibly help me

$$\int_{0}^{1} \frac{1}{x}-floor(\frac{1}{x})$$

we first had to draw the graph from 0 to 1, (which has a max of 1, a min of 0 (at 1/n for any integer n), with a negative slope at all points (more or less).

then we had to describe the integrability...

and finally, we were asked to evaluate the integral, noting that
$$\gamma = \lim_{n->\infty} \sum_{n=1}^{\infty} \frac{1}{n} - \ln (n)$$

I had no idea on what to do from here (besides the fact that there was probably going to be a substitution involving 1/x, so that the limits of integration are changed to 1-> infty)...

~dazed and confused

2. Feb 19, 2005

### fourier jr

that function $$\frac{1}{x}-\left[\frac{1}{x}\right]$$ is called the gauss function or something like that. i guess i'd start by splitting the integrable into pieces & add up the separate integrals like this:

$$\sum_{n=1}^{\infty}\int_{1/(n+1)}^{1/n} \left(\frac{1}{x}-\left[ \frac{1}{x}\right]\right) dx$$ i don't know how that helps though

re: integrability it's riemann integrable because it's continuous almost everywhere (ie. it has a countable set of discontinuities). not sure if it's also lebesgue integrable but it is if $$\int_{[0,1]} \left(\frac{1}{x}-\left[ \frac{1}{x}\right]\right) dx < \infty$$

Last edited: Feb 19, 2005
3. Feb 20, 2005

### saltydog

I've been working on it too and come out with:

$$\sum_{n=0}^{\infty}(\int_{1/(n+2)}^{1/(n+1)} \left(\frac{1}{x}-\left (n+1)\right\right)) dx$$

Solving, I get:

$$\sum_{n=0}^{\infty}(\ln\frac{n+2}{n+1}-\frac{1}{n+2})$$

This appears to converge but I can't prove it.

Last edited: Feb 20, 2005
4. Feb 20, 2005

### saltydog

I think I have a proof for convergence:

First express the sum as:

$$\ln(2)-\frac{1}{2}+\sum_{n=1}^{\infty} ( \ln[\frac{n+2}{n+1}]-\frac{1}{n+2})$$

Now the Integral Test for convergence can be used by evaluating the following integral:

$$\int_1^\infty(\ln(\frac{x+2}{x+1})-\frac{1}{x+2})dx$$

The antiderivative is:

$$(x+1)(\ln{\frac{x+2}{x+1}})$$

Since this is $(\infty*0)$ at the upper limit, I can express it as a quotient which has the indeterminate form $\frac{0}{0}$. After differentiating, I get the limit of 1. Thus the series converges.

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Last edited: Feb 20, 2005
5. Feb 22, 2005

### TheDestroyer

Well, I got another way to solve this, I hope it's correct, and i'm sorry to tell you (saltydog) the answer of your integral is not logic (0.189), the answer must be very near to 0.5 (less than 0.5 a little), we can see that from the graph

if we studied the continuity of this integral we will find that it's verified only when we are from 1/(n+1) to 1/n while n is an integer, ok, what if we tried to make a sum using the mathematical induction (i don't know if it's named like that i'm not english), let's do this:

$$\int_{\frac{1}{2}}^{1} (\frac{1}{x}-floor(\frac{1}{x})) dx = \int_{\frac{1}{2}}^{1}(\frac{1}{x}-1)dx$$
$$\int_{\frac{1}{3}}^{\frac{1}{2}}} (\frac{1}{x}-floor(\frac{1}{x})) dx = \int_{\frac{1}{3}}^{\frac{1}{2}}}(\frac{1}{x}-2)dx$$
$$\int_{\frac{1}{4}}^{\frac{1}{3}}} (\frac{1}{x}-floor(\frac{1}{x})) dx = \int_{\frac{1}{4}}^{\frac{1}{3}}}(\frac{1}{x}-3)dx$$

That will make the sum series:

$$\gamma = \sum_{n=1}^{\infty} \int_{\frac{1}{n+1}}^{\frac{1}{n}}}(\frac{1}{x}-n)dx$$

finally:
$$\gamma = \sum_{n=1}^{\infty} ln(\frac{n+1}{n}) - 1 + \frac{n}{n+1} = 0.424$$

and the sum will equal approximately 0.424 which is the area under the curves and which is a logic answer,

I'm just a second year physics student, can you discuss with me what a got? i've made all this with out studying anything about the definite integrals, i just know F(b) - F(a) from the indefinite integral?

6. Feb 22, 2005

### saltydog

I never gave a value for the integral. I only showed that it converged by showing it had a value for the upper limit. Perhaps I should have been more specific. However, the value of the integral does not help in evaluating the infinite sum. It's only a "test" to check for convergence. That's why I didn't stress it's value.

I like your analysis of the sum. It's very similar to mine. I just started the index at 0 instead of 1.

Mathematica reports the sum as 0.422784 but I suspect it does so numerically.

7. Feb 24, 2005

### TheDestroyer

I've calculated my and your sum using mathcad, do you mean the sum result (0.422784) is for your integral? or for mine? because i really got in mathcad the result 0.189

and sorry if my language disturbed you :), I love the scientific discussions ;)

with my regards, TheDestroyer

8. Feb 25, 2005

### saltydog

Hum, though I answered this yesterday. Anyway, the integral is not related to the sum. You know that right? Well, not directly, it's an approximate value. But really, it's only used as a test to determine if the sum converges. I didn't fully evaluate the integral. Perhaps I should have. However, I just wanted to see if the integral converged. If it did, that's all I needed to know to prove that the sum converged. The value of 0.422784 is for the infinite sum (not the integral) as reported by Mathematica.

9. Oct 26, 2011

### epsdelta74

Well six years late for the discussion ( I Googled 'interesting integrals' out of boredom ) but really, what in the Platonic realm does the original sum mean? In the infinite series, the limit as nāā has already been taken, so the additional limit makes no sense at all... there were a number of enthusiastic responses that completely ignored this ambiguity.