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I was playing around with $\mathcal{I}=\int_0^1\text{frac}({\frac{1}{x^n}}) dx$, where $\text{frac(.)}$ is the fractional part function, and I discovered that $$\mathcal{I}=~~~ \frac{1}{1-n}; n\leq0$$ \begin{align*} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\frac{1}{1-n}-\zeta(1/n)~~;n\\in(0,1) \end{align*} \begin{align*} ~~~~~~~~~~~1-\gamma~; n=1 \end{align*}

Where $\gamma$ is the Euler-Mascheroni constant. And $\zeta(s)$ is the Riemann Zeta function.

My questions are

1) Is anything similar known? Any other definite integrals relating the fractional part and the Riemann zeta function?

2) It is apparent from the above that $\zeta(1/n) < \frac{1}{1-n}$ for $n\\in(0,1)$. Now I've found out that the same inequality holds even when $n>1$. However, the same technique for evaluation for $\mathcal{I}$ doesnt work when $n>1$, as the computation depends on the sum $$\sum_{n=1}^\infty \frac{1}{n^p}$$ which diverges when $p\leq 1$. And if $n>1$, we have that $1/n<1$ so one of the sums in the process of evaluation becomes divergent.

I'm guessing that some complex analysis is required to overcome this difficulty. But I'm not familiar with that as of now.

I'd be grateful for any comments on this.

Thank you. :)

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By the way, you can use "\cases" to properly display your equation: "{\cal I}=\cases{{1\over 1-n} & \text{for $n\le 0$},\cr {1\over 1-n}-\zeta(1/n) & \text{for $n\in (0,1)$},\cr 1-\gamma & \text{for $n=1$}}" –  B R Mar 14 '12 at 13:10
    
Oh yeah. I was really annoyed when all I could see in the preview was a box with the code. I didnt know how to get rid of that. So I just tried something to display the formula. –  Koundinya Vajjha Mar 15 '12 at 5:58

1 Answer 1

up vote 6 down vote accepted

This really isn't particularly remarkable. By definition, $$\zeta(s) = \sum_{n = 1}^{\infty}{\frac{1}{n^s}} = s \int_{1}^{\infty}{\frac{\lfloor x \rfloor}{x^s} \frac{dx}{x}}$$ for $\Re(s) > 1$, where the second inequality follows by a partial summation argument, and $\lfloor x \rfloor$ is the floor function. We can write $\lfloor x \rfloor = x - \{x\}$, where $\{x\}$ is the fractional part of $x$, and split up the integral above in order to find that $$\zeta(s) = \frac{1}{s - 1} + 1 - s \int_{1}^{\infty}{\frac{\{ x \}}{x^s} \frac{dx}{x}}.$$ This is now defines a meromorphic function on $\Re(s) > 0$ with just a simple pole at $s = 1$ with residue $1$. In any case, this is a well-known integral representation of $\zeta(s)$.

This should explain your result for $0 < n < 1$, by making the change of variables $n = 1/s$, and then making the change of variables $y = x^{-1/n}$ in the integral. Also, the $n = 1$ case is definitional, as the Euler--Mascheroni constant is defined to be $$1 - \int_{1}^{\infty}{\frac{\{ x \}}{x^2} dx}$$ A partial summation argument should show why this is the same as $\lim_{x \to \infty} \left( \sum_{n \leq x}{\frac{1}{n}} - \log x\right)$.

You can also see why $$\zeta(\sigma) < \frac{\sigma}{\sigma - 1}$$ for all $\sigma > 0$; this is simply because $$\sigma \int_{1}^{\infty}{\frac{\{ x \}}{x^{\sigma}} \frac{dx}{x}} > 0.$$

Note that none of this really uses complex analysis, apart from the fact that $\zeta(s)$ isn't implicitly defined for $0 < \Re(s) < 1$ initially, so you need to take a leap of faith to believe that the integral representation of $\zeta(s)$ is actually valid in this strip.

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Thanks a lot. You just...took care of everything. :/ –  Koundinya Vajjha Mar 15 '12 at 5:59

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