A curious definite integral. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T10:30:26Z http://mathoverflow.net/feeds/question/91158 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91158/a-curious-definite-integral A curious definite integral. Koundinya Vajjha 2012-03-14T10:31:24Z 2012-03-14T12:49:36Z <p>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*}</p> <p>Where $\gamma$ is the Euler-Mascheroni constant. And $\zeta(s)$ is the Riemann Zeta function.</p> <p>My questions are</p> <p>1) Is anything similar known? Any other definite integrals relating the fractional part and the Riemann zeta function?</p> <p>2) It is apparent from the above that $\zeta(1/n) &lt; \frac{1}{1-n}$ 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 $1/n&lt;1$ so one of the sums in the process of evaluation becomes divergent. </p> <p>I'm guessing that some complex analysis is required to overcome this difficulty. But I'm not familiar with that as of now. </p> <p>I'd be grateful for any comments on this.</p> <p>Thank you. :)</p> http://mathoverflow.net/questions/91158/a-curious-definite-integral/91171#91171 Answer by Peter Humphries for A curious definite integral. Peter Humphries 2012-03-14T12:49:36Z 2012-03-14T12:49:36Z <p>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 $\zeta(s)$.</p> <p>This should explain your result for $0 &lt; n &lt; 1$, 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 $\lim_{x \to \infty} \left( \sum_{n \leq x}{\frac{1}{n}} - \log x\right)$.</p> <p>You can also see why $$\zeta(\sigma) &lt; \frac{\sigma}{\sigma - 1}$$ for all $\sigma > 0$; this is simply because $$\sigma \int_{1}^{\infty}{\frac{\{ x \}}{x^{\sigma}} \frac{dx}{x}} > 0.$$</p> <p>Note that none of this really uses complex analysis, apart from the fact that $\zeta(s)$ isn't implicitly defined for $0 &lt; \Re(s) &lt; 1$ 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.</p>