I am trying to compute $$\int_0^1 \left(\frac{1}{x} - \biggl\lfloor \frac{1}{x}\biggr\rfloor\right) dx$$ with no success. Any hints?
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See this. – science Apr 18 '15 at 00:13
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1It is more or less the definition of the Euler-Mascheroni constant: $$\gamma = \lim_{n\to +\infty}\left(H_n-\log n\right).$$ – Jack D'Aurizio Apr 18 '15 at 00:26
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Let $t=\dfrac1x$ and then decompose the resulting integral into a series of integrals. – Lucian Apr 18 '15 at 04:01
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$$\int_0^1 \left(\frac{1}{x} - \left\lfloor \frac{1}{x}\right\rfloor\right) dx=\lim_{k\to +\infty}\int_{\frac{1}{k}}^1 \left(\frac{1}{x} - \left\lfloor \frac{1}{x}\right\rfloor\right) dx$$
Now to compute the other integral use the decomposition: $$\int_{\frac{1}{k}}^1 \left(\frac{1}{x} - \left\lfloor \frac{1}{x}\right\rfloor\right) dx=\sum_{i=1}^{k-1} \int_{\frac{1}{i+1}}^{\frac{1}{i}}\left(\frac{1}{x} -i\right)dx=\sum_{i=1}^{k-1}\left[\ln(x)-ix\right]_{\frac{1}{i+1}}^{\frac{1}{i}}$$
and you can compute this sum and determine the limit, After computation you obtain:
$$\sum_{i=1}^{k-1}\left[\ln(x)-ix\right]_{\frac{1}{i+1}}^{\frac{1}{i}}=\ln(k)-\sum_{i=2}^{k}\frac{1}{i}$$
and this gives you:
$$\int_0^1 \left(\frac{1}{x} - \left\lfloor \frac{1}{x}\right\rfloor\right) dx=1-\gamma$$
with $\gamma$ denotes Euler–Mascheroni constant
Elaqqad
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Split it into a sum of integrals over intervals for which you know the value of the floor term. Note that this will be an infinite sum.
dalastboss
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