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Can the following integral be computed?

enter image description here

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  • $\begingroup$ "Can the following integral be computed?" $\endgroup$
    – Pedro
    Jan 22, 2014 at 5:20
  • $\begingroup$ Sorry........ :-) $\endgroup$
    – user121418
    Jan 22, 2014 at 5:21
  • $\begingroup$ You can, but it is slightly messy. Divide $[0,1]$ into intervals where $\{1/x\}$ is continuous and has an easy expression. $\endgroup$
    – Pedro
    Jan 22, 2014 at 5:23
  • $\begingroup$ A related problem. $\endgroup$ Jan 22, 2014 at 5:46

3 Answers 3

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Hint: consider $$A_n=\int_{1/(n+1)}^{1/n}\{1/x\}^{4}dx$$. This can be computed. Does the series $\sum_{i=1}^\infty A_i$ converges?

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  • $\begingroup$ Sketch $\{\frac{1}{x}\}^4$ between 0 and 1. You will find that it is bounded above by a simple function. So the integral you want is bounded above by a simple integral. $\endgroup$
    – Empy2
    Jan 22, 2014 at 6:06
  • $\begingroup$ please someone tell me the answer. $\endgroup$
    – user121418
    Jan 22, 2014 at 6:23
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With some preliminary transformations (in attachement) the integral is reduced to the integral of a polygamma function which is known (I let WolframAlpha do the known part of the job). enter image description here

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  • $\begingroup$ Numerical computations with Maple and Mathematica confirm the result. $\endgroup$
    – Lucian
    Jan 22, 2014 at 19:23
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First, let's prove that it converges:

$$\int_0^1\bigg\{\frac1x\bigg\}^4dx=\int_1^\infty\frac{\{t\}^4}{t^2}dt\color{red}<\int_1^\infty\frac{1^4}{t^2}dt=\bigg[-\frac1t\bigg]_1^\infty=1,\qquad\text{since }0\le\{t\}<1.$$


$$\int_0^1\bigg\{\frac1x\bigg\}^4dx=\int_1^\infty\frac{\{t\}^4}{t^2}dt=\sum_1^\infty\int_k^{k+1}\frac{(t-k)^4}{t^2}dt=\sum_1^\infty\int_0^1\frac{u^4}{(u+k)^2}du=$$

$$=\sum_1^\infty\bigg(\frac43-2k+4k^2-\frac1{k+1}+4k^3\ln\frac k{k+1}\bigg)=\ldots<1$$

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