Can the following integral be computed?
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$\begingroup$ "Can the following integral be computed?" $\endgroup$– Pedro ♦Jan 22, 2014 at 5:20
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$\begingroup$ Sorry........ :-) $\endgroup$– user121418Jan 22, 2014 at 5:21
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$\begingroup$ 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
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$\begingroup$ A related problem. $\endgroup$– Mhenni BenghorbalJan 22, 2014 at 5:46
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3 Answers
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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. $\endgroup$– Empy2Jan 22, 2014 at 6:06
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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).
answered Jan 22, 2014 at 12:08
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$\begingroup$ Numerical computations with Maple and Mathematica confirm the result. $\endgroup$– LucianJan 22, 2014 at 19:23
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First, let's prove that it converges:
$$\int_0^1\bigg\{\frac1x\bigg\}^4dx=\int_1^\infty\frac{\{t\}^4}{t^2}dt\color{red}<\int_1^\infty\frac{1^4}{t^2}dt=\bigg[-\frac1t\bigg]_1^\infty=1,\qquad\text{since }0\le\{t\}<1.$$
$$\int_0^1\bigg\{\frac1x\bigg\}^4dx=\int_1^\infty\frac{\{t\}^4}{t^2}dt=\sum_1^\infty\int_k^{k+1}\frac{(t-k)^4}{t^2}dt=\sum_1^\infty\int_0^1\frac{u^4}{(u+k)^2}du=$$
$$=\sum_1^\infty\bigg(\frac43-2k+4k^2-\frac1{k+1}+4k^3\ln\frac k{k+1}\bigg)=\ldots<1$$
