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So I have the following integral: $$ \int_{k - 1/2}^{k + 1/2} \log t \ \mathrm{d}t $$ And I would like to show it can be written as: $$ \int_{k - 1/2}^{k + 1/2} \log t \ \mathrm{d}t = \log k + \mathcal{O}(1/k^2) $$ How would I proceed? The first thing that comes to mind is integrating the function and simplifying of course. We end up with: $$ \left(x+\frac{1}{2}\right)\left(\log\left(x+\frac{1}{2}\right)\ -\ 1\right)\ -\left(x-\frac{1}{2}\right)\left(\log\left(x-\frac{1}{2}\right)\ -\ 1\right)\ $$ which is well and good, but I have not been able to simplify this and get a reasonable error term. Does anyone have any ideas? I would appreciate hints over full answers. Thanks

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For asymptotic analysis, it is not necessary to carry out the exact integral.

Perhaps you may want to play this magic. Note that, for $t\in\left(k-1/2,k+1/2\right)$, $$ \log t=\log\left(\frac{t}{k}k\right)=\log k+\log\frac{t}{k}. $$ Denote $$ s=\frac{t}{k}-1\in\left(-\frac{1}{2k},\frac{1}{2k}\right)\subseteq\left(-1,1\right). $$ Thus Taylor's theorem applies, and $$ \log t=\log k+\log\left(1+s\right)=\log k+s-\frac{s^2}{2}+O(s^3). $$ Thanks to this result, \begin{align} \int_{k-1/2}^{k+1/2}\log t{\rm d}t&=\log k+\int_{k-1/2}^{k+1/2}\left[s-\frac{s^2}{2}+O(s^3)\right]{\rm d}t\\ &=\log k+k\int_{-1/2k}^{1/2k}\left[s-\frac{s^2}{2}+O(s^3)\right]{\rm d}s, \end{align} which suffices to lead to your expected estimate.

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  • $\begingroup$ Beautiful! Where did you get the intuition to do such a change of variables? $\endgroup$ – rubikscube09 Jul 22 '18 at 5:18
  • $\begingroup$ @rubikscube09: This is because you hope to apply Taylor theorem to $\log\left(t/k\right)$, while $t/k\in\left(1-1/2k,1+1/2k\right)$. Thus a natural choice is to expand $\log\left(t/k\right)$ at $t/k=1$, hence the change of variable follows immediately. $\endgroup$ – hypernova Jul 22 '18 at 9:51
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You properly wrote$$\left(x+\frac{1}{2}\right)\left(\log\left(x+\frac{1}{2}\right)\ -\ 1\right) -\left(x-\frac{1}{2}\right)\left(\log\left(x-\frac{1}{2}\right)\ -\ 1\right)$$ Now rewrite $$\log(x+a)=\log(x)+\log\left(1+\frac a x\right)=\log(x)+\frac{a}{x}-\frac{a^2}{2 x^2}+\frac{a^3}{3 x^3}+O\left(\frac{1}{x^4}\right)$$ So, $$(x+a)\, (\log (x+a)-1)-(x-a)\,(\log (x-a)-1)=2a \log(x)-\frac{a^3}{3 x^2}+O\left(\frac{1}{x^4}\right)$$ and $a=\frac 12$.

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