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This is in relation to my question here. I am reading from this paper and specifically the doubt is from a statement on page 177.

Suppose $\alpha\in(0,2)$ and $t_i=ih$ for some fixed $h>0$ and every $i$. It can be shown that $\frac{1}{\Gamma(\alpha)}\int_0^{t_n}((t_{n+1}-\tau)^{\alpha-1})-(t_n-\tau)^{\alpha-1})f(\tau,x(\tau))d\tau\\ =\frac{h}{\Gamma(\alpha)(\alpha-1)}\sum_{i=0}^{t_{n-1}}\int_{t_i}^{t_{i+1}}(z_{n-i}^*(\tau))^{\alpha-2}f(\tau,x(\tau))d\tau $

where $z_j^*(\tau)\in (t_{j-1},t_{j+1})$ for all $j$.

This latter sum is labelled as equation (6) in the paper.

What I now don't understand is the next sentence in the paper:

We can note from (6) that the integrations's kernel $((t_{n+1}-\tau)^{\alpha-1})-(t_n-\tau)^{\alpha-1})$ decays algebraically by the order of $2-\alpha.$

I wish to know as to how it follows from (6) that $\frac{h}{\Gamma(\alpha)(\alpha-1)}\sum_{i=0}^{t_{n-1}}\int_{t_i}^{t_{i+1}}(z_{n-i}^*(\tau))^{\alpha-2}f(\tau,x(\tau))d\tau$ decays algebraically, i.e. the whole integral is bounded above by $Cn^{2-\alpha}$. (Here we also assume $f$ is bounded by some constant $M$). But I am not sure what role do these $z_i^*$ play.

Thanks for reading my question.

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By algebraic decay of order $2-\alpha$, the paper means that $$\int_{t_i}^{t_{i+1}}z_{n-i}^*(\tau)^{\alpha-2}f(\tau,x(\tau))\,d\tau=O(\frac{1}{(n-i)^{2-\alpha}}),\qquad n-i\to\infty.$$ Assuming that $|f|\le M$, this is clear since $z_{n-i}^*(\tau)\ge t_{n-i-1}=(n-i-1)h$. Therefore the integral is bounded in absolute value by $$M (n-i-1)^{\alpha-2} h^{\alpha-1}.$$ The paper is trying to explain that in (5), the region of the integral where $t_n-\tau>>0$ contributes relatively little to the integral since the integrand falls off as $O((t_n-\tau)^{-(2-\alpha)})$. After the integral is expanded as a sum in (6), this becomes the statement that the terms in the sum where $n-i>>0$ fall off as $O((n-i)^{-(2-\alpha)})$. This of course is using the index $i$ in your expression above.

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Thank you. Can I further say that the whole integral $\int_0^{t_n}...$ is $O(n^{\alpha-2})$? –  Shahab Feb 14 '13 at 13:21
    
It's not, no. $ $ $ $ $ $ –  David Moews Feb 14 '13 at 17:24
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