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I am trying to read the research paper here and have a small doubt in it. On page 177 the following integration is carried out:


Here $h>0$, $t_i=ih,\tau\in(0,t_n)$ and $z^*_1(\tau)\in (0,t_2)$.

It is not clear to me why this equality is true. My idea is that it has something to with the mean value theorem of integral calculus but I am not sure as to how it applies. Can someone help please?


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There is no page 127 in the paper you linked (Volume 206, Issue 1, 1 September 2007, Pages 174–188). – Stefan Hansen Feb 13 '13 at 9:07
@Stefan Hansen: My mistake. It is actually page 177. – Shahab Feb 13 '13 at 9:48
up vote 2 down vote accepted

On page 177, the paper deals with integrals of the form $$ \int_0^{t_1}\left(\int_{t_n-\tau}^{t_{n+1}-\tau}z^{\alpha-2}\,\mathrm dz\right)f(\tau,x(\tau))\,\mathrm d\tau,\quad\ldots\quad\int_{t_{n-1}}^{t_n}\left(\int_{t_n-\tau}^{t_{n+1}-\tau}z^{\alpha-2}\,\mathrm dz\right)f(\tau,x(\tau))\,\mathrm d\tau. $$ So let's pick a $k\in \{1,\ldots,n\}$ and consider the integral $$ \int_{t_{k-1}}^{t_k}\left(\int_{t_n-\tau}^{t_{n+1}-\tau}z^{\alpha-2}\,\mathrm dz\right)f(\tau,x(\tau))\,\mathrm d\tau $$ with $t_0=0$.

Let $\tau\in (t_{k-1},t_k)$ be fixed. By the mean value theorem you mention, we have the existence of a $z_k^*(\tau)\in (t_n-\tau,t_{n+1}-\tau)$ such that $$ \int_{t_n-\tau}^{t_{n+1}-\tau}z^{\alpha-2}\,\mathrm dz=h\cdot z_k^*(\tau)^{\alpha-2}, $$ with $h=t_{n+1}-\tau-(t_n-\tau)=t_{n+1}-t_n$. Let's take a closer look at where $z_k^*(\tau)$ can vary for a fixed $\tau$. Since $$ t_n-\tau< z_k^*(\tau)< t_{n+1}-\tau\quad \text{and}\quad t_{k-1}<\tau<t_k $$ we have that $$ t_n-t_k<z_k^*(\tau)<t_{n+1}-t_{k-1}. $$ Now, on page 176 it says that $t_n=nh$ and so $t_{n+1}-t_{k-1}=t_{n-k+2}$ and $t_n-t_k=t_{n-k}$. Thus $$ z_k^*(\tau)\in (t_{n-k},t_{n-k+2}). $$ Note that the $z_k^*$'s in the paper are reversed compared to mine, i.e. $z_k^*$ here is $z_{n-k}^*$ in the paper.

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