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How do I simplify the following expression? $$\displaystyle \frac{\int_q^1 w(s) \int_0^s e(\xi) d\xi ds}{2\int_q^1 w(s) ds} p$$ where $w(t)$ is nondecreasing $w(t)>0$ on $(q,1]$ , $e :(0,1)\rightarrow \mathbb{R}$ is continuous, $e\in L(0,1)$, and $\displaystyle\frac12<p<q<1$

Please help me. Thank you.

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i need to simplify this integration $$\displaystyle \frac{\int_q^1 w(s) \int_0^s e(\xi) d\xi ds}{2\int_q^1 w(s) ds} p$$ –  Vrouvrou May 16 '13 at 9:49
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In which way do you want to simplify it? –  Mhenni Benghorbal May 16 '13 at 10:28
    
i want to transform $\displaystyle \frac{\int_q^1 w(s) \int_0^s e(\xi) d\xi ds}{2\int_q^1 w(s) ds}$ into an integral $ds d\xi$ –  Vrouvrou May 16 '13 at 10:35
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1 Answer

Changing the order if integration gives $$ \int_q^1 w(s) \int_0^s e(\xi) d\xi ds = \int_0^q e(\xi) \int_q^1 w(s)\, ds d\xi + \int_q^1 e(\xi)\int_{\xi}^1 w(s)\, ds d\xi .$$

Note that, the first integral on the right hand side can be written as

$$ \int_0^q e(\xi)d\xi \int_q^1 w(s)\, ds = \int_q^1 w(s)\, ds\int_0^q e(\xi)d\xi. $$

Plot the region to see how the first equation derived.

Added: Plot the region $ \xi(s) = s $ where $ q \leq s \leq 1 $. Then to change the order of integration, consider taking a horizontal strip and notice that the horizontal strip will be bounded below by two different functions, namely, $s=q$ and $s=\xi$. Accordingly, find the limits for $\xi$.

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i don't understand , how you have done that ? –  Vrouvrou May 17 '13 at 19:44
    
Beghorbal please –  Vrouvrou May 17 '13 at 20:08
    
ok, we dont use Fubini ? –  Vrouvrou May 20 '13 at 14:46
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