# Thread: Help on multiple integrals?

1. ## Help on multiple integrals?

Dear friends,

I have a problem on multiple integrals.
Now, I explain it below:

Let $\displaystyle x=[x_{1},\ldots,x_{n}]$ be a vector, $\displaystyle E:=[a_{1},b_{1}]\times\ldots\times[a_{n},b_{n}]$ be a rectangle, where $\displaystyle a_{i},b_{i}$ are reals for $\displaystyle i=1,\ldots,n$, and $\displaystyle u:E\to\mathbb{R}$ be a continuous function, which vanishes on the boundary $\displaystyle \partial E.$
Then, the following is written
$\displaystyle \int\limits_{E}\bigg\{\big[(x_{i}-a_{i})(b_{i}-x_{i})\big]\int\limits_{a_{i}}^{b_{i}}\bigg|\frac{\partial}{\ partial s_{i}}u(x;s_{i})\bigg|d s_{i}\bigg\}dx$

$\displaystyle =$

$\displaystyle \Bigg(\int\limits_{a_{i}}^{b_{i}}\big[(x_{i}-a_{i})(b_{i}-x_{i})\big]d x_{i}\Bigg)\Bigg(\int\limits_{E}\bigg|\frac{\parti al}{\partial x_{i}}u(x)\bigg|dx\Bigg)$
for $\displaystyle i=1,\ldots,n$, where $\displaystyle u(x;s_{i}):=u(x_{1},\ldots,x_{i-1},s_{i},x_{i+1},\ldots,x_{n}).$

How can we get the second line from the first one?
By partial integration (but how)?

2. Dear friends,

let me tell the answer, I just found it, and now I think that there is nothing interesting about it.
Clearly, we have
$\displaystyle \int\limits_{E}dx=\int\limits_{a_{1}}^{b_{1}}\ldot s\int\limits_{a_{n}}^{b_{n}}dx_{1}\ldots dx_{n}=\int\limits_{a_{1}}^{b_{1}}\ldots\int\limit s_{a_{i-1}}^{b_{i-1}}\int\limits_{a_{i+1}}^{b_{i+1}}\ldots\int\limit s_{a_{n}}^{b_{n}}\int\limits_{a_{i}}^{b_{i}}dx_{1} \ldots dx_{i-1}dx_{i+1}\ldots dx_{n}dx_{i}$
since the integral bounds are constants.
Hence, it suffices to show that
$\displaystyle \int\limits_{a_{i}}^{b_{i}}\bigg\{\big[(x_{i}-a_{i})(b_{i}-x_{i})\big]\int\limits_{a_{i}}^{b_{i}}\bigg|\frac{\partial}{\ partial s_{i}}u(x;s_{i})\bigg|ds_{i}\bigg\}dx_{i}\qquad(*)$

$\displaystyle =$

$\displaystyle \Bigg(\int\limits_{a_{i}}^{b_{i}}\big[(x_{i}-a_{i})(b_{i}-x_{i})\big]d x_{i}\Bigg)\Bigg(\int\limits_{a_{i}}^{b_{i}}\bigg| \frac{\partial}{\partial s_{i}}u(x;s_{i})\bigg|ds_{i}\Bigg)$

$\displaystyle =$

$\displaystyle \Bigg(\int\limits_{a_{i}}^{b_{i}}\big[(x_{i}-a_{i})(b_{i}-x_{i})\big]d x_{i}\Bigg)\Bigg(\int\limits_{a_{i}}^{b_{i}}\bigg| \frac{\partial}{\partial x_{i}}u(x;x_{i})\bigg|dx_{i}\Bigg).$

From $\displaystyle (*)$, we see that $\displaystyle \int\limits_{a_{i}}^{b_{i}}\bigg|\frac{\partial}{\ partial s_{i}}u(x;s_{i})\bigg|ds_{i}$ is independed from the $\displaystyle i^{th}$ component $\displaystyle x_{i}$, and thus the equality is obvious.
Hence integrating both sides of $\displaystyle (*)$ by
$\displaystyle \int\limits_{a_{1}}^{b_{1}}\ldots\int\limits_{a_{i-1}}^{b_{i-1}}\int\limits_{a_{i+1}}^{b_{i+1}}\ldots\int\limit s_{a_{n}}^{b_{n}}dx_{1}\ldots dx_{i-1}dx_{i+1}\ldots dx_{n},$
we get the desired result.

Please, tell me if I am wrong, thanks.