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Let $r: \mathbb{R} \rightarrow \mathbb{R}$ be a square integrable differentiable real function with square integrable derivative and let $\alpha = \int_{-\infty}^{+\infty}[r'(x)]^2dx$, where $r'$ denotes the derivative of r.

Now let $x = \sum_{i=1}^N (y_i-k)^2$, for $k, y_i \in \mathbb{R}, i \in \mathbb{Z}$ and define $u: \mathbb{R}^N \rightarrow \mathbb{R}$ such that $u(\vec y) = r(x)$, for all $\vec y = (y_1, y_2, ..., y_N)$. I want to compute the following integral as a function of $\alpha$ and $k$:

$$\int_{\vec y \in \mathbb{R}^N}\left|\frac{d}{d \vec y}u(\vec y)\right|^2d \vec y \tag{1}$$

Note: $\frac{d}{d \vec y}$ denotes the gradient of $r(\vec y)$.

Could anyone give me a help?


PS: some answers for the comments below:

1) Do you have any specific $r$ in mind? Actually no, but it would be of great help if one could solve for $r(x) = e^{-c^2 x}$.

2) Why $i$ goes from $1$ to $N$? It's just a choice. It could be $i=0$ to $N-1$. The important is that there are finite values for $i$.

3) What if $N = 1$? If $N=1$, then $x = y_1$.

4) Where it all comes from? This is part of a multidimensional cost function I need to minimize by varying the value of $k$. Ideally, it would be very great if I could write the cost function in terms of $k$ and $\alpha$ because of computational performance. Actually, I need to minimize the euclidean norm of the gradient function of $u$ by varying the unconstrained parameter $k$.


Some work on this until now:

$$|\frac{d}{d \vec y} u(\vec y)|^2 = \sum_{j=1}^N [\frac{d}{d y_j}u(\sum_{i=1}^N (y_i-k)^2)]^2\tag{2}$$.

So, the integral can be rewritten as:

$$\sum_{j=1}^N \int_{-\infty}^{\infty} [\frac{d}{d y_j}u(\sum_{i=1}^N (y_i-k)^2)]^2 d y_j\tag{3}$$

Applying the derivative:

$$\frac{d}{d y_j}u(\sum_{i=1}^N (y_i-k)^2)=2(y_j-k) r'(\sum_{i=1}^N (y_i-k)^2)\tag{4}$$

Replacing again in the integral, we get:

$$\sum_{j=1}^N \int_{-\infty}^{\infty} [\frac{d}{d y_j}u(\sum_{i=1}^N (y_i-k)^2)]^2 d y_j = \sum_{j=1}^N \int_{-\infty}^{\infty} [2(y_j-k) r'(\sum_{i=1}^N (y_i-k)^2)]^2 d y_j \tag{5}$$

From now, I would make a change of variables $x = \sum_{i=1}^N (y_i-k)^2$

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$r:\mathbb{R}\rightarrow\mathbb{R}$ in the first line and $r:\mathbb{R}^N\rightarrow \mathbb{R}$ in the second. –  Nonliapunov Oct 5 '12 at 13:22
    
ok, ok, ok....... –  user38397 Oct 5 '12 at 13:32
    
How would it go for $N=1$? –  Berci Oct 5 '12 at 16:29
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You're going to need to give a better explanation and to put the problem in to context. Do you have a certain $r$ in mind? Why do you define $u$ the way you do? Why do you want to know the integral of $|\nabla u|^2$? In short, where has this question come from, why do you need to answer it, and what're you going to do with the answer? –  Fly by Night Oct 5 '12 at 16:29
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I can do so, but why do you need all this? It's just computation of an integral, nothing more than that. The problem is well defined. There are LOTS of answered posts in this forum just like this one. –  user38397 Oct 5 '12 at 16:34

1 Answer 1

up vote 1 down vote accepted

The answer does not depend on $k$ and is not a simple function of $\alpha$ since the integral is $$ 2\sigma_{N-1}\cdot\int_0^{+\infty}r'(x)^2x^{N/2}\mathrm dx, $$ where $\sigma_{N-1}$ denotes the surface of the unit sphere in $\mathbb R^N$.

Edit: For every $i$, $\partial_ir(x)=(\partial_ix)r'(x)=2(y_i-k)r'(x)$, where $\partial_i$ denotes the partial derivative with respect to $y_i$. Thus, $$ \|\nabla u(\vec y)\|^2=\sum\limits_{i=1}^N(\partial_ir(x))^2=4r'(x)^2\sum\limits_{i=1}^N(y_i-k)^2=4xr'(x)^2. $$ The change of variable $\vec y=\vec k+s\vec u$, where $\vec k$ is the vector whose every coordinate is $k$ and $\vec u$ is a unit vector yields $\mathrm d\vec y=s^{N-1}\mathrm ds\mathrm d\sigma_{N-1}(\vec u)$. Integrating with respect to $\mathrm d\sigma_{N-1}$ yields the value $$ 4\sigma_{N-1}\cdot\int_0^{+\infty}s^2r'(s^2)^2s^{N-1}\mathrm ds. $$ The change of variable $x=s^2$ yields the value given at the beginning of this post.

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could you give me a clue on how you got this result? You have a +1 from me. –  user38397 Oct 5 '12 at 18:03

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