I annoyingly can't justify a step in the solution of the following problem. I have the following expression at hand:
$$ \sum_{n=1}^{N}\int_{-\infty}^{\infty}{(y(x_n + \xi) - t_n})\nu(\xi)\eta(x_n + \xi)d\xi \tag{1}$$
Where $\nu(\xi)$ is a probability distribution of $\xi$. We know that each integral in the summation is well behaving and has a finite result. $y$ and $\eta$ are arbitrary, well behaving functions. Each $x_n$ is an arbitrary shifting factor.
I need to be able to show that this expression equals to :
$$ \int_{-\infty}^{\infty}\eta(x)\left(\sum_{n=1}^{N}{(y(x) - t_n})\nu(x-x_n)\right)dx \tag{2}$$
My aim is to tranform each $x_n + \xi$ term in $\eta$ to the same variable, say $x$. In $(1)$, for each integral I apply the following substitution: $k_n=x_n + \xi$. Since $\dfrac{dk_n}{d\xi}=1$, I obtain:
$$ \sum_{n=1}^{N}\int_{-\infty}^{\infty}{(y(k_n) - t_n})\nu(k_n - x_n)\eta(k_n)dk_n \tag{3}$$
Now, I am stuck with this: If I can show that I can transform each $k_n$ in each integral to the same variable $x$ as $x=k_n$, then each integral will have the same integration variable and I can proceed by interchanging summation and integral and obtain $(2)$. But I am uneasy with this operation: Each $k_n$ is equal to $k_n=\xi + x_n$. So, if I transform any single variable $k_n$, say $k_1$, to $x$, then it is clearly $x=k_1\neq k_2 \neq \dots \neq k_N$. Since it is $x=k_n - x_n + x_1$ for any $k_n$, $n\neq 1$. So, I would obtain:
$$ \int_{-\infty}^{\infty}{(y(x) - t_1})\nu(x-x_1)\eta(x)dx + \sum_{n=2}^{N}\int_{-\infty}^{\infty}{(y(x-x_n+x_1) - t_n})\nu(x - x_1)\eta(x-x_n+x_1)dx $$
This last expression is not different than $(1)$ so this causes me to just run in circles. What justification do I need here to obtain the expression in form $(2)$? I nearly wasted two days on this, I direly need help...
Hint: $\int g(u)\operatorname d u = \int g(v)\operatorname d v$
The variable of integration is a 'dummy variable'. It is defined only within the scope of the integral. You can alpha-replace it with any unbound variable.
Hence $$\begin{align} \sum_{n=1}^{N}\int_\Bbb R G(x_n, k_n)\operatorname d k_n & = \int_\Bbb R G(x_1, k_1)\operatorname d k_1+\int_\Bbb R G(x_2, k_2)\operatorname d k_2+\cdots+\int_\Bbb R G(x_N, k_N)\operatorname d k_N \\[1ex] & = \int_\Bbb R G(x_1, x)\operatorname d x+\int_\Bbb R G(x_2, x)\operatorname d x+\cdots+\int_\Bbb R G(x_N, x)\operatorname d x \\[1ex] &= \sum_{n=1}^{N} \int_\Bbb R G(x_n, x)\operatorname d x \\[1ex] &= \int_\Bbb R \sum_{n=1}^{N} G(x_n, x)\operatorname d x \end{align}$$