How do "Dummy Variables" work?
I do not understand how dummy variables work in math.
Suppose we have:
$$I_1 = \int_{0}^{\infty} e^{-x^2} dx$$
How is this equivalent to:
$$I_2 = \int_{0}^{\infty} e^{-y^2} dy$$
How does this dummy variable system work?
Since $y$ is the dependent variable for $I_1$ How can $y$ itself be and independent variable for $I_2$
??
Thanks!
$$ \sum_{j=1}^3 j^4 = 1^4 + 2^4 + 3^4 = \sum_{k=1}^3 k^4. $$ In the first term, $1^4$, we can say that $j=1$; in the second term $2^4$ we have $j=2$, and in the third term, $3^4$ we have $j=3$. But when we call the index $k$ rather than $j$, then in the first term $k=1$, in the second $k=2$, and in the third $k=3$.
$j$ or $k$ is a bound variable (also sometimes called a dummy variable). The value of the whole expression $1^4+2^4+3^4$, which is $98$, does not depend on the value of a bound variable.
If we write $\displaystyle\sum_{j=1}^3 (j \cos (j+m))^4$, then $j$ is bound and $m$ is free. The sum is $$(1\cos (1+m))^4 + (2\cos (2+m))^4 + (3\cos(3+m))^4.\tag 1$$ Its value depends on the value of the free variable $m$, but not on the value of anything called $j$. Accordingly we do not see $j$ in $(1)$. We could rename $j$ and call it $k$, and the whole thing would still be equal to the expression $(1)$ in which we also do not see $k$.
See this Wikipedia article.
Another example is expressions like $\displaystyle\lim_{h\to0}\frac{(x+h)^3-x^3}h=3x^3$. The value of this expression depends on the value of the free variable $x$, but not of the bound (or "dummy") variable $h$. We could have said $\displaystyle\lim_{k\to0}\frac{(x+k)^3-x^3}k=3x^3$ and it would still be $3x^2$.
In both cases they are the independent variable. Note in your integrals that $x$ appears nowhere in $I_2$ and $y$ occurs nowhere in $I_1$.
It's common to write $y$ as the dependent variable and $x$ the independent variable. But there's no rule that says you have to. You can write $x = f(y) = y^2 + 2\sin y$ if you wanted to, or put the $y$-axis horizontal and $x$ vertical.
edit: note in any event that $I$ is constant $\left({I = \dfrac {\sqrt{\pi}} 2}\right)$ so it isn't "dependent" in the way you may be used to. Mathematicians will say things like "$I$ is a trivial function of $x$" which means that even though it's technically true that $I = f(x)$, the relationship isn't interesting.
Here's an interesting non-trivial function:
$\displaystyle I(z) = \int_0^z e^{-t^2} \, \mathrm dt$
IMO this can be understood much better by writing it out in a proper context-free way, rather than standard mathematical notation. The integration symbol $\int_0^\infty$ is basically a higher-order-function, taking a real function1 and returning a single number. $$ \int\limits_0^\infty : (\mathbb{R}\to \mathbb{R}) \to \mathbb{R}. $$ Now, people keep saying stuff like "$\sin(x)$ is a function..." but really this is incorrect. $\sin(x)$ is, for any value of $x$, just a single real number. What's a function is $\sin$ itself, i.e. not applied to anything. For instance, $$ \int\limits_0^\pi \sin = 2 $$ would be quite a reasonable statement. OTOH, it's nonsense to write $$ \int\limits_{-\infty}^0 e^x = 1 $$ because $e^x$ is not a function.
Trouble is, most functions you need to integrate won't have a predefined name. You can't thus write "the function itself" directly, but have to specify those results of the function for general inputs, in form of some algebraic expression. You know, the equation-definitions $$ f(x) = e^{-x^2}. $$ Here it should be quite clear that the $x$ symbol is just an arbitrary choice – it's something you use to signify "anything you can put into the function", and $f(y) = e^{-y^2}$ says obviously exactly the same thing.
Always having to give functions a name before integrating would be tedious, so we have "shortcut syntax" to define an "anonymous function" and use it right away. You can thus think of the $\mathrm{d}x$ symbol much as a lambda function.
1In fact, more accurately and generally, a differential form.
2Again, that's not all there is to it since $\mathrm{d}x$ really is a differential form.