Integral of null function I am a math grad student, and I understand that from an undergrad calc students perspective that it is fine notation/convention to write
$$\int dx=\int 1 dx$$
But I don't have any idea why we should be able to say that this thing is undefined. 
If we view $\int$ as an operator, what is $\int dx$ the integral of? Where is the function? we can just throw a 1 in there with no problems? it seems like we are taking the integral of a null function, but we should be looking at functions from $\mathbb R \rightarrow \mathbb R$ so I don't know why this would be allowed.
Ay comments are welcome.
Thanks
 A: In math, nothing has any meaning or value until we define it. In this case, when we write $\int dx$ we are just using this as shorthand notation for $\int 1dx$; in other words, $\int dx=\int 1dx$ by definition.
But why do we define it this way and why does this make sense? Well, the answer is that although $dx$ is a differential and not formally a real number per se, one often intuitively thinks of it as an "infinitesimally small" real number, which must obey the usual laws of arithmetic. In particular, $1\cdot dx=dx$. So when we apply the $\int$ operator to both sides, we get $\int 1dx=\int dx$. I should however stress that this is by no means a derivation or proof of the identity, merely a motivation of why we would want to define things this way.
To make this slightly more formal, the integral $\int_a^bf(x)dx$ is defined by the limit of the Riemann sum
$$\int_a^bf(x)dx:=\lim_{\Delta x\to0}\sum_{i=0}^{h} f(x_i)\Delta x\quad\text{where }h=\frac{b-a}{\Delta x}.$$
When $f(x)=1$, this is
$$\int_a^b1\,dx=\lim_{\Delta x\to 0}\sum_{i=0}^{h}(1\cdot\Delta x)=\lim_{\Delta x\to 0}\sum_{i=0}^{h}\Delta x.$$
We would like to say intuitively that $\Delta x$ tends to $dx$ infinitesimally, and that $\sum$ tends to $\int$ as $\Delta x$ goes to $dx$. This is core to the intuition of the integral being the continuous analogue of the sum. Hence, since we can safely treat $\Delta x$ as the same thing as $1\cdot\Delta x$, it intuitively makes sense to want to treat $dx$ the same as $1dx$ as well, in the context of integration.
A: Well, it's just a notational convenience, I think it makes sense if we look at the Riemann sum that define the integral:
$$\sum f(x_i)\Delta _i$$
Here the  $\Delta_i$ is the length of the interval on the partition. I think the value $\Delta_i$ is replaced with $dx$ in the notation for the limit, so if have only $\Delta_i$ in the sum with nothing on the left it would be equivalent to multiply by 1 (and not 0).
A: I like to think of it similarly to when we use u-substitution. If we let $u=x^2$ then $\frac{du}{dx}=2x$, which we would rearrange to $dx=\frac{du}{2x}$ to continue with the problem. In cases like this we can treat $dx$ as a variable in its own right, multiplying and dividing by it.
Similarly, if I wanted to find the product of $x$ and $y$ I would say $x\times y=xy$, it is not necessary to say $1x\times 1y=1xy$. Even though it says $dx$ we can read this as $1.dx$ since $0.dx$ would have no real meaning
