I am quite confused about how the integral definition is used to prove $\ln(xy)= \ln(x)+ \ln(y) $ below, especially in how it does not treat the integrand as a strict reciprocal function; and even if it did I am not seeing how the bounds apply. Here is the proof I am reading:
$$ \begin{aligned} \ln x y & = \int_1^{x y} d t / t \\ & =\int_1^x d t / t+\int_x^{x y} d t / t \end{aligned} $$ Then, let $u= \frac {t}{x}$, so, $t=u x$ and $d t=x \cdot d u$. Therefore, $\frac{1}{t} d t=\frac{1}{u x} \cdot x d u=\frac{1}{u} d u$, so the above integral is $$ \int_1^x \frac {d t} t+\int_1^y \frac{du} u=\ln x+\ln y $$ I have three questions here:
- When we conclude that $\int_1^y \frac{du} u=\ln y$ using the definition $\ln(x)=\int_1^x\frac {dt}t$, I thought that $t$ must be a single variable, not any function, such that the integrand is exactly the reciprocal function. However, here we have a function $u$ which is not defined as just a single variable. How is claiming this integral is the $\ln$ definition justified?
- If we are going to make the argument that $\frac {du} u$ represents the reciprocal function $\frac {dt}t$, and on that account apply the definition, are we not unjustified in using the bounds of integration for $u$ when we interpret the integral as being with respect to $t$? If we say the integral definition of $\ln(x)$ applies here due to the integrand really being $\frac {dt}t$, which is strictly the reciprocal function, it seems we can only make this argument when the bounds of integration are the original $x$ and $xy$, and the definition of $\ln(y)$ would still not apply as we cannot use the bounds $1$ and $y$, which are the bounds for $u$ not $t$.
- When we have used properties of integrals to break up $\int_1^{x y} \frac {d t} t$ into 2 integrals, how does this work when $x$ and $y$ are variables? Do we treat them as constants here? I am not sure whether we can use this property of integrals for variables.
$\newcommand{\d}{\,\mathrm{d}}$In formal logic, I am under the impression (I know nothing of the subject) that there are some precise definitions of constants and logical variables.
Outside of that context, it is my opinion that discussion of constant/variable is only useful for the intuition, and that these words are convenient fictions to help us understand a situation. These words have clearly confused you greatly and I've seen them confuse my friends, very often, but there is little to no meaningful difference between them.
If you make things more precise with elements and functions, then you will not get confused (given a certain 'mathematical maturity'). "$u$" on its own is a meaningless symbol. $u=t/x$ is implicitly defining a function $u:[1,x]\to\Bbb R$ that maps $t\mapsto t/x$. In the same way, "$t$" on its own has no meaning, unless you tell me - hey, here is some $t\in [1,x]$. Then I know $t$ denotes some element of the interval $[1,x]$. I still do not care in the slightest about "$t$"'s status as a constant or a variable; I perceive that to be a meaningless question.
The definition: $$\ln(x)=\int_1^x\frac{1}{t}\d t$$Is more precisely saying: $\ln:(0,\infty)\to\Bbb R$ is the function which takes an positive real $x$ to the real number $\int_1^x\frac{1}{t}\d t$. Think about how that integral is actually defined: nowhere does the definition say, "hey! Give me '$t$' as an input, or else!" We may as well integrate $\frac{1}{\text{cucumber}}\d\,\text{cucumber}$. I might want to define $g:(0,\infty)\to\Bbb R$ by $a\mapsto1/a$ and then say, $\ln(x):=\int_1^x g$. This expression is well-defined, and suppresses any notation of the "variable" $t$. $\int_1^y\frac{1}{u}\d u$ means exactly the same thing as $\int_1^yg$. Earlier, we had defined the function $u:t\mapsto t/x$, but the notation $\int\frac{1}{u}\d u$ does not care about that function $u$; I suppose this is an abuse of notation, since the symbol $u$ is being overloaded. Maybe this 'abuse' is the main source of your confusion: when handling the "change-of-variable", you must forget the previous definition.
We can state the "change-of-variable" formula without mentioning "other variables":
We can also state it using just one symbol to denote our "variable":
It may be helpful for you to internalise this formula. I deliberately avoided attaching any specific notation to the inputs of the functions $F$ and $(F\circ G)G'$, to emphasise that such notation is placeholder notation.
Whether or not $u$ is a "single variable" is irrelevant. In isolation, what does that even mean? This is not physics, where we are modelling a moving object '$u$' whose value really does vary because, as we speak, it is rolling down a hill. No. In $\int\frac{1}{u}\d u$, $u$ is a placeholder for a real number. It is no more or less variable than the symbol $1$. The notation just indicates a little something about how the (Riemann) integral is defined: you take some interval $(u,u+h)$ and add $h\cdot 1/u$ to the Riemann sum... where $u$ may be any real number at all, in the interval $[1,x]$.
We don't need to make any argument at all. $\frac{1}{u}\d u$ and $\frac{1}{t}\d t$ are themselves both meaningless (caveat: I ignore the $\d$ notations that do have meaning in other areas of maths) outside of the integral sign $\int$ and, strictly speaking, $\frac{1}{u}$ does not represent a function any more than $\frac{1}{t}$ does. These are placeholder symbols for the function $a\mapsto1/a$, or for $t\mapsto1/t$, or for $\text{eggs}\mapsto1/\text{eggs}$... the symbol used is irrelevant. Remember, in the change-of-variables formula, when we 'switch' to $1/u\d u$ we forget the previous definition of the symbol $u$ as a function ($t\mapsto t/x$).
"Bounds for $u$ not $t$" doesn't make sense. $1,y,x,xy$ are all real numbers. The integral is well-defined irrelevant of which numbers you use as in the limit. The integral definition of $\ln$ does not apply to the integrand "really being $1/t \d t$". The integral definition of $\ln$, strictly speaking, applies to the integrand $g$. $1/t \d t$ is merely a convenient way to denote the integral, but you mustn't cling to this notation - it is, again, placeholder notation. The integral does not "care" what the letters are inside of it, nor does it care what its bounds are. $\int_1^y g$ makes equal sense to $\int_x^{xy} g$. The definitions apply, equally, and these equal $\ln(y)-\ln(1)$ and $\ln(xy)-\ln(x)$ respectively (these are in fact the same value).
$x$ and $y$ are not really "variables" because "variables" doesn't really mean anything here. In $\ln xy=\cdots$, $x$ and $y$ are implicitly assumed to denote some positive real numbers. The only purpose of giving them a "generic" symbol like "$x$" is to suggest that, when we eventually show $\ln xy=\ln x+\ln y$, that this equation will be true for all possible $x,y$. But $x$ may as well be $\pi$, or $e$, or $1$, or $2$. $\int_a^b h$ makes sense where $h$ is a function, and $a,b$ are real numbers. The 'variability' of $a,b,h$ is not meaningful and is certainly irrelevant.
You can use this property of integrals because examining any proof of this property together with any definition of integral reveals, again, that the integral "does not care" about such concerns of "variable, constant" etc.