Is the term "properly defined" equivalent to "well-defined"? In many math books I see the terms "properly defined" or/and "well-defined". Do they have the same meaning? 
For example: "......, so that the integral is properly defined", 
or, 
"....., hence the fuction $f$ is well-defined", 
... etc.
I think that "well-defined" means, at the same time, "properly defined" for mathematical objects. But what about the other implication?
Is there a difference between these two concepts in mathematics?
 A: "Well-defined" has a technical meaning though it is also used informally (but in my experience, often in a way that can be rationalized as the technical meaning). I don't think I've ever seen "properly defined" given a formal definition, and it is almost certainly an informal statement in this case. It may or may not be synonymous with "well-defined". You haven't provided enough context to say how either example is being used.
The formal definition of well-defined has to do with quotient sets (see also) or more general but related constructions. The best way to think of this in my opinion is that given a set $X$ and an equivalence relation on $X$, $\sim$, the quotient set $X/{\sim}$ has the same elements as $X$ but when $x$ and $y$ are viewed as elements of $X/{\sim}$ we treat them as equal if and only if $x\sim y$. Now the defining property of functions is that they take equal inputs to equal outputs, i.e. if $x = y$ then $f(x) = f(y)$. If $f:X/{\sim}\to Z$ is a function on the quotient set $X/{\sim}$, this becomes $x\sim y$ implies $f(x)=f(y)$, but this could fail to be true. It is not automatic that $f$ respects $\sim$. $f$ is well-defined (as a function on $X/{\sim}$) if it does respect $\sim$, i.e. if $x\sim y$ does imply $f(x)=f(y)$.
You actually have extensive experience with one set that is commonly treated in exactly this manner: the rationals, $\mathbb Q$. You were most likely taught that two rational numbers are "the same" if they are the same when reduced to lowest terms. You may have also been taught another way to see if two rational numbers are equal: $\frac{n}{d}=\frac{n'}{d'}$ if and only if $nd'=n'd$. You can show that this defines an equivalence relation on pairs of numbers (if we exclude $0$ from the denominator). This can be formalized as $\mathbb Q\cong(\mathbb Z\times\mathbb N^+)/{\sim}$ where $\sim$ is the equivalence relation just mentioned.
Consider the "function" $\frac{n}{d}\mapsto nd$. As a function on pairs of numbers $(n,d)$ it is a perfectly meaningful function, but as a "function" on rationals it is not well-defined. For example, $\frac{2n}{2d}$ would be mapped to $4nd$ which is not equal to $nd$ in general. As a far more subtle example, is $(z,q)\mapsto z^q:\mathbb C\times\mathbb Q^+\to\mathbb C$ well-defined? The answer is "no" if we want $(z^a)^b=z^{ab}$ to hold. We would get $(-1)^{\frac{2}{2}}=(-1)^1=-1$ but $((-1)^2)^\frac{1}{2}=1\neq -1$. You might say that you can choose $1^\frac{1}{2}=-1$, but if you did that consistently, then $(1^2)^\frac{1}{2}=-1\neq 1$.
There is also the notion of a (partial) function simply being "defined". If we think of $(n,d)\mapsto\frac{n}{d}$ as on operation on all pairs of integers, then it has no value when $d=0$. We say $\frac{n}{d}$ is "not defined" for $d=0$. More formally, we can view this as a relation $R(n,d,q)$ with the rough, informal idea being that $R(n,d,q)$ holds when $q=\frac{n}{d}$. A formal definition would be $R(n,d,q)\iff \frac{d}{1}q=\frac{n}{1}\land d\neq 0$. With this, a formula like $\frac{n}{d}>0$ more formally would mean $\exists q.R(n,d,q)\land q>0$. When $\exists q.R(n,d,q)$ does hold, we would say that $\frac{n}{d}$ is defined. It's possible this is what is intended by "the integral is properly defined". If we think of (definite) integration (over some given domain) as an operation mapping functions to numbers, then it may well be a partial function. That is, it may not be defined for certain functions. If this was the intent, the adverb "properly" is not doing anything formally. It would be just as correct to simply say "the integral is defined". That said, most lower-level texts are extremely unclear on what the domain and codomain of the described operations are, and they use notation which encourages confusion. So "properly defined" may be intended as emphasis and contrast against an integral that is "improperly 'defined'" as "$\infty$".
