Assumed True until proven False. The Curious Case of the Vacuous Truth
One can, correctly, assign the truth-value of true to the statement $P\implies Q$ whenever $P$ is false, or whenever $Q$ is true. $P\implies Q\,$ is false if and only if both $P$ is true and $Q$ is false. That covers all the cases. So we can say that $P\implies Q$ is true, unless "proven false", by which I mean to say:
$P \implies Q$ is true if and only if it is not the case that both $P$ is true and $Q$ is false.
What we can also say is that in classical logic and in math, it is a mistake to attribute any sort of causal relationship between $P$ and $Q$ when writing or reading an implication $P\implies Q$. Put differently, $P\implies Q$, by itself, does not imply any causal relationship between $P$ and $Q$: It is defined to convey nothing more, and nothing less, than is conveyed by the statement: $\;\lnot P \lor Q$, or if you prefer, it tells us nothing more (and nothing less) than what is conveyed by the statement: $\;\lnot(P \land \lnot Q)$.
Your concern is not trivial, nor are you alone in being "bothered" by that lack of some stronger relationship between $P$ and $Q$. There are logics, such as relevance logic which aim to capture aspects of implication that are ignored by the "material implication" operator in classical truth-functional logic, requiring some sort of relevance between antecedent and conditional of a true implication. See also the Wikipedia entry entitled: Paradoxes of material implication for more on "alternate" non-classical logics.
"... the vacuous truth is a "definition of convenience" in a sense."? No, not mere convenience. There are strong pressures that push towards making this choice of truth-values in lines 3 and 4 of the truth-table for the conditional. Here's one that others haven't mentioned.
One thing mathematicians need to be very clear about is the use of statements of generality and especially statements of multiple generality – you know the kind of thing, e.g. the definition of continuity that starts for any $\epsilon$ ... there is a $\delta$ ... And the quantifier-variable notation serves mathematicians brilliantly to regiment statements of multiple generality and make them utterly unambiguous and transparent.
Quantifiers matter to mathematicians, then: that's uncontentious. OK, so now think about restricted quantifiers that talk about only some of a domain (e.g. talk not about all numbers but just about all the even ones). How might we render Goldbach's Conjecture, say? As a first step, we might write
$(\forall n \in \mathbb{N})$(if $n$ is even and greater than 2, then $n$ is the sum of two primes)
We restrict the quantifier here by using a conditional. So now think about the embedded conditional here. What if $n$ is odd, so the antecedent of the conditional is false??? If we say this instance of the conditional lacks a truth-value, or may be false, then the quantification would have non-true instances and so would not be true! But of course we can't refute Goldbach's Conjecture by looking at odd numbers!! So, if the quantified conditional is indeed to come out true when Goldbach is right, then we'll have to say that the irrelevant instances of the conditional with a false antecedent come out true by default. In other words, the embedded conditional will have to be treated as a material conditional.
So: to put it a bit tendentiously and over-briefly, if mathematicians are to deal nicely with expressions of generality using the quantifier-variable notation they have come to know and love, they will have to get used to using material conditionals too.
You can interpret $P \Rightarrow Q$ as meaning "whenever I've written $P$ down in a proof, I can subsequently write $Q$ down in a proof." (In other words, we're defining implication as "the thing that satisfies modus ponens.") Certainly you can always write down a tautology in a proof, so $Q$ can always be $T$. Similarly, if you write down a false statement in a proof then you can prove anything, so $P$ can always be $F$. The only thing you can't do is start with true statements and end up with false statements, so $P$ can't be $T$ if $Q$ is $F$.