$\int_0^x\frac{\lfloor t\rfloor}{t^2}dt=H_x$ isn't true?

Question

I incorrectly proved that $$\int_0^x\frac{\lfloor t\rfloor}{t^2}dt=H_x$$According to Desmos, it should be $$\int_0^x\frac{\lfloor t\rfloor}{t^2}dt+1=H_x\tag{1}$$Here's my proof $$\int_0^x\frac{\lfloor t\rfloor}{t^2}dt=\sum_{k=0}^{x-1}\int_k^{k+1}\frac{\lfloor t\rfloor}{t^2}dt=\sum_{k=0}^{x-1}k\int_k^{k+1}\frac1{t^2}dt=\sum_{k=0}^{x-1}k\left(\frac1k-\frac1{k+1}\right)=\sum_{k=0}^{x-1}\frac k{k(k+1)}=\sum_{k=0}^{x-1}\frac1{k+1}=H_x$$ I think simplifying $\frac k{k(k+1)}$ to $\frac1{k+1}$ is where I got things wrong, but how do I prove that this is illegitimate?

Edit: $x$ is an integer.

Also, other proofs for $(1)$ are appreciated.

Answer 1

When you do this:

$$ \sum_{k=0}^{x-1}\int_k^{k+1}\frac{\lfloor t\rfloor}{t^2}dt=\sum_{k=0}^{x-1}k\int_k^{k+1}\frac1{t^2}dt $$

you assume that all the integrals in the RHS converge. But this is not true for $k=0$, because $\int_0^1 1/t^2dt$ does not converge - it's infinity, so the first term in the sum (at $k=0$) is actually $$k\cdot \int_0^1 1/t^2dt=0\cdot \infty,$$ which is undefined because you have to precisely measure "exactly how" infinite the integral is. Instead, you can take the first term in front of the sum:

$$ \sum_{k=0}^{x-1}\int_k^{k+1}\frac{\lfloor t\rfloor}{t^2}dt =\underbrace{\int_0^1\frac{\lfloor t\rfloor}{t^2}dt}_0 + \sum_{k=1}^{x-1}\int_k^{k+1}\frac{\lfloor t\rfloor}{t^2}dt. $$

So the first term is actually 0 (because the function inside the integral is a constant 0 a.e.) and you have to evaluate the sum only for $k\ge 1$.

Answer 2

This all looks correct except for the $ k = 0 $ term. In this term, you're right that you can't simplify $ \frac k { k ( k + 1 ) } $ to $ \frac 1 { k + 1 } $, but the error comes before that; it comes where you turn $ \int _ { t = k } ^ { k + 1 } \, \frac { \lfloor t \rfloor } { t ^ 2 } \, \mathrm d t $ into $ k \, \int _ { t = k } ^ { k + 1 } \frac 1 { t ^ 2 } \, \mathrm d t $, because (when $ k = 0 $) the left-hand side is $ 0 $ while the right-hand side is undefined (of the form $ 0 \cdot \infty $). The intermediate step $ \int _ { t = k } ^ { k + 1 } \, \frac k { t ^ 2 } \, \mathrm d t $ (which you didn't write down) would still be correct, but pulling a constant $ k $ out of an integral is only valid if you know that $ k \ne 0 $ or the resulting integral converges.

So the solution is to treat the $ k = 0 $ term separately. Since this term is $ 0 $, you end up with $ 0 + \sum \limits _ { k = 1 } ^ { x - 1 } \frac 1 { k + 1 } = \sum \limits _ { n = 2 } ^ x \frac 1 n $, which is $ H _ x - \frac 1 1 = H _ x - 1 $.

Incidentally, there is a minority philosphy that, at least when dealing with variables taking discrete values, we should allow an expression such as $ 0 \cdot \infty $ (or more generally $ 0 $ multiplied by an undefined expression) and evaluate it to $ 0 $. (See the discussion of strong $ 0 $ in Donald Knuth's Two Notes on Notation, where it's not even one of the two notes from the title; there's a lot in that little paper.) In that case, your manipulation is correct until you simplify $ \frac k { k ( k + 1 ) } $ to $ \frac 1 { k + 1 } $, like you suspected.