What is the meaning of $\int_{0}^{k}\pi(x)dx?$

Question

I saw this post about the prime counting function.

Let's look at this intgeral: $$I(k)=\int_{0}^{k}\pi(x)\mathrm dx$$

where $I(k)=0,0,1,3,5,8,...$ for $k=1,2,3,4,5,6...$

What is this integral trying to show me? All these values of $I(k)$ and its relation, what does it mean?

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Let $$S(k)=\sum_{j=2}^{k-1}\pi(j)$$ I saw these relations: $$S(2k+1)-2S(2k+2)+S(2k+3)=0$$

$$S(6k+1)-3S(6k+2)+3S(6k+3)-S(6k+4)=0$$

$$S(6k+1)-4S(6k+2)+6S(6k+3)-4S(6k+4)+S(6k+5)=0$$

Meaning? Is there some usefulness to understand prime numbers distribution?

Answer 1

The integral is equal to the sum $$\sum_{j=3}^k \pi(j-1)=\sum_{j=2}^{k-1} \pi(j)$$ hence it is a formula for the sum of the prime-counting-function values upto $\ k-1\ $.

Answer 2

Since the prime counting function $\pi(x)$ is, at least when discussing it normally, a discrete function (i.e. we generally have $x$ an integer, a positive one at that), we can think of the integral as being a sum as well.

In particular,

$$I(k) = \int_0^k \pi(x)dx = \sum_{n=0}^{k-1} \pi(n)$$

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Let's not take this without proof. I won't go into the full rigor of the proof - I'd rather look at a more intuitive approach to deriving the relation.

Recall the prime counting function's definition:

$$\pi(x) = \sum_{\text{primes} \; p \leq x } 1 = \text{the number of primes less than or equal to x}$$

If we were to graph this function as if $x$ were a real number, we'd effectively see what looks almost like a bar graph, making a jump of one unit every time $x$ is prime as $x$ grows. The integral of $\pi(x)$ can then be envisioned as the area under this curve.

Well, since $x$ is not a prime if it's not an integer, on the interval $[n,n+1)$ for $n$ an integer, we can always be assured that $\pi(x)$ is constant - one of the "bars" in the bar graph analogy. The width of this interval is $1$ and its value is $\pi(x)$ so the integral over the interval is $1\cdot \pi(x) = \pi(x)$ - the area of one of the "bars".

We repeat this over each interval up to $k$ - we take the integral over $[0,1),$ over $[1,2)$, and so on, up to $[k-1,k)$. This gives us the integral over $[0,k)$ (which is effectively the same as the interval over $[0,k]$ since that one point has infinitesimal contribution to the integral when you think of the area analogue).

But then this construction also showed

$$\int_n^{n+1} \pi(x)dx = \pi(n)$$

so in reality

$$\int_0^1 \pi(x)dx = \pi(0) \;\;\; \int_1^2 \pi(x)dx = \pi(1) \;\;\; ... \;\;\; \int_{k-1}^k \pi(x)dx = \pi(k-1)$$

Thus, we can sum these integrals and combine them, as well as their results, and obtain

$$\begin{align} \int_0^1 \pi(x)dx + \int_1^2 \pi(x)dx + ... + \int_{k-1}^k \pi(x)dx &= \int_0^k \pi(x)dx \\ &= \sum_{n=0}^{k-1} \pi(n) \\ &= \pi(0) + \pi(1) + ... + \pi(k-1) \end{align}$$

And thus,

$$\int_0^k \pi(x)dx = \sum_{n=0}^{k-1} \pi(n)$$

If we want to be pedantic, we can also note that $\pi(0) = \pi(1) = 0$ and reindex the sum a bit to begin at $n=2$:

$$\int_0^k \pi(x)dx = \sum_{n=2}^{k-1} \pi(n)$$

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So, in conclusion, what is the integral of the prime counting function on an interval $[0,k]$, assuming $k$ is an integer?

It is simply the sum of the values of the prime counting function for arguments $0,1,2,...,k-1$.

Of course this approach in deriving that fact also permits one to look a bit at cases when the lower bound is nonzero (and possibly not an integer), and when the upper bound, too, is not an integer. Whether such considerations would prove useful is beyond me.

I do wonder why the integral notation was used as opposed to the summation, though. But that would be a topic for another question.