# Asymptotic bounds on $\pi^{-1}(x)$ (inverse prime counting function)

What are the current best asymptotic bounds on $\pi^{-1}(x)$, where $\pi(x)$ denotes the prime counting function (number of primes at most $x$)?

In other words, I am curious about the state of the art for estimating the $n^{th}$ prime. From the prime number theorem, it seems clear that $\pi^{-1}(x)=\Theta(x \log x)$. Can someone point me in the direction of literature that answers the question? Please excuse my inability to find such literature myself...

-
– Ricky Demer May 24 '13 at 5:24
Thanks, should I go ahead and delete this question? – pre-kidney May 24 '13 at 5:59

Here is how we invert the prime counting function $\pi(x)$ to estimate the n-th prime $p_n$. Let $g(u),u \ge 2$ be positive, continuous increasing function and let $f(x)$ be defined by

$$x = \int_{2}^{f(x)} \frac{du}{g(u)}$$

Then

$$f(x) = \int_{2}^{x} g(f(u))du + f(2)$$

Take $g(u) = \ln u$ so our first estimate of $f(x) is$that $f(x) \sim x\ln x$. Now use this estimate in the above integral to obtain the second estimate for $f(x)$. You can repeat this iterative process $k$ times to obtain the asymptotic expansion of $f(x)$ with an error $O(x/(\ln x)^{-k})$.

Now from the Prime number theorem, we have

$$\pi(x) \sim \int_{2}^{x} \frac{du}{\ln(u)}$$

Taking $x = p_n$, the n-th prime, the above method says that the asymptotic expansion of $p_n$ is $f(n)$. Thus we obtain

$$p_n \sim n\ln n + n\ln\ln n - n + \ldots$$

-

Assuming the Riemann hypothesis $$|p_n -\text{li}^{-1}(n) |\le \pi^{-1} \sqrt{n} (\log n)^{5/2}, \qquad n \ge 11$$ This follows from known bounds for $\pi(x)$ due to Schoenfeld.

You may see the details in the paper http://front.math.ucdavis.edu/1203.5413

-
I like this answer the best, but I'm keeping Nilotpal's answer as the "accepted" one because his is more informative. – pre-kidney Feb 8 '14 at 6:08

[I think this is morally the same as Nilotpal Sinha's answer, but it only uses asymptotic estimates and no integrals, so I thought I'd post it here in case someone found it useful or interesting.]

Start with the version of the Prime Number Theorem which says that $\pi(x) \sim x / \ln(x)$. Let $p_n$ denote the $n$-th prime, so in particular $\pi(p_n) = n$, and substitute in $p_n$ in the asymptotic estimate for $\pi(x)$ to get $n = \pi(p_n) \sim p_n / \ln(p_n)$, or equivalently

$p_n \sim n \ln p_n$

(If you're being nitpicky this works because the sequence $p_n$ is monotonic and unbounded.) Now take logs of both sides (again, if you're being nitpicky, this works because both $n$ and $p_n / \ln(p_n)$ are unbounded) and multiply by $n$ to get:

$n \ln p_n \sim n \ln(n) + n\ln\ln(p_n)$.

At this point, if you want just the leading term, you can show easily that $\lim_{n \to \infty} n \ln\ln(p_n) / n\ln(n) = 0$, so that in fact $p_n \sim n \ln p_n \sim n \ln n$, or you can probably keep substituting in to get finer and finer error terms.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9813448190689087, "perplexity": 183.9396420543689}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049275412.55/warc/CC-MAIN-20160524002115-00037-ip-10-185-217-139.ec2.internal.warc.gz"}