natural log of pi to sixty digits, a reference needed

I have found a table of the logs of gamma functions at basic fractions, accurate to 60 decimal digits. It omits $\ln(\Gamma(1/2)) = \ln(\pi)/2$ and I want that number.

Where is a cit-able source that contains this this number to high accuracy?

• Mathematical software can do this very quickly - Wolfram Alpha for example. – Jair Taylor Nov 17 '12 at 20:26
• Yes, but can I trust the output, and is there a guaranteed error bound? – Terry Loring Nov 17 '12 at 21:12
• If it's very important to you, code it yourself and make the source available. I guess you can't include it in the article because it would take up too much space in a journal, but I'm sure you could work it out somehow. – Max Morin Nov 17 '12 at 22:52
• I found a nice table of values of the gamma function,not the log of the gamma function, and it includes the value at 1/2 (so sqrt(pi)). It is in "High-precision values of the gamma function and of some related coefficients" by A Fransén, S Wrigge, from 1980. – Terry Loring Nov 19 '12 at 6:28
• I suppose it would be sufficient if the quantity is calculated to $n$ digits and then to $2n$ digits on different machines and software and the results compared. If they agree to $n$ digits then IMO that would be more trustworthy than an average math paper. – Maesumi Aug 4 '13 at 10:23

(In know this question is old and answered, but...) For $\ln(\pi)$ the Online Encycolpedia of Integer Sequences has a value to 105 decimal places. It is also citable.

• Thanks. My need for this is gone, but someone else might need it. – Terry Loring Mar 14 '15 at 3:10

Project Gutenburg has published a compilation of 'Miscellaneous Mathematical Constants' which includes the natural log of pi to 2000 decimal places as:

log(Pi) natural logarithm of Pi to 2000 places.

1.1447298858494001741434273513530587116472948129153115715136230714721377698848 260797836232702754897077020098122286979891590482055279234565872790810788102868 252763939142663459029024847733588699377892031196308247567940119160282172273798 881265631780498236973133106950036000644054872638802232700964335049595118150662 372524683433912698965797514047770385779953998258425660228485014813621791592525 056707638686028076345688975051233436078143991414426429596712897781136526452345 041059007160818570824981188183186897672845928110257656875172422338337189273043 288217348651042761532375161028392221340143696717585616442473718780506046692056 283377310133621627451589875201512996545465739691528252391695852453793594601400 379956519666036538000112659858500129765699060744667455472671045084950668558743 390774251341592412652317771784917799588095767880510296444750901508911403278080 768337337938949488075152890091875363766086707435833345108139232535574067684327 431198049633999761803046221286361595859836404758009861799938264629277646275948 484896414107483132593462053635073046055030768215494444154778884559535228440047 850918217255915179900785243523837112867132342905566964492585582623118824223244 661476739136153339414264534600881979155478967757529878307593230499751706785370 666315222134751026417324918906534257373051835228316776877311442944368108997522 287634554909933469253981028398378467695079971965163008386496663274223886761392 944112379606529081463545502415193643368404005225615575618053680459613160686367 226297126848055518038239624057983138433955882483556816617339018195508924667782 042898879384623081953507082523699065543916029676565349509487102686726405036344 889957813954840804697878603723560031033518890166410542245140400821480026071893 924502077785635698810693233664357379481092927781936265980614204270094398298364 733767922501305495445975380037647617519082652294857728828349379913418698964043 483457091550460629912859614271432256377699794328889523074041463529466113313641 884192574888189320796571991444939402534883228262813

• Thanks. This is not a refereed paper in an established journal, but it is close. – Terry Loring Nov 17 '12 at 21:20
• @TerryLoring: Why would an established journal want to use its pages to publish digits of an easily computed mathematical constant? – Nate Eldredge Nov 17 '12 at 21:35
• I need to compute a bunch on constants that depend on eight values of the Gamma function. I want to document how I got those, with error bounds. – Terry Loring Nov 17 '12 at 21:47
• @theo: Is it normal? Or more generally: Is log(x) normal if x is? – Marlo Jun 1 '18 at 17:16

Here it is, to 80 digits.

0.57236494292470008707171367567652935582364740645765578575681153573606\ 888494241304

Youcan cite me, "personal communication".

• Thanks. I may cite you. Physics journals often frown on citing personal communication, but this may end up in a math journal where such references are common. – Terry Loring Nov 17 '12 at 21:29
• I guess I will just insist. – Terry Loring Nov 17 '12 at 21:35
• This is not correct to 80 digits. You already have a mistake at digit 16. – Winther Mar 26 '15 at 21:55
• Ahh... you're right, this is $\log \sqrt{\pi}$. I just fixed it. – Hans Engler Mar 28 '15 at 2:56

I know that you are asking for a table or software where you can just look this up, but this number is quickly computable to high accuracy with the right approach. (Assuming that you have an engine that can keep track of high precision in the first place.)

To speed up computations, use that $$\ln(\pi)=\ln\left(\frac{\pi}{22/7}\right)+\ln(2)+\ln(11)-\ln(7)$$ The latter three terms can be looked up to high precision, and now you have to find $\ln(7\pi/22)$, where $7\pi/22$ is very close to $1$. In fact $|1-7\pi/22|<0.00041$. So even the slowly converging Taylor series for $\ln$ can output a result quickly, assuming that you again use a table, this time to look up many digits of $\pi$.

\begin{align} \ln(1+(7\pi/22-1)) & = -\sum_{n=1}^\infty\frac{(-1)^n}{n}(7\pi/22-1)^n \end{align} Since $(7\pi/22-1)<0$, we do not really have an alternating series, and we'll have to think about Taylor's error bound. The $n+1$st derivative of $\ln(1+x)$ is $(-1)^{n}n!/(1+x)^{n+1}$. On the interval $[-0.00041,0.00041]$, this is bounded in absolute value by $n!/0.99959^{n+1}$. So an error bound for the $n$th partial sum would be $$\frac{n!/0.99959^{n+1}}{(n+1)!}0.00041^{n+1}=\frac{1}{n+1}\left(\frac{0.00041}{0.99959}\right)^{n+1}$$ Using $n=17$ brings this under $10^{-62}$.

At present, I do not have access to an engine that could keep track of enough decimal precision, or else I would conclude my answer with the end result of this process. But here is the formula it yields, with an absolute error bound of less than $10^{-62}$:$$\ln(\pi)\approx-\sum_{n=1}^{17}\frac{(-1)^n}{n}(7\pi/22-1)^n+\ln(2)+\ln(11)-\ln(7)$$

• Oops - I had omitted the leading minus sign from the Taylor series. Fixed now. – alex.jordan Nov 17 '12 at 22:45
• +1; This is certainly a smart way to go about computing $\ln(\pi)$. Also, you should probably add an error term in your last line, else that line is incorrect (though its meaning is clear). – JavaMan Nov 17 '12 at 22:56
• @JavaMan Thanks! I actually just changed the = to \approx. An explicit error bound is a good idea. – alex.jordan Nov 17 '12 at 22:57