Euler's constant

γ is related to the digamma function Ψ, and hence the derivative of the gamma function Γ, when both functions are evaluated at 1. Thus:

$${\displaystyle -\gamma =\Gamma '(1)=\Psi (1).}$$

This is equal to the limits:

$${\displaystyle {\begin{aligned}-\gamma &=\lim _{z\to 0}\left(\Gamma (z)-{\frac {1}{z}}\right)\\&=\lim _{z\to 0}\left(\Psi (z)+{\frac {1}{z}}\right).\end{aligned}}}$$

Further limit results are:[16]

$${\displaystyle {\begin{aligned}\lim _{z\to 0}{\frac {1}{z}}\left({\frac {1}{\Gamma (1+z)}}-{\frac {1}{\Gamma (1-z)}}\right)&=2\gamma \\\lim _{z\to 0}{\frac {1}{z}}\left({\frac {1}{\Psi (1-z)}}-{\frac {1}{\Psi (1+z)}}\right)&={\frac {\pi ^{2}}{3\gamma ^{2}}}.\end{aligned}}}$$

A limit related to the beta function (expressed in terms of gamma functions) is

$${\displaystyle {\begin{aligned}\gamma &=\lim _{n\to \infty }\left({\frac {\Gamma \left({\frac {1}{n}}\right)\Gamma (n+1)\,n^{1+{\frac {1}{n}}}}{\Gamma \left(2+n+{\frac {1}{n}}\right)}}-{\frac {n^{2}}{n+1}}\right)\\&=\lim \limits _{m\to \infty }\sum _{k=1}^{m}{m \choose k}{\frac {(-1)^{k}}{k}}\log {\big (}\Gamma (k+1){\big )}.\end{aligned}}}$$

γ can also be expressed as an infinite sum whose terms involve the Riemann zeta function evaluated at positive integers:

$${\displaystyle {\begin{aligned}\gamma &=\sum _{m=2}^{\infty }(-1)^{m}{\frac {\zeta (m)}{m}}\\&=\log {\frac {4}{\pi }}+\sum _{m=2}^{\infty }(-1)^{m}{\frac {\zeta (m)}{2^{m-1}m}}.\end{aligned}}}$$

Other series related to the zeta function include:

$${\displaystyle {\begin{aligned}\gamma &={\tfrac {3}{2}}-\log 2-\sum _{m=2}^{\infty }(-1)^{m}\,{\frac {m-1}{m}}{\big (}\zeta (m)-1{\big )}\\&=\lim _{n\to \infty }\left({\frac {2n-1}{2n}}-\log n+\sum _{k=2}^{n}\left({\frac {1}{k}}-{\frac {\zeta (1-k)}{n^{k}}}\right)\right)\\&=\lim _{n\to \infty }\left({\frac {2^{n}}{e^{2^{n}}}}\sum _{m=0}^{\infty }{\frac {2^{mn}}{(m+1)!}}\sum _{t=0}^{m}{\frac {1}{t+1}}-n\log 2+O\left({\frac {1}{2^{n}\,e^{2^{n}}}}\right)\right).\end{aligned}}}$$

The error term in the last equation is a rapidly decreasing function of n. As a result, the formula is well-suited for efficient computation of the constant to high precision.

Other interesting limits equaling Euler's constant are the antisymmetric limit:[17]

$${\displaystyle {\begin{aligned}\gamma &=\lim _{s\to 1^{+}}\sum _{n=1}^{\infty }\left({\frac {1}{n^{s}}}-{\frac {1}{s^{n}}}\right)\\&=\lim _{s\to 1}\left(\zeta (s)-{\frac {1}{s-1}}\right)\\&=\lim _{s\to 0}{\frac {\zeta (1+s)+\zeta (1-s)}{2}}\end{aligned}}}$$

and the following formula, established in 1898 by de la Vallée-Poussin:

$${\displaystyle \gamma =\lim _{n\to \infty }{\frac {1}{n}}\,\sum _{k=1}^{n}\left(\left\lceil {\frac {n}{k}}\right\rceil -{\frac {n}{k}}\right)}$$

where ⌈ ⌉ are ceiling brackets. This formula indicates that when taking any positive integer n and dividing it by each positive integer k less than n, the average fraction by which the quotient n/k falls short of the next integer tends to γ (rather than 0.5) as n tends to infinity.

Closely related to this is the rational zeta series expression. By taking separately the first few terms of the series above, one obtains an estimate for the classical series limit:

$${\displaystyle \gamma =\lim _{n\to \infty }\left(\sum _{k=1}^{n}{\frac {1}{k}}-\log n-\sum _{m=2}^{\infty }{\frac {\zeta (m,n+1)}{m}}\right),}$$

where ζ(s, k) is the Hurwitz zeta function. The sum in this equation involves the harmonic numbers, H_n. Expanding some of the terms in the Hurwitz zeta function gives:

$${\displaystyle H_{n}=\log(n)+\gamma +{\frac {1}{2n}}-{\frac {1}{12n^{2}}}+{\frac {1}{120n^{4}}}-\varepsilon ,}$$

where 0 < ε < 1/252n⁶.

γ can also be expressed as follows where A is the Glaisher–Kinkelin constant:

$${\displaystyle \gamma =12\,\log(A)-\log(2\pi )+{\frac {6}{\pi ^{2}}}\,\zeta '(2)}$$

γ can also be expressed as follows, which can be proven by expressing the zeta function as a Laurent series:

$${\displaystyle \gamma =\lim _{n\to \infty }\left(-n+\zeta {\Bigl (}{\frac {n+1}{n}}{\bigr )}\right)}$$

γ equals the value of a number of definite integrals:

$${\displaystyle {\begin{aligned}\gamma &=-\int _{0}^{\infty }e^{-x}\log x\,dx\\&=-\int _{0}^{1}\log \left(\log {\frac {1}{x}}\right)dx\\&=\int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {1}{x\cdot e^{x}}}\right)dx\\&=\int _{0}^{1}{\frac {1-e^{-x}}{x}}\,dx-\int _{1}^{\infty }{\frac {e^{-x}}{x}}\,dx\\&=\int _{0}^{1}\left({\frac {1}{\log x}}+{\frac {1}{1-x}}\right)dx\\&=\int _{0}^{\infty }\left({\frac {1}{1+x^{k}}}-e^{-x}\right){\frac {dx}{x}},\quad k>0\\&=2\int _{0}^{\infty }{\frac {e^{-x^{2}}-e^{-x}}{x}}\,dx,\\&=\int _{0}^{1}H_{x}\,dx,\end{aligned}}}$$

where H_x is the fractional harmonic number.

The third formula in the integral list can be proved in the following way:

$${\displaystyle {\begin{aligned}&\int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {1}{xe^{x}}}\right)dx=\int _{0}^{\infty }{\frac {e^{-x}+x-1}{x[e^{x}-1]}}dx=\int _{0}^{\infty }{\frac {1}{x[e^{x}-1]}}\sum _{m=1}^{\infty }{\frac {(-1)^{m+1}x^{m+1}}{(m+1)!}}dx\\[2pt]&=\int _{0}^{\infty }\sum _{m=1}^{\infty }{\frac {(-1)^{m+1}x^{m}}{(m+1)![e^{x}-1]}}dx=\sum _{m=1}^{\infty }\int _{0}^{\infty }{\frac {(-1)^{m+1}x^{m}}{(m+1)![e^{x}-1]}}dx=\sum _{m=1}^{\infty }{\frac {(-1)^{m+1}}{(m+1)!}}\int _{0}^{\infty }{\frac {x^{m}}{e^{x}-1}}dx\\[2pt]&=\sum _{m=1}^{\infty }{\frac {(-1)^{m+1}}{(m+1)!}}m!\zeta (m+1)=\sum _{m=1}^{\infty }{\frac {(-1)^{m+1}}{m+1}}\zeta (m+1)=\sum _{m=1}^{\infty }{\frac {(-1)^{m+1}}{m+1}}\sum _{n=1}^{\infty }{\frac {1}{n^{m+1}}}=\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }{\frac {(-1)^{m+1}}{m+1}}{\frac {1}{n^{m+1}}}\\[2pt]&=\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }{\frac {(-1)^{m+1}}{m+1}}{\frac {1}{n^{m+1}}}=\sum _{n=1}^{\infty }\left[{\frac {1}{n}}-\log \left(1+{\frac {1}{n}}\right)\right]=\gamma \end{aligned}}}$$

The integral on the second line of the equation stands for the Debye function value of +∞, which is m! ζ(m + 1).

Definite integrals in which γ appears include:

$${\displaystyle {\begin{aligned}\int _{0}^{\infty }e^{-x^{2}}\log x\,dx&=-{\frac {(\gamma +2\log 2){\sqrt {\pi }}}{4}}\\\int _{0}^{\infty }e^{-x}\log ^{2}x\,dx&=\gamma ^{2}+{\frac {\pi ^{2}}{6}}\end{aligned}}}$$

One can express γ using a special case of Hadjicostas's formula as a double integral[10][18] with equivalent series:

$${\displaystyle {\begin{aligned}\gamma &=\int _{0}^{1}\int _{0}^{1}{\frac {x-1}{(1-xy)\log xy}}\,dx\,dy\\&=\sum _{n=1}^{\infty }\left({\frac {1}{n}}-\log {\frac {n+1}{n}}\right).\end{aligned}}}$$

An interesting comparison by Sondow[18] is the double integral and alternating series

$${\displaystyle {\begin{aligned}\log {\frac {4}{\pi }}&=\int _{0}^{1}\int _{0}^{1}{\frac {x-1}{(1+xy)\log xy}}\,dx\,dy\\&=\sum _{n=1}^{\infty }\left((-1)^{n-1}\left({\frac {1}{n}}-\log {\frac {n+1}{n}}\right)\right).\end{aligned}}}$$

It shows that log 4/π may be thought of as an "alternating Euler constant".

The two constants are also related by the pair of series[19]

$${\displaystyle {\begin{aligned}\gamma &=\sum _{n=1}^{\infty }{\frac {N_{1}(n)+N_{0}(n)}{2n(2n+1)}}\\\log {\frac {4}{\pi }}&=\sum _{n=1}^{\infty }{\frac {N_{1}(n)-N_{0}(n)}{2n(2n+1)}},\end{aligned}}}$$

where N₁(n) and N₀(n) are the number of 1s and 0s, respectively, in the base 2 expansion of n.

We also have Catalan's 1875 integral[20]

$${\displaystyle \gamma =\int _{0}^{1}\left({\frac {1}{1+x}}\sum _{n=1}^{\infty }x^{2^{n}-1}\right)\,dx.}$$

In general,

$${\displaystyle \gamma =\lim _{n\to \infty }\left({\frac {1}{1}}+{\frac {1}{2}}+{\frac {1}{3}}+\ldots +{\frac {1}{n}}-\log(n+\alpha )\right)\equiv \lim _{n\to \infty }\gamma _{n}(\alpha )}$$

for any α > −n. However, the rate of convergence of this expansion depends significantly on α. In particular, γ_n(1/2) exhibits much more rapid convergence than the conventional expansion γ_n(0).[21][22] This is because

$${\displaystyle {\frac {1}{2(n+1)}}<\gamma _{n}(0)-\gamma <{\frac {1}{2n}},}$$

while

$${\displaystyle {\frac {1}{24(n+1)^{2}}}<\gamma _{n}(1/2)-\gamma <{\frac {1}{24n^{2}}}.}$$

Even so, there exist other series expansions which converge more rapidly than this; some of these are discussed below.

Euler showed that the following infinite series approaches γ:

$${\displaystyle \gamma =\sum _{k=1}^{\infty }\left({\frac {1}{k}}-\log \left(1+{\frac {1}{k}}\right)\right).}$$

The series for γ is equivalent to a series Nielsen found in 1897:[16][23]

$${\displaystyle \gamma =1-\sum _{k=2}^{\infty }(-1)^{k}{\frac {\left\lfloor \log _{2}k\right\rfloor }{k+1}}.}$$

In 1910, Vacca found the closely related series[24][25][26][27][28][16][29]

$${\displaystyle {\begin{aligned}\gamma &=\sum _{k=2}^{\infty }(-1)^{k}{\frac {\left\lfloor \log _{2}k\right\rfloor }{k}}\\[5pt]&={\tfrac {1}{2}}-{\tfrac {1}{3}}+2\left({\tfrac {1}{4}}-{\tfrac {1}{5}}+{\tfrac {1}{6}}-{\tfrac {1}{7}}\right)+3\left({\tfrac {1}{8}}-{\tfrac {1}{9}}+{\tfrac {1}{10}}-{\tfrac {1}{11}}+\cdots -{\tfrac {1}{15}}\right)+\cdots ,\end{aligned}}}$$

where log₂ is the logarithm to base 2 and ⌊ ⌋ is the floor function.

In 1926 he found a second series:

$${\displaystyle {\begin{aligned}\gamma +\zeta (2)&=\sum _{k=2}^{\infty }\left({\frac {1}{\left\lfloor {\sqrt {k}}\right\rfloor ^{2}}}-{\frac {1}{k}}\right)\\[5pt]&=\sum _{k=2}^{\infty }{\frac {k-\left\lfloor {\sqrt {k}}\right\rfloor ^{2}}{k\left\lfloor {\sqrt {k}}\right\rfloor ^{2}}}\\[5pt]&={\frac {1}{2}}+{\frac {2}{3}}+{\frac {1}{2^{2}}}\sum _{k=1}^{2\cdot 2}{\frac {k}{k+2^{2}}}+{\frac {1}{3^{2}}}\sum _{k=1}^{3\cdot 2}{\frac {k}{k+3^{2}}}+\cdots \end{aligned}}}$$

From the Malmsten–Kummer expansion for the logarithm of the gamma function[30] we get:

$${\displaystyle \gamma =\log \pi -4\log \left(\Gamma ({\tfrac {3}{4}})\right)+{\frac {4}{\pi }}\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {\log(2k+1)}{2k+1}}.}$$

An important expansion for Euler's constant is due to Fontana and Mascheroni

$${\displaystyle \gamma =\sum _{n=1}^{\infty }{\frac {|G_{n}|}{n}}={\frac {1}{2}}+{\frac {1}{24}}+{\frac {1}{72}}+{\frac {19}{2880}}+{\frac {3}{800}}+\cdots ,}$$

where G_n are Gregory coefficients.[16][29][31] This series is the special case k = 1 of the expansions

$${\displaystyle {\begin{aligned}\gamma &=H_{k-1}-\log k+\sum _{n=1}^{\infty }{\frac {(n-1)!|G_{n}|}{k(k+1)\cdots (k+n-1)}}&&\\&=H_{k-1}-\log k+{\frac {1}{2k}}+{\frac {1}{12k(k+1)}}+{\frac {1}{12k(k+1)(k+2)}}+{\frac {19}{120k(k+1)(k+2)(k+3)}}+\cdots &&\end{aligned}}}$$

convergent for k = 1, 2, ...

A similar series with the Cauchy numbers of the second kind C_n is[29][32]

$${\displaystyle \gamma =1-\sum _{n=1}^{\infty }{\frac {C_{n}}{n\,(n+1)!}}=1-{\frac {1}{4}}-{\frac {5}{72}}-{\frac {1}{32}}-{\frac {251}{14400}}-{\frac {19}{1728}}-\ldots }$$

Blagouchine (2018) found an interesting generalisation of the Fontana–Mascheroni series

$${\displaystyle \gamma =\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{2n}}{\Big \{}\psi _{n}(a)+\psi _{n}{\Big (}-{\frac {a}{1+a}}{\Big )}{\Big \}},\quad a>-1}$$

where ψ_n(a) are the Bernoulli polynomials of the second kind, which are defined by the generating function

$${\displaystyle {\frac {z(1+z)^{s}}{\log(1+z)}}=\sum _{n=0}^{\infty }z^{n}\psi _{n}(s),\qquad |z|<1,}$$

For any rational a this series contains rational terms only. For example, at a = 1, it becomes[33][34]

$${\displaystyle \gamma ={\frac {3}{4}}-{\frac {11}{96}}-{\frac {1}{72}}-{\frac {311}{46080}}-{\frac {5}{1152}}-{\frac {7291}{2322432}}-{\frac {243}{100352}}-\ldots }$$

Other series with the same polynomials include these examples:

$${\displaystyle \gamma =-\log(a+1)-\sum _{n=1}^{\infty }{\frac {(-1)^{n}\psi _{n}(a)}{n}},\qquad \Re (a)>-1}$$

and

$${\displaystyle \gamma =-{\frac {2}{1+2a}}\left\{\log \Gamma (a+1)-{\frac {1}{2}}\log(2\pi )+{\frac {1}{2}}+\sum _{n=1}^{\infty }{\frac {(-1)^{n}\psi _{n+1}(a)}{n}}\right\},\qquad \Re (a)>-1}$$

where Γ(a) is the gamma function.[31]

A series related to the Akiyama–Tanigawa algorithm is

$${\displaystyle \gamma =\log(2\pi )-2-2\sum _{n=1}^{\infty }{\frac {(-1)^{n}G_{n}(2)}{n}}=\log(2\pi )-2+{\frac {2}{3}}+{\frac {1}{24}}+{\frac {7}{540}}+{\frac {17}{2880}}+{\frac {41}{12600}}+\ldots }$$

where G_n(2) are the Gregory coefficients of the second order.[31]

Series of prime numbers:

$${\displaystyle \gamma =\lim _{n\to \infty }\left(\log n-\sum _{p\leq n}{\frac {\log p}{p-1}}\right).}$$

γ equals the following asymptotic formulas (where H_n is the nth harmonic number):

• ${\textstyle \gamma \sim H_{n}-\log n-{\frac {1}{2n}}+{\frac {1}{12n^{2}}}-{\frac {1}{120n^{4}}}+\cdots }$ (Euler)
• ${\textstyle \gamma \sim H_{n}-\log \left({n+{\frac {1}{2}}+{\frac {1}{24n}}-{\frac {1}{48n^{2}}}+\cdots }\right)}$ (Negoi)
• ${\textstyle \gamma \sim H_{n}-{\frac {\log n+\log(n+1)}{2}}-{\frac {1}{6n(n+1)}}+{\frac {1}{30n^{2}(n+1)^{2}}}-\cdots }$ (Cesàro)

The third formula is also called the Ramanujan expansion.

Alabdulmohsin derived closed-form expressions for the sums of errors of these approximations.[32] He showed that (Theorem A.1):

$${\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }\log n+\gamma -H_{n}+{\frac {1}{2n}}&={\frac {\log(2\pi )-1-\gamma }{2}}\\\sum _{n=1}^{\infty }\log {\sqrt {n(n+1)}}+\gamma -H_{n}&={\frac {\log(2\pi )-1}{2}}-\gamma \\\sum _{n=1}^{\infty }(-1)^{n}{\Big (}\log n+\gamma -H_{n}{\Big )}&={\frac {\log \pi -\gamma }{2}}\end{aligned}}}$$

The constant e^γ is important in number theory. Some authors denote this quantity simply as γ′. e^γ equals the following limit, where p_n is the nth prime number:

$${\displaystyle e^{\gamma }=\lim _{n\to \infty }{\frac {1}{\log p_{n}}}\prod _{i=1}^{n}{\frac {p_{i}}{p_{i}-1}}.}$$

This restates the third of Mertens' theorems.[35] The numerical value of e^γ is:[36]

Other infinite products relating to e^γ include:

$${\displaystyle {\begin{aligned}{\frac {e^{1+{\frac {\gamma }{2}}}}{\sqrt {2\pi }}}&=\prod _{n=1}^{\infty }e^{-1+{\frac {1}{2n}}}\left(1+{\frac {1}{n}}\right)^{n}\\{\frac {e^{3+2\gamma }}{2\pi }}&=\prod _{n=1}^{\infty }e^{-2+{\frac {2}{n}}}\left(1+{\frac {2}{n}}\right)^{n}.\end{aligned}}}$$

These products result from the Barnes G-function.

In addition,

$${\displaystyle e^{\gamma }={\sqrt {\frac {2}{1}}}\cdot {\sqrt[{3}]{\frac {2^{2}}{1\cdot 3}}}\cdot {\sqrt[{4}]{\frac {2^{3}\cdot 4}{1\cdot 3^{3}}}}\cdot {\sqrt[{5}]{\frac {2^{4}\cdot 4^{4}}{1\cdot 3^{6}\cdot 5}}}\cdots }$$

where the nth factor is the (n + 1)th root of

$${\displaystyle \prod _{k=0}^{n}(k+1)^{(-1)^{k+1}{n \choose k}}.}$$

This infinite product, first discovered by Ser in 1926, was rediscovered by Sondow using hypergeometric functions.[37]

It also holds that[38]

$${\displaystyle {\frac {e^{\frac {\pi }{2}}+e^{-{\frac {\pi }{2}}}}{\pi e^{\gamma }}}=\prod _{n=1}^{\infty }\left(e^{-{\frac {1}{n}}}\left(1+{\frac {1}{n}}+{\frac {1}{2n^{2}}}\right)\right).}$$

The continued fraction expansion of γ begins [0; 1, 1, 2, 1, 2, 1, 4, 3, 13, 5, 1, 1, 8, 1, 2, 4, 1, 1, 40, ...],[1] which has no apparent pattern. The continued fraction is known to have at least 475,006 terms,[8] and it has infinitely many terms if and only if γ is irrational.