Editing Euler's constant (section)

=== Integrals ===
{{mvar|γ}} equals the value of a number of definite [[integral]]s:

<math display="block">\begin{align}
\gamma &= - \int_0^\infty e^{-x} \log x \,dx \\
 &= -\int_0^1\log\left(\log\frac 1 x \right) dx \\ 
 &= \int_0^\infty \left(\frac1{e^x-1}-\frac1{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(\frac1{\log x} + \frac1{1-x}\right)dx\\
 &= \int_0^\infty \left(\frac1{1+x^k}-e^{-x}\right)\frac{dx}{x},\quad k>0\\
 &= 2\int_0^\infty \frac{e^{-x^2}-e^{-x}}{x} \, dx ,\\
&= \log\frac{\pi}{4}-\int_0^\infty \frac{\log x}{\cosh^2x} \, dx ,\\
 &= \int_0^1 H_x \, dx, \\
 &= \frac{1}{2}+\int_{0}^{\infty}\log\left(1+\frac{\log\left(1+\frac{1}{t}\right)^{2}}{4\pi^{2}}\right)dt \\
&= 1-\int_0^1 \{1/x\} dx 
 \end{align}
 </math>
where {{math|{{var|H}}{{sub|{{var|x}}}}}} is the [[Harmonic number#Harmonic numbers for real and complex values|fractional harmonic number]], and <math>\{1/x\}</math> is the [[fractional part]] of <math>1/x</math>.

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

<math display="block">\begin{align}
&\int_0^{\infty} \left(\frac{1}{e^x - 1} - \frac{1}{x e^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{align}</math>

The integral on the second line of the equation is the definition of the [[Riemann zeta function]], which is {{math|{{var|m}}!{{var|ζ}}({{var|m}} + 1)}}.

Definite integrals in which {{mvar|γ}} appears include:<ref name=":3">{{Cite web |last=Weisstein |first=Eric W. |title=Euler-Mascheroni Constant |url=https://mathworld.wolfram.com/Euler-MascheroniConstant.html |access-date=2024-10-19 |website=mathworld.wolfram.com |language=en}}</ref><ref name=":1">{{Cite journal |last=Blagouchine |first=Iaroslav V. |date=2014-10-01 |title=Rediscovery of Malmsten's integrals, their evaluation by contour integration methods and some related results |url=https://link.springer.com/article/10.1007/s11139-013-9528-5 |journal=The Ramanujan Journal |language=en |volume=35 |issue=1 |pages=21–110 |doi=10.1007/s11139-013-9528-5 |issn=1572-9303}}</ref>

<math display="block">\begin{align}
\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}
\\ \int_0^\infty \frac{e^{-x}\log x}{e^x +1} \,dx &= \frac12 \log^2 2 - \gamma \end{align}</math>

We also have [[Eugène Charles Catalan|Catalan]]'s 1875 integral{{r|SondowZudilin2006}}

<math display="block">\gamma = \int_0^1 \left(\frac1{1+x}\sum_{n=1}^\infty x^{2^n-1}\right)\,dx.</math>

One can express {{mvar|γ}} using a special case of [[Hadjicostas's formula]] as a [[Multiple integral#Double integral|double integral]]{{r|Sondow2003a|Sondow2005}} with equivalent series:

<math display="block">\begin{align}
\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{align}</math>

An interesting comparison by Sondow{{r|Sondow2005}} is the double integral and alternating series

<math display="block">\begin{align}
\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{align}</math>

It shows that {{math|log {{sfrac|4|π}}}} may be thought of as an "alternating Euler constant".

The two constants are also related by the pair of series{{r|Sondow2005a}}

<math display="block">\begin{align}
\gamma &= \sum_{n=1}^\infty \frac{N_1(n) + N_0(n)}{2n(2n+1)} \\
\log\frac4{\pi} &= \sum_{n=1}^\infty \frac{N_1(n) - N_0(n)}{2n(2n+1)} ,
\end{align}</math>

where {{math|{{var|N}}{{sub|1}}({{var|n}})}} and {{math|{{var|N}}{{sub|0}}({{var|n}})}} are the number of 1s and 0s, respectively, in the [[Binary number|base 2]] expansion of {{mvar|n}}.