카탈랑 상수

1\에 대한 내용은 카탈랑 수 문서 참고하십시오.

수학상수 Mathematical Constants
{{{#!wiki style="margin: 0 -10px -5px; min-height: calc(1.5em + 5px)" {{{#!folding [ 펼치기 · 접기 ] {{{#!wiki style="margin: -5px -1px -11px"	[math(^\ast)] 초월수임이 증명됨.
[math(0)] (덧셈의 항등원)	[math(1)] (곱셈의 항등원)	[math(sqrt{2})] (최초로 증명된 무리수)	[math(495)], [math(6174)] (카프리카 상수)	[math(0)], [math(1)], [math(3435)], [math(438579088)] (뮌하우젠 수)
[math(pi)] (원주율)[math(^\ast)]	[math(tau)] (새 원주율)[math(^\ast)]	[math(e)] (자연로그의 밑)[math(^\ast)]	[math(varphi)] (황금수)	[math(i)] (허수단위)
[math(G)] (카탈랑 상수)	[math(zeta(3))] (아페리 상수)	[math({rm Si}(pi))] (윌브레이엄-기브스 상수)	[math(gamma)] (오일러-마스케로니 상수)	[math(gamma_n)] (스틸체스 상수)
[math(Omega)] (오메가 상수)[math(^\ast)]	[math(2^{sqrt{2}})] (겔폰트-슈나이더 상수)[math(^\ast)]	[math(C_n,)] (챔퍼나운 상수)[math(^\ast)]	[math(A,)] (글레이셔-킨켈린 상수)	[math(A_k,)] (벤더스키-아담칙 상수)
[math(-e, {rm Ei}(-1))] (곰페르츠 상수)	[math(mu)] (라마누잔-졸트너 상수)	[math(B_{2})], [math(B_{4})] (브룬 상수)	[math(rho)] (플라스틱 상수)	[math(delta)], [math(alpha)] (파이겐바움 상수)

}}}}}}}}} ||

1. 개요2. 항등식3. 값4. 관련 문서

1. 개요

카탈랑 常數 / Catalan's constant

카탈랑 상수는 벨기에의 수학자 외젠 샤를 카탈랑에 의해 정의된 상수로, 아래와 같은 식으로 정의된다. [math(C)], [math(K)] 등으로 표기하는 경우도 매우 흔하다.

[math(\displaystyle \begin{aligned}
G &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} = \frac1{1^2} -\frac1{3^2} +\frac1{5^2} -\frac1{7^2} +\cdots \\
&\approx 0.9159655942
\end{aligned} )]

카탈랑 상수의 값은 소숫점 아래 육천억 자리까지 계산되었으나, 이 수가 유리수인지 무리수인지는 아직 밝혀지지 않았다. 따라서 이 상수에 디리클레 함수 [1]를 취한 [math(\bold1_{\mathbb Q}(G))]의 값은 부정이다.

카탈랑 상수는 디리클레 베타 함수 [math(\displaystyle \beta(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s})]에서 [math(s=2)]인 경우이다. 참고로, 디리클레 베타 함수는 리만 제타 함수와 관련이 있고, 결국 리만 가설로 연결된다.

2. 항등식

[math(\displaystyle \int_0^1 \frac{\arctan{x}}x \,{\rm d}x = G)]
[math(\displaystyle \int_1^\infty \frac{\operatorname{arccot}x}x \,{\rm d}x = G)]

위쪽 식에 [math(\arctan x)]의 매클로린 급수 전개식 [math(\displaystyle \arctan x = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} x^{2n+1})]을 대입한다.

[math(\displaystyle \begin{aligned}
\int_0^1 \frac{\arctan{x}}x \,{\rm d}x &= \int_0^1 \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} x^{2n} \,{\rm d}x \\
&= \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \int_0^1 x^{2n} \,{\rm d}x \\
&= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} = G \qquad \blacksquare
\end{aligned} )]

[math(\displaystyle \int_0^1 \frac{\arctan{x}}x \,{\rm d}x = G)]에서 [math(x=\dfrac1t)]로 치환한 후 [math(arctan)] 함수와 [math(operatorname{arccot})] 함수의 관계식을 사용하면 아래쪽 식의 적분 항을 얻을 수 있다.

[math(\displaystyle \begin{aligned}
G = \int_0^1 \frac{\arctan{x}}x \,{\rm d}x &= \int_\infty^1 \frac{\arctan(1/t)}{1/t} \!\left( -\frac1{t^2} \,{\rm d}t \right) \\
&= \int_1^\infty \frac{\arctan(1/t)}t \,{\rm d}t \\
&= \int_1^\infty \frac{\operatorname{arccot}t}t \,{\rm d}t \qquad \blacksquare
\end{aligned} )]

}}}||

[math(\displaystyle \int_0^\infty \frac1x \operatorname{erf}\biggl(\frac1x\biggr) \!\operatorname{erfc}\biggl(\frac1x\biggr) {\rm d}x = \frac{2G}\pi \approx 0.5831218081 \quad)] ([math(\operatorname{erf}(x))]는 오차함수, [math(\operatorname{erfc}(x))]는 여오차함수)

[math(x=\dfrac1t)]로 치환하면

[math(\displaystyle \begin{aligned}
\int_0^\infty \frac1x \operatorname{erf}\biggl(\frac1x\biggr) \!\operatorname{erfc}\biggl(\frac1x\biggr) {\rm d}x &= \int_\infty^0 t \operatorname{erf}(t) \operatorname{erfc}(t) \biggl( -\frac{{\rm d}t}{t^2} \biggr) \\
&= \int_0^\infty \frac1t \operatorname{erf}(t) \operatorname{erfc}(t) \,{\rm d}t
\end{aligned} )]

오차함수 및 여오차함수의 적분 표현식을 그대로 대입하면

[math(\displaystyle \begin{aligned}
\int_0^\infty \frac1x \operatorname{erf}\biggl(\frac1x\biggr) \!\operatorname{erfc}\biggl(\frac1x\biggr) {\rm d}x &= \int_0^\infty \frac1t \operatorname{erf}(t) \operatorname{erfc}(t) \,{\rm d}t \\
&= \int_0^\infty \frac1t \biggl( \frac{2t}{\sqrt\pi} \int_0^1 e^{-t^2y^2} \,{\rm d}y \biggr) \!\biggl( \frac{2t}{\sqrt\pi} \int_1^\infty e^{-t^2z^2} \,{\rm d}z \biggr) {\rm d}t \\
&= \frac2\pi \int_0^1 \int_1^\infty \int_0^\infty e^{-(y^2+z^2)t^2} \cdot 2t \,{\rm d}t \,{\rm d}z \,{\rm d}y
\end{aligned} )]

[math(t^2=x)]로 치환하면

[math(\displaystyle \begin{aligned}
\int_0^\infty \frac1x \operatorname{erf}\biggl(\frac1x\biggr) \!\operatorname{erfc}\biggl(\frac1x\biggr) {\rm d}x &= \frac2\pi \int_0^1 \int_1^\infty \int_0^\infty e^{-(y^2+z^2)t^2} \cdot 2t \,{\rm d}t \,{\rm d}z \,{\rm d}y \\
&= \frac2\pi \int_0^1 \int_1^\infty \int_0^\infty e^{-(y^2+z^2)x} \,{\rm d}x \,{\rm d}z \,{\rm d}y \\
&= \frac2\pi \int_0^1 \int_1^\infty \biggl[ -\frac{e^{-(y^2+z^2)x}}{y^2+z^2} \biggr]_{x\to0}^{x\to\infty} {\rm d}z \,{\rm d}y \\
&= \frac2\pi \int_0^1 \int_1^\infty \frac1{y^2+z^2} \,{\rm d}z \,{\rm d}y \\
&= \frac2\pi \int_0^1 \biggl[ \frac1y \arctan \biggl( \frac zy \biggr) \biggr]_{z\to1}^{z\to\infty} {\rm d}y \\
&= \frac2\pi \int_0^1 \frac1y \biggl[ \frac\pi2 -\arctan \biggl( \frac1y \biggr) \biggr] \,{\rm d}y \\
&= \frac2\pi \int_0^1 \frac1y \arctan y \,{\rm d}y \\
&= \frac{2G}\pi
\end{aligned} )]

참고로, 위 적분의 후반부에서 [math(arctan y +arctan biggl( dfrac1y biggr) != dfracpi2)] (단, [math(y>0)])임을 사용하였다.
}}}||

[math(\displaystyle \int_{-\infty}^0 \arctan{e^x} \,{\rm d}x = G)]
[math(\displaystyle \int_0^\infty \operatorname{arccot}e^x \,{\rm d}x = G)]

[math(\displaystyle \int_0^1 \frac{\arctan{x}}x \,{\rm d}x = G)]에서 [math(x=e^t)]로 치환하면 위쪽 식을 증명할 수 있다.

[math(\displaystyle \begin{aligned}
G = \int_0^1 \frac{\arctan{x}}x \,{\rm d}x &= \int_{-\infty}^0 \frac{\arctan{e^t}}{e^t} \cdot e^t \,{\rm d}t \\
&= \int_{-\infty}^0 \arctan{e^t} \,{\rm d}t \qquad \blacksquare
\end{aligned} )]

여기서 [math(t=-x)]로 치환한 후 [math(arctan)] 함수와 [math(operatorname{arccot})] 함수의 관계식을 사용하면 아래쪽 식을 증명할 수 있다.

[math(\displaystyle \begin{aligned}
G = \int_{-\infty}^0 \arctan{e^t} \,{\rm d}t &= \int_\infty^0 \arctan{e^{-x}} \,(-{\rm d}x) \\
&= \int_0^\infty \arctan \frac1{e^x} \,{\rm d}x \\
&= \int_0^\infty \operatorname{arccot} e^x \,{\rm d}x \qquad \blacksquare
\end{aligned} )]

}}}||

[math(\displaystyle \int_0^{\pi/2} \frac x{\sin x} \,{\rm d}x = 2G \approx 1.8319311884)]

탄젠트 반각 치환을 사용하자. [math(t=\tan \dfrac x2)]로 치환하면 [math(x=2\arctan t)]이고 [math(\sin x = \dfrac{2t}{1+t^2})]이고 [math({\rm d}x = \dfrac{2\,{\rm d}t}{1+t^2})]이므로 다음과 같이 증명할 수 있다.

[math(\displaystyle \begin{aligned}
\int_0^{\pi/2} \frac x{\sin x} \,{\rm d}x &= \int_0^1 2\arctan t \cdot \frac{1+t^2}{2t} \cdot \frac{2\,{\rm d}t}{1+t^2} \\
&= 2\int_0^1 \frac{\arctan t}t \,{\rm d}t \\
&= 2G
\end{aligned} )]

}}}||

[math(\displaystyle \int_0^{1/2} \Gamma(1+x)\Gamma(1-x) \,{\rm d}x = \frac{2G}\pi \quad)] ([math(\Gamma(x))]는 감마 함수)

감마 함수의 반사 공식 [math(\Gamma(x)\Gamma(1-x) = \dfrac\pi{\sin\pi x})]를 사용하여 증명할 수 있다.

[math(\displaystyle \begin{aligned}
\int_0^{1/2} \Gamma(1+x)\Gamma(1-x) \,{\rm d}x &= \int_0^{1/2} x\Gamma(x)\Gamma(1-x) \,{\rm d}x \\
&= \int_0^{1/2} \frac{\pi x}{\sin\pi x} \,{\rm d}x
\end{aligned} )]

[math(\pi x=t)]로 치환하면

[math(\displaystyle \begin{aligned}
\int_0^{1/2} \Gamma(1+x)\Gamma(1-x) \,{\rm d}x &= \int_0^{1/2} \frac{\pi x}{\sin\pi x} \,{\rm d}x \\
&= \int_0^{\pi/2} \frac t{\sin t} \frac{{\rm d}t}\pi \\
&= \frac1\pi \int_0^{\pi/2} \frac t{\sin t} \,{\rm d}t \\
&= \frac{2G}\pi
\end{aligned} )]

}}}||

[math(\displaystyle \int_0^1 K(k) \,{\rm d}k = 2G)]
[math(\displaystyle \int_0^1 E(k) \,{\rm d}k = G+\frac12 \approx 1.4159655942 \quad)] ([math(K(k))], [math(E(k))]는 각각 완전 제1, 2종 타원적분)
증명은 타원 적분 문서의 관련 공식 문단 참고.

[math(\displaystyle \int_0^{\pi/2} \operatorname{igd}(x) \,{\rm d}x = 2G \quad)] ([math(\operatorname{igd}(x))]는 구데르만 역함수)
증명은 구데르만 함수 문서의 극한값 및 미적분 문단 참고.

[math(\displaystyle \int_1^\infty \frac{\ln x}{1+x^2} \,{\rm d}x = G)]
[math(\displaystyle \int_0^1 \frac{\ln x}{1+x^2} \,{\rm d}x = -G)]

위쪽 식부터 증명하자. 우선 [math(x=e^t)]로 치환한 후 분모와 분자에 [math(e^{-2t})]를 곱하면

[math(\displaystyle \begin{aligned}
\int_1^\infty \frac{\ln x}{1+x^2} \,{\rm d}x &= \int_0^\infty \frac{te^t}{1+e^{2t}} \,{\rm d}t \\
&= \int_0^\infty \frac{te^{-t}}{1+e^{-2t}} \,{\rm d}t
\end{aligned} )]

[math(\displaystyle \frac1{1+e^{-2t}} = \sum_{n=0}^\infty (-1)^n e^{-2nt})]임을 이용하면

[math(\displaystyle \begin{aligned}
\int_1^\infty \frac{\ln x}{1+x^2} \,{\rm d}x &= \int_0^\infty \frac{te^{-t}}{1+e^{-2t}} \,{\rm d}t \\
&= \int_0^\infty t \sum_{n=0}^\infty (-1)^n e^{-(2n+1)t} \,{\rm d}t \\
&= \sum_{n=0}^\infty (-1)^n \int_0^\infty t e^{-(2n+1)t} \,{\rm d}t
\end{aligned} )]

여기서 이 정적분의 값은

[math(\displaystyle \begin{aligned}
\int_0^\infty t e^{-(2n+1)t} \,{\rm d}t &= \biggl. \frac{t e^{-(2n+1)t}}{-(2n+1)} \biggr|_0^\infty -\int_0^\infty \frac{e^{-(2n+1)t}}{-(2n+1)} \,{\rm d}t \\
&= \int_0^\infty \frac{e^{-(2n+1)t}}{2n+1} \,{\rm d}t = \biggl. \frac{e^{-(2n+1)t}}{-(2n+1)^2} \biggr|_0^\infty \\
&= \frac1{(2n+1)^2}
\end{aligned} )]

이므로 위쪽 식을 증명할 수 있다.

[math(\displaystyle \begin{aligned}
\therefore \int_1^\infty \frac{\ln x}{1+x^2} \,{\rm d}x &= \sum_{n=0}^\infty (-1)^n \int_0^\infty t e^{-(2n+1)t} \,{\rm d}t \\
&= \sum_{n=0}^\infty (-1)^n \frac1{(2n+1)^2} \\
&= G \qquad \blacksquare
\end{aligned} )]

한편, [math(\displaystyle \int_1^\infty \frac{\ln x}{1+x^2} \,{\rm d}x = G)]에서 [math(x=\dfrac1t)]로 치환하면 아래쪽 식을 증명할 수 있다.

[math(\displaystyle \begin{aligned}
G = \int_1^\infty \frac{\ln x}{1+x^2} \,{\rm d}x &= \int_\infty^1 \frac{\ln t^{-1}}{1+t^{-2}} \!\left( -\frac1{t^2} \,{\rm d}t \right) \\
&= \int_1^\infty \frac{\ln t^{-1}}{t^2+1} \,{\rm d}t \\
&= -\int_1^\infty \frac{\ln t}{t^2+1} \,{\rm d}t \qquad \blacksquare
\end{aligned} )]

}}}||

[math(\displaystyle \int_0^{\pi/4} \ln{\tan x} \,{\rm d}x = -G)]
[math(\displaystyle \int_0^{\pi/4} \ln{\operatorname{cot}x} \,{\rm d}x = G)]

[math(\displaystyle \quad \int_0^1 \frac{\ln x}{1+x^2} \,{\rm d}x = -G)]에서

[math(x=\tan t)]로 치환하면 위쪽 식을 증명할 수 있다.
{{{#!wiki style="text-align:center"

[math(\displaystyle \begin{aligned}
-G = \int_0^1 \frac{\ln x}{1+x^2} \,{\rm d}x &= \int_0^{\pi/4} \frac{\ln \tan t}{1+\tan^2 t} \,\sec^2 t \,{\rm d}t \\
&= \int_0^{\pi/4} \ln \tan t \,{\rm d}t \qquad \blacksquare \\
\end{aligned} )]}}}

[math(x=\cot t)]로 치환하면 아래쪽 식을 증명할 수 있다.
{{{#!wiki style="text-align:center"

[math(\displaystyle \begin{aligned}
-G = \int_0^1 \frac{\ln x}{1+x^2} \,{\rm d}x &= \int_0^{\pi/4} \frac{\ln \cot t}{1+\cot^2 t} \,(-\csc^2 t \,{\rm d}t) \\
&= -\int_0^{\pi/4} \ln \cot t \,{\rm d}t \\
\therefore \int_0^{\pi/4} \ln \cot t \,{\rm d}t &= G \qquad \blacksquare
\end{aligned} )]}}}
}}}||

[math(\displaystyle \iint_{[0,1]^2} \frac1{1+x^2y^2} \,{\rm d}x\,{\rm d}y = G)]

[math(\displaystyle \frac1{1+x^2y^2} = \sum_{n=0}^\infty (-1)^n x^{2n} y^{2n})]임을 이용한다.

[math(\displaystyle \begin{aligned}
\iint_{[0,1]^2} \frac1{1+x^2y^2} \,{\rm d}x\,{\rm d}y &= \int_0^1 \int_0^1 \sum_{n=0}^\infty (-1)^n x^{2n} y^{2n} \,{\rm d}x \,{\rm d}y \\
&= \sum_{n=0}^\infty (-1)^n \int_0^1 x^{2n} \,{\rm d}x \int_0^1 y^{2n} \,{\rm d}y \\
&= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} \\
&= G
\end{aligned} )]

}}}||

3. 값

소수점 이하 10000자리 [접기 · 펼치기]: 0.9159655941 7721901505 4603514932 3841107741 4937428167 2134266498 1196217630 1977625476 9479356512 9261151062
4857442261 9196199579 0358988033 2585905943 1594737481 1584069953 3202877331 9460519038 7274781640 8786590902
4706484152 1630002287 2764094238 8259957741 5088163974 7025248201 1560707644 8838078733 7048990086 4775113225
9971343407 4854075532 3076856533 5768095835 2602193823 2395080072 0680355761 0482357339 4231914982 9836189977
0690364041 8086217941 1019175327 4314997823 3976105512 2477953032 4875371878 6658280823 6057022559 4194818097
5350971131 5712615804 2427236364 3985001738 2875977976 5306837009 2980873887 4956108936 5977194096 8726844441
6680462162 4339864838 9162804482 8150627302 2742073884 3117221827 2190472255 8705319086 8573542349 8539498309
9191159673 8846450861 5152499624 2370437451 7773723517 7544070853 8464401321 7483929999 4757244619 9754961975
8706400747 4870701490 9376788730 4586997986 0644874974 6438720623 8513712392 7363049985 0353922392 8787979063
3644032354 7845358519 2777778727 0906083031 9943013323 1671247615 8709792455 4791190921 2620185480 3963934243

4956537596 7394943547 3001438518 0705051250 7488613285 6412934495 9502298722 9831628948 1646162257 3989476231
8195420066 0718814275 9497559958 9836373037 6753385338 1354503127 6817240118 1407215346 8831683568 1686393272
9367758667 3925839540 6180333878 3068706490 1433486017 2981069921 7995653095 8187157911 5539560366 8903699049
3966753843 7758104931 8995538551 6262196253 3168040162 7375213012 0940604538 7950760538 2712319746 7900882369
1786155733 8912441722 3833938148 1207759942 9849172439 7668575632 7180688082 7998297937 8849432724 9346576074
9054387481 9526813074 4370462946 3589281027 6531705076 5479744948 3994895947 7092788591 1958487241 2786608408
8554597823 8124922605 0561009458 4486698958 5768716111 7178666233 6847409949 3855413210 9375528181 5525881591
5022282444 5444171860 9946588151 7664960782 2367897051 9269711312 5713754543 7012432967 3057246845 0158193130
1608776621 5650957554 6796667866 1708234768 2558133518 6819377456 5001456526 1704096074 6889539302 3479198060
0084245562 1751084234 7173638787 9369577878 4409337922 1989457534 0961647424 5546224787 8800292291 4803690711

5270795545 5054147826 8849818524 6005814466 5178681423 1541148785 5409966516 7385397276 1469701690 4391511490
0893330791 8457465762 0996775481 2313820154 3601098852 7216297701 0876157478 1735641636 9857035534 0672649351
9631695547 6721150777 2315900448 3382605161 1638343086 5139797225 1617413853 8129324801 1946362518 8008403981
9455390551 8210424606 2921852175 6024654860 1929767239 7405110395 2645692429 7864212424 0375189267 8729602717
7337873837 9978326676 2086119520 6791215126 3821192523 2940406920 5994386427 4693215338 8566711733 0827142408
3326592032 6075316592 8042310230 9973584003 9594034263 2227688070 1186819617 6780905631 5815978453 7637578356
3735902771 6488313102 8876937950 5350732080 1807581022 3823080317 6250432942 4722268391 2297129553 5135510431
4761886655 4743676921 8412018877 1617992285 6205635220 5470320069 1808688066 1211742040 6099241234 8760515406
8202262559 5048124858 9411873583 4682290423 0836155547 6947777083 1940874812 4916748929 0065936961 6416623436
8370754396 3838945144 0119556487 3813429212 2982001302 1079961922 4249244930 5199923585 8158082603 5249799850

5918669722 0123164897 1048307017 9352811222 8966355128 3174373523 9301140279 2389808744 5696483090 1320787765
8785362301 3542800016 2905587729 5006795876 1782473748 7137806004 2208445346 0450647024 4325808516 4777173903
1960286555 3832828141 5915248735 2633071505 1314788284 4999238663 2431981063 3651524331 1321463900 9333621591
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3899302958 4492586921 8978864438 8819358114 5267705631 6060973768 4654083694 2302038168 2639245857 9107404870
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1084480513 1449637450 4507208211 7288523799 3949048626 1381992522 0523930672 7369305935 2176372166 1889041942
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9329069964 7498379824 1227697609 4331079408 5840310549 1213134469 6275752331 3373211080 3099242578 8565804104

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4. 관련 문서

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1. 개요

2. 항등식

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4. 관련 문서

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