定义

卡塔兰常数（Catalan’s Constant）常用字母 G 表示, 我们先来看看它的定义：

级数定义式： G:=∑n=0∞(−1)n(2n+1)2G:=\sum_{n=0}^{∞}{\dfrac{(-1)^{n}}{(2n+1)^{2}}}，积分定义式：G:=−∫0π/4ln⁡tan⁡x dxG:=-\int_{0}^{\pi/{4}}\ln \tan x \ \mathrm{d}x

这两种定义可以互相转化：作换元 t=tan⁡xt=\tan x, 有

G=−∫0π/4ln⁡tan⁡x dx=−∫01ln⁡t darctan⁡t=[−ln⁡t⋅arctan⁡t]01+∫01arctan⁡tt dt=0+∫01∑n=0∞(−1)nt2n2n+1 dt=∑n=0∞(−1)n∫01t2n2n+1 dt=∑n=0∞(−1)n[t2n+1(2n+1)2]01=∑n=0∞(−1)n(2n+1)2.\begin{aligned} G&=-\int_{0}^{\pi/4}\ln\tan x\ \mathrm{d}x\\ &=-\int_{0}^{1}\ln t\ \mathrm{d}\arctan t\\ &=\left[-\ln t\cdot\arctan t \right]_{0}^{1}+\int_{0}^{1}\frac{\arctan t}{t}\ \mathrm{d}t\\ &=0+\int_{0}^{1}\sum_{n=0}^{∞}{\frac{(-1)^{n}t^{2n}}{2n+1}}\ \mathrm{d}t\\ &=\sum_{n=0}^{∞}{(-1)^{n}}\int_{0}^{1}\frac{t^{2n}}{2n+1}\ \mathrm{d}t\\ &=\sum_{n=0}^{∞}{(-1)^{n}\left[\frac{t^{2n+1}}{(2n+1)^{2}} \right]_{0}^{1}}\\ &=\sum_{n=0}^{∞}{\frac{(-1)^{n}}{(2n+1)^{2}}}. \end{aligned}

由它的定义和上述推导过程, 有
（1）

∫0π/4ln⁡tan⁡x dx=−G.\begin{aligned} \int_{0}^{\pi/{4}}\ln \tan x \ \mathrm{d}x=-G. \end{aligned}

（2）

∫0π/4ln⁡cot⁡x dx=G.\begin{aligned} \int_{0}^{\pi/{4}}\ln \cot x \ \mathrm{d}x=G. \end{aligned}

（3）

∫01arctan⁡tt dt=G.\begin{aligned} \int_{0}^{1}\frac{\arctan t}{t} \ \mathrm{d}t=G. \end{aligned}

这三个积分表达式的任何一个都可以作为卡塔兰常数的积分定义式.

相关积分

先证明两个常见的不含卡塔兰常数的积分等式：
①

∫0π/2ln⁡sin⁡x dx=−π2ln⁡2.\begin{aligned} \int_{0}^{\pi/2}\ln \sin x \ \mathrm{d}x=-\frac{\pi}{2}\ln2. \end{aligned}

②

∫0π/4ln⁡(1+tan⁡x) dx=π8ln⁡2.\begin{aligned} \int_{0}^{\pi/4}\ln(1+\tan x) \ \mathrm{d}x=\frac{\pi}{8}\ln2. \end{aligned}

证：

∫0π/2ln⁡sin⁡x dx=∫0π/2ln⁡cos⁡x dx=12∫0π/2(ln⁡sin⁡x+ln⁡cos⁡x) dx=12∫0π/2(ln⁡sin⁡2x+ln⁡12) dx=14∫0πln⁡sin⁡x dx+12⋅(ln⁡12)⋅π2=12∫0π/2ln⁡sin⁡x dx−π4ln⁡2=−π2ln⁡2.\begin{aligned} &\int_{0}^{\pi/2}\ln \sin x \ \mathrm{d}x\\ =&\int_{0}^{\pi/2}\ln \cos x \ \mathrm{d}x\\ =&\frac{1}{2}\int_{0}^{\pi/2}(\ln \sin x+\ln \cos x) \ \mathrm{d}x\\ =&\frac{1}{2}\int_{0}^{\pi/2}(\ln \sin 2x+\ln\frac{1}{2}) \ \mathrm{d}x\\ =&\frac{1}{4}\int_{0}^{\pi}\ln \sin x \ \mathrm{d}x+\frac{1}{2}\cdot(\ln\frac{1}{2})\cdot\frac{\pi}{2}\\ =&\frac{1}{2}\int_{0}^{\pi/2}\ln \sin x \ \mathrm{d}x-\frac{\pi}{4}\ln2\\ =&-\frac{\pi}{2}\ln2. \end{aligned}

∫0π/4ln⁡(1+tan⁡x) dx=∫0π/4ln⁡(1+tan⁡(π4−x)) dx=∫0π/4ln⁡(1+1−tan⁡x1+tan⁡x) dx=∫0π/4(ln⁡2−ln⁡(1+tan⁡x)) dx=π4ln⁡2−∫0π/4(ln⁡(1+tan⁡x)) dx=π8ln⁡2.\begin{aligned} &\int_{0}^{\pi/4}\ln(1+\tan x) \ \mathrm{d}x\\ =&\int_{0}^{\pi/4}\ln(1+\tan(\frac{\pi}{4}-x)) \ \mathrm{d}x\\ =&\int_{0}^{\pi/4}\ln(1+\frac{1-\tan x}{1+\tan x}) \ \mathrm{d}x\\ =&\int_{0}^{\pi/4}(\ln2-\ln(1+\tan x)) \ \mathrm{d}x\\ =&\frac{\pi}{4}\ln2-\int_{0}^{\pi/4}(\ln(1+\tan x)) \ \mathrm{d}x\\ =&\frac{\pi}{8}\ln2. \end{aligned}

接下来推导 19 个跟卡塔兰常数有关的积分等式：
（4）

∫0π/4ln⁡sin⁡x dx=−12G−π4ln⁡2.\begin{aligned} \int_{0}^{\pi/4}\ln \sin x \ \mathrm{d}x=-\frac{1}{2}G-\frac{\pi}{4}\ln2. \end{aligned}

（5）

∫0π/4ln⁡cos⁡x dx=12G−π4ln⁡2.\begin{aligned} \int_{0}^{\pi/4}\ln \cos x \ \mathrm{d}x=\frac{1}{2}G-\frac{\pi}{4}\ln2. \end{aligned}

证：

∫0π/4ln⁡sin⁡x dx+∫0π/4ln⁡cos⁡x dx=∫0π/4ln⁡(sin⁡xcos⁡x) dx=∫0π/4ln⁡sin⁡2x dx+∫0π/4ln⁡12 dx=12∫0π/2ln⁡sin⁡x dx−π4ln⁡2=12(−π2ln⁡2)−π4ln⁡2=−π2ln⁡2.\begin{aligned} &\int_{0}^{\pi/4}\ln \sin x \ \mathrm{d}x+\int_{0}^{\pi/4}\ln \cos x \ \mathrm{d}x\\ =&\int_{0}^{\pi/4}\ln(\sin x\cos x) \ \mathrm{d}x\\ =&\int_{0}^{\pi/4}\ln \sin2x \ \mathrm{d}x+\int_{0}^{\pi/4}\ln\frac{1}{2} \ \mathrm{d}x\\ =&\frac{1}{2}\int_{0}^{\pi/2}\ln \sin x \ \mathrm{d}x-\frac{\pi}{4}\ln2=\frac{1}{2}(-\frac{\pi}{2}\ln2)-\frac{\pi}{4}\ln2\\ =&-\frac{\pi}{2}\ln2. \end{aligned}

∫0π/4ln⁡cos⁡x dx−∫0π/4ln⁡sin⁡x dx=∫0π/4ln⁡cot⁡xdx=G.\begin{aligned} \int_{0}^{\pi/4}\ln \cos x \ \mathrm{d}x-\int_{0}^{\pi/4}\ln \sin x \ \mathrm{d}x =\int_{0}^{\pi/4}\ln \cot x\mathrm{d}x=G. \end{aligned}

（6）

∫0π/2ln⁡(1+sin⁡x) dx=2G−π2ln⁡2.\begin{aligned} \int_{0}^{\pi/2}\ln(1+\sin x) \ \mathrm{d}x=2G-\frac{\pi}{2}\ln2. \end{aligned}

（7）

∫0π/2ln⁡(1+cos⁡x) dx=2G−π2ln⁡2.\begin{aligned} \int_{0}^{\pi/2}\ln(1+\cos x) \ \mathrm{d}x=2G-\frac{\pi}{2}\ln2. \end{aligned}

证：

∫0π/2ln⁡(1+cos⁡x) dx=∫0π/2ln⁡(2cos⁡2x2) dx=π2ln⁡2+2∫0π/2ln⁡cos⁡x2 dx=π2ln⁡2+4∫0π/4ln⁡cos⁡x dx=π2ln⁡2+4(12G−π4ln⁡2)=2G−π2ln⁡2.\begin{aligned} &\int_{0}^{\pi/2}\ln(1+\cos x) \ \mathrm{d}x\\ =&\int_{0}^{\pi/2}\ln(2\cos^2\frac{x}{2}) \ \mathrm{d}x\\ =&\frac{\pi}{2}\ln2+2\int_{0}^{\pi/2}\ln \cos\frac{x}{2} \ \mathrm{d}x\\ =&\frac{\pi}{2}\ln2+4\int_{0}^{\pi/4}\ln \cos x \ \mathrm{d}x\\ =&\frac{\pi}{2}\ln2+4(\frac{1}{2}G-\frac{\pi}{4}\ln2)\\ =&2G-\frac{\pi}{2}\ln2. \end{aligned}

由区间再现得（6）=（7）.
（8）

∫0π/2xsin⁡x1+cos⁡xdx=2G−π2ln⁡2.\begin{aligned} \int_{0}^{\pi/2}\frac{x\sin x}{1+\cos x}\mathrm{d}x=2G-\frac{\pi}{2}\ln2. \end{aligned}

（9）

∫0π/2xcos⁡x1+sin⁡xdx=−2G+πln⁡2.\begin{aligned} \int_{0}^{\pi/2}\frac{x\cos x}{1+\sin x}\mathrm{d}x=-2G+\pi\ln2. \end{aligned}

证：
由

∫0π/2ln⁡(1+cos⁡x) dx=[xln⁡(1+cos⁡x)]0π2−∫0π/2−xsin⁡x1+cos⁡xdx=∫0π/2xsin⁡x1+cos⁡x dx.\begin{aligned} \int_{0}^{\pi/2}\ln(1+\cos x) \ \mathrm{d}x=\left[ x\ln(1+\cos x) \right]_{0}^{\frac{\pi}{2}}- \int_{0}^{\pi/2}\frac{-x\sin x}{1+\cos x}\mathrm{d}x=\int_{0}^{\pi/2}\frac{x\sin x}{1+\cos x} \ \mathrm{d}x. \end{aligned}

得

∫0π/2xsin⁡x1+cos⁡x dx=2G−π2ln⁡2.\begin{aligned} \int_{0}^{\pi/2}\frac{x\sin x}{1+\cos x} \ \mathrm{d}x=2G-\frac{\pi}{2}\ln2. \end{aligned}

由

∫0π/2ln⁡(1+sin⁡x) dx==[xln⁡(1+sin⁡x)]0π2−∫0π/2xcos⁡x1+sin⁡xdx=π2ln⁡2−∫0π/2xcos⁡x1+sin⁡x dx.\begin{aligned} \int_{0}^{\pi/2}\ln(1+\sin x) \ \mathrm{d}x==\left[ x\ln(1+\sin x) \right]_{0}^{\frac{\pi}{2}}- \int_{0}^{\pi/2}\frac{x\cos x}{1+\sin x}dx=\frac{\pi}{2}\ln2-\int_{0}^{\pi/2}\frac{x\cos x}{1+\sin x} \ \mathrm{d}x. \end{aligned}

得

∫0π/2xcos⁡x1+sin⁡x dx=−2G+πln⁡2.\begin{aligned} \int_{0}^{\pi/2}\frac{x\cos x}{1+\sin x} \ \mathrm{d}x=-2G+\pi\ln2. \end{aligned}

（10）

∫0π/2ln⁡(1−cos⁡x) dx=−2G−π2ln⁡2.\begin{aligned} \int_{0}^{\pi/2}\ln(1-\cos x) \ \mathrm{d}x=-2G-\frac{\pi}{2}\ln2. \end{aligned}

（11）

∫0π/2ln⁡(1−sin⁡x) dx=−2G−π2ln⁡2.\begin{aligned} \int_{0}^{\pi/2}\ln(1-\sin x) \ \mathrm{d}x=-2G-\frac{\pi}{2}\ln2. \end{aligned}

证：

∫0π/2ln⁡(1−cos⁡x) dx=∫0π/2ln⁡(2sin⁡2x2) dx=π2ln⁡2+2∫0π/2ln⁡sin⁡x2 dx=π2ln⁡2+4∫0π/4ln⁡sin⁡x dx=π2ln⁡2+4(−12G−π4ln⁡2)=−2G−π2ln⁡2.\begin{aligned} &\int_{0}^{\pi/2}\ln(1-\cos x) \ \mathrm{d}x\\ =&\int_{0}^{\pi/2}\ln(2\sin^2\frac{x}{2}) \ \mathrm{d}x\\ =&\frac{\pi}{2}\ln2+2\int_{0}^{\pi/2}\ln \sin\frac{x}{2} \ \mathrm{d}x\\ =&\frac{\pi}{2}\ln2+4\int_{0}^{\pi/4}\ln \sin x \ \mathrm{d}x\\ =&\frac{\pi}{2}\ln2+4(-\frac{1}{2}G-\frac{\pi}{4}\ln2)\\ =&-2G-\frac{\pi}{2}\ln2. \end{aligned}

由区间再现得（11）=（10）.