Editing Catalan's constant (section)

==Integral identities==
As Seán Stewart writes, "There is a rich and seemingly endless source of definite integrals that
can be equated to or expressed in terms of Catalan's constant."<ref>{{citation | last = Stewart | first = Seán M. | doi = 10.1017/mag.2020.99 | issue = 561 | journal = [[The Mathematical Gazette]] | mr = 4163926 | pages = 449–459 | title = A Catalan constant inspired integral odyssey | volume = 104 | year = 2020| s2cid = 225116026 }}</ref> Some of these expressions include:
<math display="block">\begin{align}
G &= -\frac{1}{\pi i}\int_{0}^{\frac{\pi}{2}} \ln\ln \tan x \ln \tan x \,dx \\[3pt]
G &= \iint_{[0,1]^2} \! \frac{1}{1+x^2 y^2} \,dx\, dy \\[3pt] 
G &= \int_0^1\int_0^{1-x} \frac{1}{1 -x^2-y^2} \,dy\,dx \\[3pt]
G &= \int_1^\infty \frac{\ln t}{1 + t^2} \,dt \\[3pt]
G &= -\int_0^1 \frac{\ln t}{1 + t^2} \,dt \\[3pt]
G &= \frac{1}{2} \int_0^\frac{\pi}{2} \frac{t}{\sin t} \,dt \\[3pt]
G &= \int_0^\frac{\pi}{4} \ln \cot t \,dt \\[3pt]
G &= \frac{1}{2} \int_0^\frac{\pi}{2} \ln \left( \sec t +\tan t \right) \,dt \\[3pt]
G &= \int_0^1 \frac{\arccos t}{\sqrt{1+t^2}} \,dt \\[3pt]
G &= \int_0^1 \frac{\operatorname{arcsinh} t}{\sqrt{1-t^2}} \,dt \\[3pt]
G &= \frac{1}{2} \int_0^\infty \frac{\operatorname{arctan} t}{t\sqrt{1+t^2}} \,dt \\[3pt]
G &= \frac{1}{2} \int_0^1 \frac{\operatorname{arctanh} t}{\sqrt{1-t^2}} \,dt \\[3pt]
G &= \int_0^\infty \arccot e^{t} \,dt \\[3pt]
G &= \frac{1}{4} \int_0^{{\pi^2}/{4}} \csc \sqrt{t} \,dt \\[3pt]
G &= \frac{1}{16} \left(\pi^2 + 4\int_1^\infty \arccsc^2 t \,dt\right) \\[3pt]
G &= \frac{1}{2} \int_0^\infty \frac{t}{\cosh t} \,dt  \\[3pt]
G &= \frac{\pi}{2} \int_1^\infty \frac{\left(t^4-6t^2+1\right)\ln\ln t}{\left(1+t^2\right)^3} \,dt \\[3pt]
G &= \frac{1}{2} \int_0^\infty \frac{\arcsin \left(\sin t\right)}{t} \,dt \\[3pt]
G &= 1 + \lim_{\alpha\to{1^-}}\!\left\{\int_0^{\alpha}\!\frac{\left(1+6t^2+t^4\right)\arctan{t}}{t\left(1-t^2\right)^2}\, dt 
+  2\operatorname{artanh}{\alpha} - \frac{\pi\alpha}{1-\alpha^2} \right\} \\[3pt]
G &= 1 - \frac18 \iint_{\R^2}\!\!\frac{x\sin\left(2xy/\pi\right)}{\,\left(x^2+\pi^2\right)\cosh x\sinh y\,} \,dx\,dy \\[3pt]
G &= \int_{0}^{\infty}\int_{0}^{\infty}\frac{\sqrt[4]{x} \left(\sqrt{x} \sqrt{y}-1\right)}{(x+1)^2 \sqrt[4]{y} (y+1)^2 \log (x y)}dxdy
\end{align}</math>

where the last three formulas are related to [[Carl Johan Malmsten|Malmsten's]] integrals.<ref>{{Cite journal| first1=Iaroslav| last1=Blagouchine| title=Rediscovery of Malmsten's integrals, their evaluation by contour integration methods and some related results| year=2014| doi=10.1007/s11139-013-9528-5| url=https://iblagouchine.perso.centrale-marseille.fr/publications/Blagouchine-Malmsten-integrals-and-their-evaluation-by-contour-integration-methods-(Ramanujan-J-2014).pdf| volume=35| journal=The Ramanujan Journal| pages=21–110| s2cid=120943474| access-date=2018-10-01| archive-url=https://web.archive.org/web/20181002020243/https://iblagouchine.perso.centrale-marseille.fr/publications/Blagouchine-Malmsten-integrals-and-their-evaluation-by-contour-integration-methods-(Ramanujan-J-2014).pdf| archive-date=2018-10-02| url-status=dead }}</ref>

If {{math|K(''k'')}} is the [[Elliptic integral#Complete elliptic integral of the first kind|complete elliptic integral of the first kind]], as a function of the elliptic modulus {{math|''k''}}, then
<math display="block"> G = \tfrac{1}{2} \int_0^1 \mathrm{K}(k)\,dk </math>

If {{math|E(''k'')}} is the [[Elliptic integral#Complete elliptic integral of the second kind|complete elliptic integral of the second kind]], as a function of the elliptic modulus {{math|''k''}}, then
<math display="block"> G = -\tfrac{1}{2}+\int_0^1 \mathrm{E}(k)\,dk </math>

With the [[gamma function]] {{math|1=Γ(''x'' + 1) = ''x''!}}
<math display="block">\begin{align}
G &= \frac{\pi}{4} \int_0^1 \Gamma\left(1+\frac{x}{2}\right)\Gamma\left(1-\frac{x}{2}\right)\,dx \\
&= \frac{\pi}{2} \int_0^\frac12\Gamma(1+y)\Gamma(1-y)\,dy
\end{align}</math>

The integral
<math display="block"> G = \operatorname{Ti}_2(1)=\int_0^1 \frac{\arctan t}{t}\,dt </math>
is a known special function, called the [[inverse tangent integral]], and was extensively studied by [[Srinivasa Ramanujan]].