Calculus – Sequences and series – Catalan's constant

Catalan's constant \(G\) is defined as the sum of the series \(\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\).

Note that this series is absolutely convergent. However, it is not easy to find its sum. We have \(G:=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\approx0.916\).

There are several alternative (integral) representations for this constant. Since \(\arctan(x)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}\) for \(|x| < 1\) we have

\[\int_0^1\frac{\arctan(x)}{x}\,dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1x^{2n}\,dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\cdot\frac{x^{2n+1}}{2n+1}\bigg|_0^1 =\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}=G.\]

Using \(x=e^{-t}\) we obtain

\[G=\int_0^1\frac{\arctan(x)}{x}\,dx=\int_{\infty}^0\frac{\arctan\left(e^{-t}\right)}{e^{-t}}\,d e^{-t} =\int_0^{\infty}\arctan\left(e^{-t}\right)\,dt.\]

The substitution \(\arctan{x}=t\) or \(x=\tan(t)\) leads to

\[\int_0^1\frac{\arctan(x)}{x}\,dx=\int_0^{\frac{1}{4}\pi}\frac{t}{\tan(t)}d\tan(t)=\int_0^{\frac{1}{4}\pi}\frac{t}{\tan(t)}\cdot\frac{1}{\cos^2(t)}\,dt =\int_0^{\frac{1}{4}\pi}\frac{t}{\sin(t)\cos(t)}\,dt.\]

This can be rewritten as

\[G=\int_0^{\frac{1}{4}\pi}\frac{t}{\sin(t)\cos(t)}\,dt=\int_0^{\frac{1}{4}\pi}\frac{2t}{2\sin(t)\cos(t)}\,dt =\int_0^{\frac{1}{4}\pi}\frac{2t}{\sin(2t)}\,dt=\frac{1}{2}\int_0^{\frac{1}{2}\pi}\frac{x}{\sin(x)}\,dx.\]

Previously, we showed that \(\displaystyle\int_0^1\sqrt{\frac{x}{1-x}}\ln\left(\frac{x}{1-x}\right)\,dx=\pi\). In the proof we used

\[\int_1^{\infty}\frac{\ln(t)}{1+t^2}\,dt=-\int_0^1\frac{\ln(t)}{1+t^2}\,dt\quad\Longrightarrow\quad\int_0^{\infty}\frac{\ln(t)}{1+t^2}\,dt=0.\]

Using integration by parts we have

\begin{align*} \int_1^{\infty}\frac{\ln(x)}{1+x^2}\,dt&=-\int_0^1\frac{\ln(x)}{1+x^2}\,dx=-\int_0^1\ln(x)\,d\arctan(x)\\[2.5mm] &=-\ln(x)\arctan(x)\bigg|_0^1+\int_0^1\arctan(x)\,d\ln(x)=\int_0^1\frac{\arctan(x)}{x}\,dx=G, \end{align*}

since \(\arctan(1)\ln(1)=\frac{1}{4}\pi\cdot0=0\) and \(\arctan(x)\ln(x)\sim x\ln(x)\to0\) for \(x\downarrow0\).

An alternative derivation is as follows. Using integration by parts we obtain

\begin{align*} \int_0^1x^n\ln(x)\,dx&=\frac{1}{n+1}\ln(x)\,dx^{n+1}=\frac{1}{n+1}x^{n+1}\ln(x)\bigg|_0^1-\frac{1}{n+1}\int_0^1x^{n+1}\,d\ln(x)\\[2,5mm] &=0-\frac{1}{n+1}\int_0^1x^n\,dx=-\frac{1}{(n+1)^2}x^{n+1}\bigg|_0^1=-\frac{1}{(n+1)^2},\quad n=0,1,2,\ldots. \end{align*}

With \(\displaystyle\frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-x^2)^n\) for \(|x|<1\) this implies that

\[\int_0^1\frac{\ln(x)}{1+x^2}\,dx=\sum_{n=0}^{\infty}(-1)^n\int_0^1x^{2n}\ln(x)\,dx=-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}=-G.\]

Alternative forms of these integrals are obtained using certain substitutions.

If we set \(t=\tan(\theta)\), then we have \(dt=\left(1+\tan^2(\theta)\right)\,d\theta\) and therefore

\[G=-\int_0^1\frac{\ln(t)}{1+t^2}\,dt=-\int_0^{\frac{1}{4}\pi}\frac{\ln\left(\tan(\theta)\right)}{1+\tan^2(\theta)}\,d\tan(\theta) =-\int_0^{\frac{1}{4}\pi}\ln\left(\tan(\theta)\right)\,d\theta\quad\text{or}\quad G=\int_1^{\infty}\frac{\ln(t)}{1+t^2}\,dt=\int_{\frac{1}{4}\pi}^{\frac{1}{2}\pi}\ln\left(\tan(\theta)\right)\,d\theta.\]

Setting \(\ln(t)=x\) or \(t=e^x\) we obtain

\[G=\int_1^{\infty}\frac{\ln(t)}{1+t^2}\,dt=\int_0^{\infty}\frac{x}{1+e^{2x}}\cdot e^x\,dx=\int_0^{\infty}\frac{x}{e^{-x}+e^x}\,dx =\frac{1}{2}\int_0^{\infty}\frac{x}{\cosh(x)}\,dx.\]

The relation

\[\tan^2(\theta)=\frac{\sin^2(\theta)}{\cos^2(\theta)}=\frac{2\sin^2(\theta)}{2\cos^2(\theta)}=\frac{1-\cos(2\theta)}{1+\cos(2\theta)}\]

leads to another interesting representation

\[G=-\int_0^{\frac{1}{4}\pi}\ln\left(\tan(\theta)\right)\,d\theta=-\frac{1}{2}\int_0^{\frac{1}{4}\pi}\ln\left(\tan^2(\theta)\right)\,d\theta =-\frac{1}{2}\int_0^{\frac{1}{4}\pi}\ln\left(\frac{1-\cos(2\theta)}{1+\cos(2\theta)}\right)\,d\theta,\]

which is equivalent to

\[G=-\frac{1}{4}\int_0^{\frac{1}{2}\pi}\ln\left(\frac{1-\cos(x)}{1+\cos(x)}\right)\,dx\quad\text{or}\quad G=\frac{1}{4}\int_0^{\frac{1}{2}\pi}\ln\left(\frac{1+\cos(x)}{1-\cos(x)}\right)\,dx.\]

When we studied Serret's integral we used the fact that \(\displaystyle\int_0^{\frac{1}{4}\pi}\ln\left(\cos(\tfrac{1}{4}\pi-\theta)\right)\,d\theta=\int_0^{\frac{1}{4}\pi}\ln(\cos(t))\,dt\). This integral can be evaluated in terms of Catalan's constant. Using the fact that \(\cos(x)=\sin(\frac{1}{2}\pi-x)\) we obtain that

\[\int_0^{\frac{1}{2}\pi}\ln(\sin(x))\,dx=\int_0^{\frac{1}{2}\pi}\ln(\sin(\tfrac{1}{2}\pi-x))\,dx=\int_0^{\frac{1}{2}\pi}\ln(\cos(x))\,dx.\]

Note that this implies that

\[\int_0^{\frac{1}{2}\pi}\ln(\tan(x))\,dx=\int_0^{\frac{1}{2}\pi}\ln(\sin(x))\,dx-\int_0^{\frac{1}{2}\pi}\ln(\cos(x))\,dx=0.\]

Earlier we have seen that

\[\int_0^{\frac{1}{4}\pi}\ln(\tan(x))\,dx=-G\quad\text{and}\quad\int_{\frac{1}{4}\pi}^{\frac{1}{2}\pi}\ln(\tan(x))\,dx=G.\]

If we apply the substitution \(2x=t\) or we find that

\[\int_0^{\frac{1}{2}\pi}\ln(\sin(2x))\,dx=\frac{1}{2}\int_0^{\pi}\ln(\sin(t))\,dt=\frac{1}{2}\int_0^{\frac{1}{2}\pi}\ln(\sin(t))\,dt+\frac{1}{2}\int_{\frac{1}{2}\pi}^{\pi}\ln(\sin(t))\,dt.\]

Then we apply the substitution \(t=\pi-x\) to the latter integral to find that

\[\int_{\frac{1}{2}\pi}^{\pi}\ln(\sin(t))\,dt=-\int_{\frac{1}{2}\pi}^0\ln(\sin(\pi-x))\,dx=\int_0^{\frac{1}{2}\pi}\ln(\sin(x))\,dx.\]

Hence we have \(\displaystyle\int_0^{\frac{1}{2}\pi}\ln(\sin(2x))\,dx=\int_0^{\frac{1}{2}\pi}\ln(\sin(x))\,dx\).

Finally, we derive for \(I=\displaystyle\int_0^{\frac{1}{2}\pi}\ln(\sin(x))\,dx=\int_0^{\frac{1}{2}\pi}\ln(\cos(x))\,dx\) that

\begin{align*} I&=\int_0^{\frac{1}{2}\pi}\ln(\sin(x))\,dx=\int_0^{\frac{1}{2}\pi}\ln(\sin(2x))\,dx=\int_0^{\frac{1}{2}\pi}\ln(2\sin(x)\cos(x))\,dx =\int_0^{\frac{1}{2}\pi}\left(\ln(2)+\ln(\sin(x))+\ln(\cos(x))\right)\,dx\\[2.5mm] &=\int_0^{\frac{1}{2}\pi}\ln(2)\,dx+\int_0^{\frac{1}{2}\pi}\ln(\sin(x))\,dx+\int_0^{\frac{1}{2}\pi}\ln(\cos(x))\,dx =\frac{1}{2}\pi\ln(2)+2I, \end{align*}

which implies that \(I=-\frac{1}{2}\pi\ln(2)\). The substitution \(x=\frac{1}{2}\pi-t\) now leads to

\[\int_{\frac{1}{4}\pi}^{\frac{1}{2}\pi}\ln(\sin(x))\,dx=-\int_{\frac{1}{4}\pi}^0\ln(\sin(\tfrac{1}{2}\pi-t)\,dt=\int_0^{\frac{1}{4}\pi}\ln(\cos(t))\,dt\]

and

\[\int_{\frac{1}{4}\pi}^{\frac{1}{2}\pi}\ln(\cos(x))\,dx=-\int_{\frac{1}{4}\pi}^0\ln(\cos(\tfrac{1}{2}\pi-t)\,dt=\int_0^{\frac{1}{4}\pi}\ln(\sin(t))\,dt.\]

Hence we have

\[\int_0^{\frac{1}{4}\pi}\ln(\sin(x))\,dx+\int_0^{\frac{1}{4}\pi}\ln(\cos(x))\,dx=\int_0^{\frac{1}{2}\pi}\ln(\sin(x))\,dx =\int_0^{\frac{1}{2}\pi}\ln(\cos(x))\,dx=-\frac{1}{2}\pi\ln(2)\]

and

\[\int_0^{\frac{1}{4}\pi}\ln(\sin(x))\,dx-\int_0^{\frac{1}{4}\pi}\ln(\cos(x))\,dx=\int_0^{\frac{1}{4}\pi}\ln(\tan(x))\,dx=-G.\]

This implies that

\[\int_0^{\frac{1}{4}\pi}\ln(\sin(x))\,dx=-\frac{1}{2}G-\frac{1}{4}\pi\ln(2)\quad\text{and}\quad\int_0^{\frac{1}{4}\pi}\ln(\cos(x))\,dx=\frac{1}{2}G-\frac{1}{4}\pi\ln(2).\]

Note that this implies that

\[\int_0^{\frac{1}{4}\pi}\ln(2\sin(x))\,dx=-\frac{1}{2}G\quad\text{and}\quad\int_0^{\frac{1}{4}\pi}\ln(2\cos(x))\,dx=\frac{1}{2}G.\]

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Last modified on March 15, 2021