An arccotangent integral

Let $\zeta$ denote the Riemann zeta function. Prove that

$\int_0^\infty \left ( \arccot x \right )^3\; \mathrm{d}x = \frac{3 \pi^2 \ln 2}{4} - \frac{21\zeta(3)}{8}$

Solution

Background: I have always been of the opinion that integrals of this form diverged when integrated from something to infinity. Apparently, this is not the case since the above not only converges but it has a nice closed form involving the Riemann zeta function. The technique to break it down , is pretty simple actually. Integration by parts! Yep! We combine IBP along with Fourier series and voilà. However, somewhere in the middle of this evaluation process we will come across a famous constant , $\mathcal{G}$ , coming literally out of nowhere. But this is what to expect when reducing trigonometric integrals to logarithmic ones.

We begin our evaluation by stating two lemmata:

Lemma 1: It holds that $\displaystyle \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )}{1+x^2} \, \mathrm{d}x = \frac{\pi \ln 2}{2} - \mathcal{G}$ .

Proof: Successively, we have that:

$\begin{align*} \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )}{1+x^2} \, \mathrm{d}x &\overset{x=\tan \theta}{=\! =\! =\! =\! =\!} \int_{0}^{\pi/4} \frac{\ln \left ( 1+\tan^2 \theta \right )}{1+\tan^2 \theta} \cdot \sec^2 \theta \, \mathrm{d}\theta\\ &=\int_{0}^{\pi/4} \ln \left ( 1+ \tan^2 \theta \right ) \, \mathrm{d}\theta \\ &= \int_{0}^{\pi/4} \ln \sec^2 \theta \, \mathrm{d} \theta \\ &=2 \int_{0}^{\pi/4} \ln \sec \theta \, \mathrm{d} \theta \\ &=-2 \int_{0}^{\pi/4} \ln \cos \theta \, \mathrm{d} \theta \\ &=-2 \left ( -\int_{0}^{\pi/4} \left (\sum_{n=1}^{\infty} (-1)^n \frac{\cos 2n \theta}{n} - \ln 2 \right ) \, \mathrm{d}\theta \right )\\ &=2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \int_{0}^{\pi/4} \cos 2n\theta \, \mathrm{d} \theta +2 \int_{0}^{\pi/4} \ln 2 \, \mathrm{d}\theta \\ &= 2 \sum_{n=1}^{\infty} \frac{(-1)^n \sin \frac{n \pi}{2}}{2n^2} +\frac{\pi \ln 2}{2} \\ &= \sum_{n=1}^{\infty} \frac{(-1)^n \sin \frac{n \pi}{2}}{2n^2} + \frac{\pi \ln 2}{2} \\ &=\frac{\pi \ln 2}{2} + \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}\\ &= \frac{\pi \ln 2}{2} - \mathcal{G} \end{align*}$

where $\mathcal{G}$ is the Catalan’s constant. Also,

$\ln 2\cos x = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{\cos 2n \theta}{n}$

which is the known Fourier series expansion of the $\ln \cos$ function.

Lemma 2: It holds that $\displaystyle \int_{0}^{1} \frac{\ln x \arctan x}{1+x^2} \, \mathrm{d}x = \frac{7 \zeta(3)}{16} - \frac{\pi \mathcal{G}}{4}$ .

Proof: Successively, we have that:

$\begin{align*} \int_{0}^{1} \frac{\ln x \arctan x}{1+x^2} \, \mathrm{d}x &\overset{x=\tan \theta}{=\! =\! =\! =\!} \int_{0}^{\pi/4} \theta \ln \tan \theta \, \mathrm{d} \theta \\ &=\int_{0}^{\pi/4} \theta \sum_{n \; \text{odd}} \frac{\cos 2n \theta}{n} \, \mathrm{d} \theta \\ &=\sum_{n \; \text{odd}} \frac{1}{n} \int_{0}^{\pi/4} \theta \cos 2 n \theta \, \mathrm{d} \theta \\ &= \frac{7 \zeta(3)}{16} - \frac{\pi \mathcal{G}}{4} \end{align*}$

Hence,

$\begin{align*} \int_{0}^{\infty} \left ( \mathrm{arccot}x \right )^3 \, \mathrm{d}x &= \int_{0}^{\infty} \left ( x \right ) ' \left ( \mathrm{arccot}x \right )^3 \, \mathrm{d}x \\ &=\left [ x \left ( \mathrm{arccot}x \right )^3 \right ]_0^{\infty} +3 \int_{0}^{\infty} \frac{x\left( \mathrm{arccot }x \right )^2}{ x^2+1} \, \mathrm{d}x \\ &=3 \left [ \frac{\ln \left ( 1+x^2 \right ) \, \left (\mathrm{arccot}(x) \right )^2}{2} \right ]_0^{\infty} + 3 \int_{0}^{\infty} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x^2+1} \, \mathrm{d}x \\ &=3 \left ( \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x^2+1} \, \mathrm{d}x + \int_1^{\infty} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x^2+1} \, \mathrm{d}x \right ) \\ &=3 \bigg( \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x^2+1} \, \mathrm{d}x + \\ &\quad \quad \quad + \int_{0}^{1} \frac{\ln \left ( 1+\frac{1}{x^2} \right ) \arctan x}{1+\frac{1}{x^2}} \cdot \frac{1}{x^2} \, \mathrm{d}x \bigg ) \\ &=3 \bigg ( \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x^2+1} \, \mathrm{d}x + \\ &\quad \quad \quad + \int_{0}^{1} \frac{\left (\ln \left ( 1+x^2 \right ) - 2 \ln x \right ) \arctan x}{1+x^2} \, \mathrm{d}x \bigg )\\ &=3\bigg ( \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )\left ( \mathrm{arccot} x + \arctan x \right )}{x^2+1} \, \mathrm{d}x - \\ &\quad \quad \quad - 2 \int_{0}^{1} \frac{\ln x \arctan x}{1+x^2} \, \mathrm{d}x \bigg ) \\ &=3\left ( \frac{\pi}{2} \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )}{1+x^2} \, \mathrm{d}x - 2 \int_{0}^{1} \frac{\ln x \arctan x}{1+x^2} \, \mathrm{d}x \right )\\ &=\frac{3 \pi^2 \ln 2}{4} - \frac{21\zeta(3)}{8} \end{align*}$

Not Lebesgue integrable function

Invertible matrix

"Upper" bound

Existence of c

Recursive formula of volume of sphere

Root inequality

Definite parametric integral

Is f necessarily constant?

Double inequality

Let’s expand on some things here.

1. Let $\mathcal{H}_n$ denote the $n$ -th harmonic number. It holds that:

$\frac{x \arctan x}{x^2+1} = \sum_{n=1}^{\infty} (-1)^{n-1} \left ( \mathcal{H}_{2n} - \frac{\mathcal{H}_n}{2} \right ) x^{2n}$

( the proof is left to the reader )

2. Following $(1)$ it is easy to see that:

$\int_{0}^{1} \frac{\ln x \arctan x}{1+x^2} \, \mathrm{d}x = \frac{1}{4} \sum_{n=1}^{\infty} (-1)^n \cdot \frac{\mathcal{H}_{2n} -\frac{\mathcal{H}_n}{2}}{n^2}$

The second series is an old chestnut.

3. It holds that:

$\sum_{n=1}^{\infty} (-1)^n \frac{\mathcal{H}_n}{n^2} = -\frac{5 \zeta(3)}{8}$

Proof: Successively we have that:

$\begin{align*} \sum_{n=1}^\infty\frac{\mathcal{H}_n}{n^2} &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{n^2}\left(\frac1k-\frac1{k+n}\right)\\ &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{nk(k+n)} \\ &=\sum_{k=1}^\infty\sum_{n=k+1}^\infty\frac1{nk(n-k)}\\ &=\sum_{n=2}^\infty\sum_{k=1}^{n-1}\frac1{nk(n-k)}\\ &=\sum_{n=2}^\infty\sum_{k=1}^{n-1}\frac1{n^2}\left(\frac1k+\frac1{n-k}\right)\\ &=2\sum_{n=1}^\infty\frac{\mathcal{H}_{n-1}}{n^2}\\ &=2\sum_{n=1}^\infty\frac{\mathcal{H}_n}{n^2}-2\zeta(3) \\ \sum_{n=1}^\infty\frac{\mathcal{H}_n}{n^2} &=2\zeta(3) \end{align*}$

and similarly,

$\begin{align*} \sum_{n=1}^\infty\frac{(-1)^n \mathcal{H}_n}{n^2} &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(-1)^n}{n^2}\left(\frac1k-\frac1{k+n}\right)\\ &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(-1)^n}{nk(k+n)} \\ \sum_{n=1}^\infty\frac{(-1)^n \mathcal{H}_n}{n^2}&=-\frac34\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^n \mathcal{H}_{n-1}}{n^2}\\ &=-\frac34\zeta(3)+\frac12\sum_{n=1}^\infty\sum_{k=1}^{n-1}\frac{(-1)^n}{n^2}\left(\frac1k+\frac1{n-k}\right)\\ &=-\frac34\zeta(3)+\frac12\sum_{k=1}^\infty\sum_{n=k+1}^\infty\frac{(-1)^n}{nk(n-k)}\\ &=-\frac34\zeta(3)+\frac12\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^{n+k}}{(n+k)kn} \end{align*}$

Hence,

$\begin{align*} \zeta(3) &= \frac{1}{2} \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{nk(n+k)}\\ &= \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1+(-1)^k + (-1)^n + (-1)^{n+k}}{nk(n+k)}\\ &= 2\zeta(3) + \sum_{n=1}^{\infty} \frac{(-1)^n \mathcal{H}_n}{n^2} + 2 \sum_{n=1}^{\infty}\frac{(-1)^n \mathcal{H}_n}{n^2} + \frac{3\zeta(3)}{2}\\ \\ \sum_{n=1}^{\infty} (-1)^n \frac{\mathcal{H}_n}{n^2}&= - \frac{5 \zeta(3)}{8} \end{align*}$

since $1+(-1)^k+(-1)^n+(-1)^{n+k}=4$ if-f both $k$ and $n$ are even and $0$ otherwise.

4. On the other hand we can get the generating function of the previous series. It is known that

$\begin{align*} \sum_{n=1}^{\infty} \mathcal{H}_n^{(q)} x^n = \frac{\mathrm{Li}_q(x)}{1-x} &\overset{q=1}{\implies} \sum_{n=1}^{\infty} \mathcal{H}_n x^n = -\frac{\ln (1-x)}{1-x} \\ &\Rightarrow \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n} x^n = \mathrm{Li}_2(x) + \frac{\log^2 \left ( 1-x \right )}{2} \end{align*}$

Integrating once more we get the generating function!

2 thoughts on “An arccotangent integral”

Tolaso says:

June 8, 2019 at 8:57 am

Let’s expand on some things here.

1. Let $\mathcal{H}_n$ denote the $n$ -th harmonic number. It holds that:

$\frac{x \arctan x}{x^2+1} = \sum_{n=1}^{\infty} (-1)^{n-1} \left ( \mathcal{H}_{2n} - \frac{\mathcal{H}_n}{2} \right ) x^{2n}$

( the proof is left to the reader )

2. Following $(1)$ it is easy to see that:

$\int_{0}^{1} \frac{\ln x \arctan x}{1+x^2} \, \mathrm{d}x = \frac{1}{4} \sum_{n=1}^{\infty} (-1)^n \cdot \frac{\mathcal{H}_{2n} -\frac{\mathcal{H}_n}{2}}{n^2}$

The second series is an old chestnut.

3. It holds that:

$\sum_{n=1}^{\infty} (-1)^n \frac{\mathcal{H}_n}{n^2} = -\frac{5 \zeta(3)}{8}$

Proof: Successively we have that:

$\begin{align*} \sum_{n=1}^\infty\frac{\mathcal{H}_n}{n^2} &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{n^2}\left(\frac1k-\frac1{k+n}\right)\\ &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{nk(k+n)} \\ &=\sum_{k=1}^\infty\sum_{n=k+1}^\infty\frac1{nk(n-k)}\\ &=\sum_{n=2}^\infty\sum_{k=1}^{n-1}\frac1{nk(n-k)}\\ &=\sum_{n=2}^\infty\sum_{k=1}^{n-1}\frac1{n^2}\left(\frac1k+\frac1{n-k}\right)\\ &=2\sum_{n=1}^\infty\frac{\mathcal{H}_{n-1}}{n^2}\\ &=2\sum_{n=1}^\infty\frac{\mathcal{H}_n}{n^2}-2\zeta(3) \\ \sum_{n=1}^\infty\frac{\mathcal{H}_n}{n^2} &=2\zeta(3) \end{align*}$

and similarly,

$\begin{align*} \sum_{n=1}^\infty\frac{(-1)^n \mathcal{H}_n}{n^2} &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(-1)^n}{n^2}\left(\frac1k-\frac1{k+n}\right)\\ &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(-1)^n}{nk(k+n)} \\ \sum_{n=1}^\infty\frac{(-1)^n \mathcal{H}_n}{n^2}&=-\frac34\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^n \mathcal{H}_{n-1}}{n^2}\\ &=-\frac34\zeta(3)+\frac12\sum_{n=1}^\infty\sum_{k=1}^{n-1}\frac{(-1)^n}{n^2}\left(\frac1k+\frac1{n-k}\right)\\ &=-\frac34\zeta(3)+\frac12\sum_{k=1}^\infty\sum_{n=k+1}^\infty\frac{(-1)^n}{nk(n-k)}\\ &=-\frac34\zeta(3)+\frac12\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^{n+k}}{(n+k)kn} \end{align*}$

Hence,

$\begin{align*} \zeta(3) &= \frac{1}{2} \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{nk(n+k)}\\ &= \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1+(-1)^k + (-1)^n + (-1)^{n+k}}{nk(n+k)}\\ &= 2\zeta(3) + \sum_{n=1}^{\infty} \frac{(-1)^n \mathcal{H}_n}{n^2} + 2 \sum_{n=1}^{\infty}\frac{(-1)^n \mathcal{H}_n}{n^2} + \frac{3\zeta(3)}{2}\\ \\ \sum_{n=1}^{\infty} (-1)^n \frac{\mathcal{H}_n}{n^2}&= - \frac{5 \zeta(3)}{8} \end{align*}$

since $1+(-1)^k+(-1)^n+(-1)^{n+k}=4$ if-f both $k$ and $n$ are even and $0$ otherwise.

4. On the other hand we can get the generating function of the previous series. It is known that

$\begin{align*} \sum_{n=1}^{\infty} \mathcal{H}_n^{(q)} x^n = \frac{\mathrm{Li}_q(x)}{1-x} &\overset{q=1}{\implies} \sum_{n=1}^{\infty} \mathcal{H}_n x^n = -\frac{\ln (1-x)}{1-x} \\ &\Rightarrow \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n} x^n = \mathrm{Li}_2(x) + \frac{\log^2 \left ( 1-x \right )}{2} \end{align*}$

Integrating once more we get the generating function!

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Tolaso says:

August 2, 2019 at 11:20 pm

Another way to prove that $\displaystyle \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n^2} = 2\zeta(3)$ .

Proof: Successively we have:

$\begin{align*} \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n^2} &= \sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{m=1}^{\infty} \left ( \frac{1}{m} - \frac{1}{m+n} \right ) \\ &=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{mn\left ( m+n \right )} \\ &=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{mn} \int_{0}^{1} x^{m+n-1} \, \mathrm{d}x \\ &=\int_{0}^{1} \frac{1}{x} \sum_{n=1}^{\infty} \frac{x^n}{n} \sum_{m=1}^{\infty} \frac{x^m}{m} \,\mathrm{d}x \\ &= \int_{0}^{1} \frac{1}{x} \left ( \sum_{n=1}^{\infty} \frac{x^n}{n} \right )^2 \, \mathrm{d}x \\ &=\int_{0}^{1} \frac{\ln^2(1-x)}{x} \, \mathrm{d}x\\ &\!\!\!\!\overset{u=1-x}{=\! =\! =\! =\!} \int_{0}^{1} \frac{\ln^2 x}{1-x} \, \mathrm{d}x \\ &=2\zeta(3) \end{align*}$

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