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Inverse tangent integral

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Post Mon Jul 22, 2013 9:24 pm
zaidalyafey Global Moderator
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Lewin in his book gives the following definition

\(\displaystyle \text{Ti}_2(x) = \int^x_0 \frac{\arctan(t)}{t}\, dt\)

I only have the value

\(\displaystyle \text{Ti}_2(1) = G\)

where \(\displaystyle G\) represents the Catalan's constant .

Do we have other known values ?
Wanna learn what we discuss , see Book

Post Tue Jul 23, 2013 9:39 am
Shobhit Site Admin
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Ramanujan gave another special value:

\(\displaystyle \text{Ti}_2 \left( 2-\sqrt{3}\right)= \frac{\pi}{12}\log\left( 2-\sqrt{3}\right)+\frac{2}{3}G\)

Post Tue Jul 23, 2013 10:24 am
galactus User avatar
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Shobhit wrote:
Ramanujan gave another special value:

\(\displaystyle \text{Ti}_2 \left( 2-\sqrt{3}\right)= \frac{\pi}{12}\log\left( 2-\sqrt{3}\right)+\frac{2}{3}G\)


We have done this one in the contest at some point.

\(\displaystyle \int_{0}^{2-\sqrt{3}}\frac{\tan^{-1}(x)}{x}dx\)

An interesting, but apparently little known, identity that can be used is

\(\displaystyle 2\int_{0}^{\frac{\pi}{12}}log(\tan(3x))dx=\int_{0}^{\frac{\pi}{12}}log(\tan(x))dx\)

Isn't there some sort of general closed form for this arctan integral in Lewin's book?.

Post Tue Jul 23, 2013 1:08 pm
Shobhit Site Admin
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Using integration by parts and Fourier series, we have

\(\displaystyle \begin{align*}
\int_0^{2-\sqrt{3}}\frac{\arctan x}{x}dx &= \frac{\pi}{12}\log \left( 2-\sqrt{3}\right)- \int_0^{2-\sqrt{3}}\frac{\log x}{1+x^2}dx \\
&= \frac{\pi}{12}\log \left( 2-\sqrt{3}\right)- \int_0^{\frac{\pi}{12}}\log \left(\tan \theta \right) d\theta \quad x=\tan(\theta) \\
&= \frac{\pi}{12}\log \left( 2-\sqrt{3}\right) +2 \int_0^{\frac{\pi}{12}} \sum_{n \text{ odd}}^\infty \frac{\cos(2 n\theta)}{n} \ d\theta \\
&= \frac{\pi}{12}\log \left( 2-\sqrt{3}\right) + \sum_{n \text{ odd}}^\infty \frac{\sin \left(\frac{\pi n}{6} \right)}{n^2} \\
&= \frac{\pi}{12}\log \left(2-\sqrt{3} \right) + \frac{1}{2} \left( \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} +\sum_{n=0}^\infty \frac{(-1)^n}{(6n+3)^2}\right)+\frac{G}{9} \\
&= \frac{\pi}{12}\log \left(2-\sqrt{3} \right) +\frac{2}{3}G
\end{align*}\)

This proves that \(\displaystyle \text{Ti}_2 \left( 2-\sqrt{3}\right) = \frac{\pi}{12}\log \left(2-\sqrt{3} \right) +\frac{2}{3}G \tag{1}\)

Also, it can be verified by differentiation that

\(\displaystyle \text{Ti}_2 (y)-\text{Ti}_2 \left(\frac{1}{y} \right)=\frac{\pi}{2}\log y \quad y>0 \tag{2}\)

Putting \(y=2-\sqrt{3}\) and using (1) in equation (2), we obtain

\(\displaystyle \text{Ti}_2 \left( 2+\sqrt{3}\right) = \frac{5\pi }{12} \log \left( 2+\sqrt{3}\right)+\frac{2}{3}G \tag{3}\)

Post Sat Sep 07, 2013 1:28 pm

Posts: 138
Location: North Londinium, UK
galactus wrote:

Isn't there some sort of general closed form for this arctan integral in Lewin's book?.


You're undoubtedly right there, Galactus, although I've not yet bought said book (it's on my very, very long list). Regardless, the Inverse Tangent Integral is variously expressible - in closed form - in terms of a number of different transcendental functions. The easiest and most straightforward to deduce is in terms of the Clausen function, namely:

\(\displaystyle \text{Ti}_2(\tan \phi)=\phi\log(\tan\phi)+\tfrac{1}{2}\text{Cl}_2(2\phi)+\tfrac{1}{2}\text{Cl}_2(\pi-2\phi)\)

where

\(\displaystyle \text{Cl}_2(\theta)=-\int_0^{\theta}\log\bigr| 2\sin\tfrac{x}{2}\bigr| \,dx=\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^2}\)

is the Clausen function (of second order). Starting from the standard definition of the Inverse Tangent Integral, perform an integration by parts

\(\displaystyle \text{Ti}_2(z)=\int_0^z\frac{\tan^{-1}t}{t}\,dt=\tan^{-1}t\log t\, \biggr|_0^z-\int_0^z\frac{\log t}{(1+t^2)}\,dt =\)

\(\displaystyle \phi\log(\tan\phi) - \int_0^{\phi=\tan^{-1}z}\log(\tan x)\,dx\)

Next, we split the logtan integral as follows:

\(\displaystyle \int_0^{\phi}\log(\tan x)\,dx=\int_0^{\phi}\log(\sin x)\,dx-\int_0^{\phi}\log(\cos x)\,dx=\)

\(\displaystyle \int_0^{\phi}\log(2\sin x)\,dx-\int_0^{\phi}\log(2\cos x)\,dx\)

On the logsine integral we apply x --> y/2, and on the logcosine integral we apply x --> pi/2 - y/2, to obtain:

\(\displaystyle \tfrac{1}{2}\int_0^{2\phi}\log(2\sin\tfrac{y}{2})\,dy+\tfrac{1}{2}\int_{\pi}^{\pi-2\phi}\log[2\cos(\tfrac{\pi}{2}-\tfrac{y}{2})]\,dy=\)

\(\displaystyle \tfrac{1}{2}\int_0^{2\phi}\log(2\sin\tfrac{y}{2})\,dy+\tfrac{1}{2}\int_{\pi}^{\pi-2\phi}\log(2\sin\tfrac{y}{2})\,dy=\)

\(\displaystyle -\tfrac{1}{2}\text{Cl}_2(2\phi)+\tfrac{1}{2}\text{Cl}_2(0)-\tfrac{1}{2}\text{Cl}_2(\pi-2\phi)+\tfrac{1}{2}\text{Cl}_2(\pi)=\)

\(\displaystyle -\tfrac{1}{2}\text{Cl}_2(2\phi)-\tfrac{1}{2}\text{Cl}_2(\pi-2\phi)\)

Since the (Fourier-type) series definition of \(\displaystyle \text{Cl}_2(\theta)\) makes it clear that

\(\displaystyle \text{Cl}_2(\pi)=\text{Cl}_2(0)=0\)

Hence,

\(\displaystyle \text{Ti}_2(\tan \phi)=\phi\log(\tan\phi)+\tfrac{1}{2}\text{Cl}_2(2\phi)+\tfrac{1}{2}\text{Cl}_2(\pi-2\phi)\)

Post Sat Sep 07, 2013 2:58 pm

Posts: 138
Location: North Londinium, UK
From the closed form above, with \(\displaystyle \phi=\tfrac{p\pi}{q}\) and \(\displaystyle p, \,q \in \mathbb{N}\), it is possible to obtain a few closed form expressions for \(\displaystyle \text{Ti}_2(\tan\tfrac{p\pi}{q})\) that invlove a single Clausen function, rather than a combination of two Clausen functions with differing arguments. In part, this is due to the Clausen function duplication formula:

\(\displaystyle \text{Cl}_2(2\theta)=2\text{Cl}_2(\theta)-2\text{Cl}_2(\pi-\theta)\)

To prove this, start from

\(\displaystyle \text{I}(\theta) = \int_0^{\theta}\log(\sin x)\,dx\)

Using the duplication formula for the sine function, we get

\(\displaystyle \text{I}(\theta) = \int_0^{\theta}\log(2\sin\tfrac{x}{2}\cos\tfrac{x}{2})\,dx=\)

\(\displaystyle \int_0^{\theta}\log(2\sin\tfrac{x}{2})\,dx+\int_0^{\theta}\log(2\cos\tfrac{x}{2})\,dx-\log 2\int_0^{\theta}\,dx=\)

\(\displaystyle \int_0^{\theta}\log(2\sin\tfrac{x}{2})\,dx+\int_0^{\theta}\log(2\cos\tfrac{x}{2})\,dx-\theta\log 2\)

Next, we perform the substitution \(\displaystyle x \to \pi-y\) on the second integral to get

\(\displaystyle \text{I}(\theta) = \int_0^{\theta}\log(2\sin\tfrac{x}{2})\,dx-\int_{\pi}^{\pi-\theta}\log(2\sin\tfrac{x}{2})\,dx-\theta\log 2\)

And thus, from the previous (integral) definition of the Clausen function, we get

\(\displaystyle \text{I}(\theta) = -\text{Cl}_2(\theta)+\text{Cl}_2(\pi-\theta)-\theta\log 2\)

Alternatively, we ca re-write \(\displaystyle I(\theta)\) as follows:

\(\displaystyle I(\theta)= \int_0^{\theta}\log(\sin x)\,dx= \int_0^{\theta}\log(2\sin x)\,dx -\log 2\int_0^{\theta}\,dx=\)

\(\displaystyle \int_0^{\theta}\log(2\sin x)\,dx -\theta\log 2\)

And then perform the substitution \(\displaystyle x \to y/2\) to get

\(\displaystyle I(\theta) = \tfrac{1}{2}\int_0^{2\theta}\log(2\sin\tfrac{y}{2})\,dy-\theta\log 2=\)

\(\displaystyle -\tfrac{1}{2}\text{Cl}_2(2\theta)-\theta\log 2\)

Combining the two, we therefore arrive at the duplication formula

\(\displaystyle \text{Cl}_2(2\theta)=2\text{Cl}_2(\theta)-2\text{Cl}_2(\pi-\theta)\)





With the Clausen function duplication formula proven, we then return to the closed form

\(\displaystyle \text{Ti}_2(\tan\phi)+\tfrac{1}{2}\text{Cl}_2(2\phi)+\tfrac{1}{2}\text{Cl}_2(\pi-2\phi)\)


Setting \(\displaystyle \phi=\tfrac{\pi}{3}\) in the above, we get

\(\displaystyle \text{Ti}_2(\sqrt{3})=\tfrac{\pi}{3}\log\sqrt{3}+\tfrac{1}{2}\left[\text{Cl}_2(\tfrac{2\pi}{3})+\text{Cl}_2(\tfrac{\pi}{3}) \right]\)

The Clausen function duplication formula yields

\(\displaystyle \text{Cl}_2(\tfrac{2\pi}{3})=\tfrac{2}{3}\text{Cl}_2(\tfrac{\pi}{3})\)

The upshot of which is

\(\displaystyle \text{Ti}_2(\sqrt{3})=\tfrac{\pi}{6}\log 3+\tfrac{5}{6}\text{Cl}_2(\tfrac{\pi}{3})\)

In a similar manner, setting \(\displaystyle \phi=\tfrac{\pi}{4}\) yields

\(\displaystyle \text{Cl}_2(\tfrac{\pi}{2})=G=2\text{Cl}_2(\tfrac{\pi}{4})-2\text{Cl}_2(\tfrac{3\pi}{4})\)

and thus

\(\displaystyle \text{Cl}_2(\tfrac{3\pi}{4})=-\tfrac{G}{2}+\text{Cl}_2(\tfrac{\pi}{4})\)


Therefore, if we set \(\displaystyle \phi=\tfrac{\pi}{8}\) in

\(\displaystyle \text{Ti}_2(\tan\phi)+\tfrac{1}{2}\text{Cl}_2(2\phi)+\tfrac{1}{2}\text{Cl}_2(\pi-2\phi)\)

we get

\(\displaystyle \text{Ti}_2(\tan\tfrac{\pi}{8})=\text{Ti}_2(\sqrt{2}-1)=\)

\(\displaystyle \tfrac{\pi}{8}\log(\sqrt{2}-1)+\tfrac{1}{2}\left[\text{Cl}_2(\tfrac{\pi}{4})+\text{Cl}_2(\tfrac{3\pi}{4})\right]\)

Or

\(\displaystyle \text{Ti}_2(\sqrt{2}-1)=\tfrac{\pi}{8}\log(\sqrt{2}-1)+\text{Cl}_2(\tfrac{\pi}{4})-\tfrac{G}{4}\)

Post Sat Sep 07, 2013 3:15 pm

Posts: 138
Location: North Londinium, UK
Further results are easily obtained via the reciprocation formula, as previously posted by Shobhit:


Shobhit wrote:

Also, it can be verified by differentiation that

\(\displaystyle \text{Ti}_2 (y)-\text{Ti}_2 \left(\frac{1}{y} \right)=\frac{\pi}{2}\log y \quad y>0 \tag{2}\)




Taking the previous two examples


\(\displaystyle \text{Ti}_2(\sqrt{3})=\tfrac{\pi}{6}\log 3+\tfrac{5}{6}\text{Cl}_2(\tfrac{\pi}{3})\)

\(\displaystyle \text{Ti}_2(\sqrt{2}-1)=\tfrac{\pi}{8}\log(\sqrt{2}-1)+\text{Cl}_2(\tfrac{\pi}{4})-\tfrac{G}{4}\)


One can then obtain

\(\displaystyle \text{Ti}_2(\tfrac{1}{\sqrt{3}})=-\tfrac{\pi}{12}\log 3+\tfrac{5}{6}\text{Cl}_2(\tfrac{\pi}{3})\)

\(\displaystyle \text{Ti}_2(\sqrt{2} +1)=\tfrac{3\pi}{8}\log(\sqrt{2}+1)+\text{Cl}_2(\tfrac{\pi}{4})-\tfrac{G}{4}\)

Post Sat Sep 07, 2013 4:07 pm
Random Variable Integration Guru
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Posts: 381
Very nice, DreamWeaver. Some of that is definitely going in my notebook.

And not that it matters, but those \(\displaystyle Cl_{2}\) values should be expressible in terms of the trigamma function.

Post Sat Sep 07, 2013 6:54 pm

Posts: 138
Location: North Londinium, UK
Thanks RV... :D (Nice forum you guys and gals have here, btw :mrgreen: )

And you're quite right about the interchangeability - for want of a better word - between values of the Clausen function and polygamma function.

Incidentally, if you fancy another transcendental approach, the Inverse Tangent Integral is also expressible in terms of the Barnes' G-Function (reciprocal of the double gamma function - not to be confused with the digamma function).

Define multiple gamma function as follows:

\(\displaystyle \Gamma_n(1) = 1\)

\(\displaystyle \Gamma_1(z) = \Gamma(z)\)

\(\displaystyle \Gamma_{n+1}(z+1)=\frac{\Gamma_{n+1}(z)}{\Gamma_n(z)}\); for \(\displaystyle n \in \mathbb{Z}\) and \(\displaystyle z \in \mathbb{C}\)

If the condition of convexity is added, then

\(\displaystyle (-1)^{n+1}\frac{d^{n+1}}{dz^{n+1}}\log\Gamma_n(z) \ge 0,\,\,\, n > 0\)

has a unique solution.

Then the Barnes' G-Function - \(\displaystyle \text{G}(z)\) - satisfies

\(\displaystyle \text{G}(1)=\text{G}(2)=\text{G}(3)=1\)

\(\displaystyle \text{G}(1+z)=\Gamma(z)\text{G}(z)\)

\(\displaystyle \frac{d^3}{dz^3}\text{G}(z)\ge 0,\,\,\, z>0\)

Furthermore, the Barnes' G-function can be expressed in terms of the Clausen function - and vice versa - via the identity

\(\displaystyle \log\left(\frac{\text{G}(1+z)}{\text{G}(1-z)}\right)=-z\log\left(\frac{\sin \pi z}{\pi}\right)-\frac{1}{2\pi}\text{Cl}_2(2\pi z)\)

for 0 < z < 1. And so, within certain limits

\(\displaystyle \text{Ti}_2(\tan\tfrac{p\pi}{q})=\pi\log\left[\frac{\text{G}(1-\tfrac{p}{q})\text{G}(\tfrac{1}{2}-\tfrac{p}{q})}{\text{G}(1+\tfrac{p}{q})\text{G}(\tfrac{3}{2}+\tfrac{p}{q})}\right]-\frac{\pi}{2}\log\left(\cos\frac{p\pi}{q}\right)+\frac{\pi(2p+q)}{2q}\log\pi\)


I've not seen this result in any books or papers, but it's easy enough to deduce, so I can't imagine it's new.... ;)


Incidentally, going back to what you said about the polygamma function, it's worth noting the following functional relation for the the Barne's G-function (for Re(z) > 0):

\(\displaystyle \log\text{G}(z+1)-z\log\Gamma(z) = \zeta'(-1)-\zeta(-1, z)\)

In a very loose sense, this identity, in conjunction with previous results above, makes the connection between the Inverse Tangent Integral and derivatives of the Hurwitz Zeta function, Barnes G-function, Clause function, loggamma function - and much more besides - abundantly clear (it's an easy step to wander into the zone of Dirichlet Beta functions, Zeta functions, Polylogarithms, Lerch Transcendents, et al).


If you fancy a good read about the Barne's G-function, I'd highly recommend "Contributions to the theory of the Barnes' Function", by Victor S. Adamchik. It's widely available on-line in PDF form (http://repository.cmu.edu/cgi/viewconte ... xt=compsci)

And if that link doesn't work, just Google "Contributions to the theory of the Barnes' Function"... 8-)

Gethin


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