Inverse Sine/Tangent Integrals and related functions

[DreamWeaver]

Within certain branches of analysis - both real and complex - the Inverse Tangent Integral (and its generalizations) can be quite useful. Similarly, it's much less well-known (= uglier? lol) cousin, the Inverse Sine Integral can be used to solve many problems.

To that end, this is not really a tutorial, but rather a random and somewhat arbitrary collection of related results that I've evaluated over the years, and found to be quite useful. Who know's, eh? With a bit of gentle coaxing I might - just might! - manage to convert a few transcendental bipeds into true believers... [Sorry, I just love these functions. Begging your pardon... (Rofl) ]Let's start with a closed form for the Inverse Sine Integral in terms of Clausen Functions. For convenience, let's set \(\displaystyle z=\sin\varphi\,\), and I'll use the two equivalent forms as and when most appropriate. And so, starting from the integral definition of the Inverse Sine Integral

\(\displaystyle \text{Si}_2(z)=\text{Si}_2(\sin\varphi)=\int_0^z \frac{ \sin^{-1}x}{x}\,dx=\int_0^{\sin\varphi}\frac{\sin^{-1}x}{x}\,dx\)

integrate that final form by parts to obtain

\(\displaystyle \text{Si}_2(\sin\varphi)=\log x \sin^{-1}x\Biggr|_0^{\sin\varphi}-\int_0^{\sin\varphi}\frac{\log x}{\sqrt{1-x^2}}\,dx=\)

\(\displaystyle \varphi\log(\sin \varphi)-\int_0^{\sin\varphi}\frac{\log x}{\sqrt{1-x^2}}\,dx\)

Next, apply the trigonometric substitution \(\displaystyle x=\sin y\,\) on that last integral to get the equivalent form

\(\displaystyle \varphi\log(\sin \varphi)-\int_0^{\varphi}\log(\sin y)\,dy=\)

\(\displaystyle \varphi\log(\sin \varphi)-\int_0^{\varphi}\log(2\sin y)\,dy+\log 2\int_0^{\varphi}\,dy=\)

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

Finally, set \(\displaystyle x=y/2\,\) in that last logsine integral

\(\displaystyle \Rightarrow\)

\(\displaystyle \varphi\log(2\sin \varphi)-\frac{1}{2}\int_0^{2\varphi}\log\left(2\sin \frac{y}{2}\right)\,dx\)-----------------------------

Interlude:

The Clausen Function of order 2 has the integral / series representation

\(\displaystyle \text{Cl}_2(\theta)=-\int_0^{\theta}\log\Biggr|2\sin\frac{x}{2}\Biggr| \,dx=\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^2}\)Within the range \(\displaystyle 0 < \varphi < 2\pi\,\), we can drop the absolute value sign in the integral and simply write\(\displaystyle \text{Cl}_2(\theta)=-\int_0^{\theta}\log\left(2\sin\frac{x}{2}\right)\,dx=\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^2}\)

-----------------------------Returning to the problem of finding a closed form for the Inverse Sine Integral, we can now re-write\(\displaystyle \varphi\log(2\sin \varphi)-\frac{1}{2}\int_0^{2\varphi}\log\left(2\sin \frac{y}{2}\right)\,dx\)

as\(\displaystyle \varphi\log(2\sin \varphi)+\frac{1}{2}\text{Cl}_2(2\varphi)\)And so, for \(\displaystyle 0 < \varphi < \pi\,\) we have the closed form\(\displaystyle \text{Si}_2(\sin\varphi)=\varphi\log(2\sin \varphi)+\frac{1}{2}\text{Cl}_2(2\varphi)\)

Comments and questions should be posted here:

http://mathhelpboards.com/showthread.php?t=6485

 

[MarkFL]

This topic is for comments/questions pertaining to this topic:

http://mathhelpboards.com/math-notes-49/inverse-sine-tangent-integrals-related-functions-6483.html

 

[alyafey22]

Hey Gethin , try proving

\(\displaystyle \int^1_0 \frac{1-a^2 t^2}{(1+a^2t^2)^2}\,\log^2(t) \, dt \,=\frac{2\text{Ti}_2(a)}{a}\)

 

[DreamWeaver]

Incidentally, although in and of itself the following doesn't constitute a proof, we can at least check for consistency...

Consider the infamous log-trig integrals attributed to Euler:

\(\displaystyle \int_0^{\pi/2}\log(\sin x)\,dx=-\frac{\pi}{2}\log 2\)

\(\displaystyle \int_0^{\pi/2}\log(\cos x)\,dx=-\frac{\pi}{2}\log 2\)Set \(\displaystyle \varphi=\pi/2\,\) in the previous result:

\(\displaystyle \text{Si}_2(\sin\varphi)=\varphi\log(2\sin \varphi)+\frac{1}{2}\text{Cl}_2(2\varphi)\)

and you should get

\(\displaystyle \text{Si}_2\left(\sin\frac{\pi}{2}\right)=\text{Si}_2(1)=\frac{\pi}{2}\log 2+\frac{1}{2}\text{Cl}_2(\pi)\)Then set \(\displaystyle \varphi=\pi\,\) in the series definition for the Clausen function:\(\displaystyle \text{Cl}_2(\pi)=\sum_{k=1}^{\infty}\frac{\sin k\pi}{k^2}\equiv 0\)If our closed form for the Inverse Sine Integral is correct, then at the very least we would expect it to agree with Euler's classic log-trig integrals.

What we have so far is

\(\displaystyle \text{Si}_2(1)=\frac{\pi}{2}\log 2\)But we can also integrate by parts - exactly as before - from the integral definition of the Inverse Sine Integral, to obtain

\(\displaystyle \text{Si}_2(1)=\int_0^1\frac{\sin^{-1}x}{x}\,dx=\log x\sin^{-1}x\Biggr|_0^1-\int_0^1\frac{\log x}{\sqrt{1-x^2}}\,dx=\)

\(\displaystyle -\int_0^1\frac{\log x}{\sqrt{1-x^2}}\,dx=\)

\(\displaystyle -\int_0^{\pi/2}\log(\sin x)\,dx=-\int_0^{\pi/2}\log(\cos x)\,dx=-\left(\frac{\pi}{2}\log 2\right)=\frac{\pi}{2}\log 2\)Done.

- - - Updated - - -EDIT:

Sorry Mark... I asked Jameson about where to post threads like this, but then couldn't find the "notes" board he mentioned. Should really have bugged him again to make sure I posted in the right place...

 

[DreamWeaver]

A closed form for the Inverse Tangent function will follow shortly, but before we get to that, let's find a relation that connects the Inverse Sine Integral to the Inverse Tangent integral.I'm going to assume a basic knowledge of hypergeometric functions here - on the part of the reader - but if uncertain, or you'd like more info/explanation, please do feel free to post questions either on here or direct to me via PM. It'd be no bother whatsoever... :D :D :D

Also, the immediate result below is just a confirmation of the trig identity

\(\displaystyle \tan^{-1}x=\sin\frac{x}{\sqrt{1+x^2}}\)

So if you aren't all that familiar with hypergeometric function, but feel comfortable assuming said trig identity, then no need to worry...The inverse tangent resp. sine functions have the hypergeometric series representations:\(\displaystyle \sin^{-1}x= x\,_2F_1\left(\frac{1}{2}\, , \frac{1}{2}\, ; \frac{3}{2}\, ; x^2\right)=\sum_{k=0}^{\infty}\frac{(1/2)_k(1/2)_k}{(3/2)_k\, k!}x^{2k+1}\)\(\displaystyle \tan^{-1}x= x\,_2F_1\left(\frac{1}{2}\, , 1\, ; \frac{3}{2}\, ; -x^2\right)=\sum_{k=0}^{\infty}(-1)^k\frac{(1/2)_k(1)_k}{(3/2)_k\, k!}x^{2k+1}\)We can apply Pfaff's hypergeometric transform here:

\(\displaystyle _2F_1(a\,, b\,; c\,; x)=(1-x)^{-a}{}_2F_1\left(a\,, c-b\,; c\,; \frac{x}{x-1}\right)\)Pfaff's transform is valid for \(\displaystyle Re(c) > Re(b) > 0\,\). It can be extended via the principle of analytic continuation, but no such exotic concepts are required here.

Apply Pfaff's transform to the hypergeometric series definition for \(\displaystyle \tan^{-1}x\,\) to obtain\(\displaystyle \tan^{-1}x= x\,_2F_1\left(\frac{1}{2}\, , 1\, ; \frac{3}{2}\, ; -x^2\right)=\)

\(\displaystyle \frac{x}{\sqrt{1+x^2}}\,{}_2F_1\left( \frac{1}{2}\, , \frac{1}{2}\, ; \frac{3}{2}\, ; \frac{x^2}{1+x^2}\right)=\sin^{-1}\frac{x}{\sqrt{1+x^2}}\)Done.----------------------------Now, returning to the problem of finding a relation between the Inverse Sine Integral and the Inverse Tangent Integral, we can use\(\displaystyle \tan^{-1}x=\sin\frac{x}{\sqrt{1+x^2}}\)to re-write \(\displaystyle \text{Ti}_2(z)\,\) as follows:\(\displaystyle \text{Ti}_2(z)=\int_0^z\frac{\tan^{-1}x}{x}\,dx=\int_0^z\frac{\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)}{x}\,dx\)In that last integral, apply the substitution \(\displaystyle y=\frac{x}{\sqrt{1+x^2}}\)

\(\displaystyle \Rightarrow\)

\(\displaystyle x=\frac{y}{\sqrt{1-y^2}}\,\quad\, dx=\frac{dy}{\sqrt{1-y^2}}+\frac{y^2\,dy}{(1-y^2)^{3/2}}\)And so\(\displaystyle \text{Ti}_2(z)=\int_0^z\frac{\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)}{x}\,dx=\)\(\displaystyle \int_0^{z/\,\sqrt{1+z^2}}\frac{\sin^{-1}y}{\left(\frac{y}{\sqrt{1-y^2}}\right)}\frac{dy}{\sqrt{1-y^2}}-\int_0^{z/\,\sqrt{1+z^2}}\frac{\sin^{-1}y}{\left(\frac{y}{\sqrt{1-y^2}}\right)}\frac{y^2\,dy}{\sqrt{1-y^2}(1-y^2)}=\)\(\displaystyle \int_0^{z/\,\sqrt{1+z^2}}\frac{\sin^{-1}y}{y}-\int_0^{z/\,\sqrt{1+z^2}}\frac{y\,\sin^{-1}y}{(1-y^2)}\,dy=\)\(\displaystyle \text{Si}_2\left(\frac{z}{\sqrt{1+z^2}}\right)-\int_0^{z/\,\sqrt{1+z^2}}\frac{y\,\sin^{-1}y}{(1-y^2)}\,dy\)We're nearly there! (Heidy)----------------------------The final step is to evaluate that last integral, namely\(\displaystyle \int_0^{z/\,\sqrt{1+z^2}}\frac{y\,\sin^{-1}y}{(1-y^2)}\,dy\)Firstly, note that the derivative of \(\displaystyle \log(1-y^2)\,\) is

\(\displaystyle \frac{d}{dy}\, \log(1-y^2) =-\frac{2y}{(1-y^2)}\)So we can perform an integration by parts\(\displaystyle \int_0^{z/\,\sqrt{1+z^2}}\frac{y\,\sin^{-1}y}{(1-y^2)}\,dy=\)\(\displaystyle -\frac{1}{2}\sin^{-1}y\log(1-y^2)\, \Biggr|_0^{z/\,\sqrt{1+z^2}}+\frac{1}{2}\int_0^{z/\,\sqrt{1+z^2}}\frac{\log(1-y^2)}{\sqrt{1-y^2}}\,dy=
\)\(\displaystyle \frac{1}{2}\sin^{-1}\left(\frac{z}{\sqrt{1+z^2}}\right) \log \left(\frac{1}{1+z^2}\right) +\frac{1}{2}\int_0^{z/\,\sqrt{1+z^2}}\frac{\log(1-y^2)}{\sqrt{1-y^2}}\,dy\equiv\)\(\displaystyle -\frac{1}{2}\tan^{-1}z \log(1+z^2)+\frac{1}{2}\int_0^{z/\,\sqrt{1+z^2}}\frac{\log(1-y^2)}{\sqrt{1-y^2}}\,dy\)So, just to sum up, thus far we have...\(\displaystyle \text{Ti}_2(z)=\text{Si}_2\left(\frac{z}{\sqrt{1+z^2}}\right)-\int_0^{z/\,\sqrt{1+z^2}}\frac{y\,\sin^{-1}y}{(1-y^2)}\,dy=\)\(\displaystyle \text{Si}_2\left(\frac{z}{\sqrt{1+z^2}}\right)
+\frac{1}{2}\tan^{-1}z \log(1+z^2)-\frac{1}{2}\int_0^{z/\,\sqrt{1+z^2}}\frac{\log(1-y^2)}{\sqrt{1-y^2}}\,dy\)Next, let's a aplly the simple trigonometric substitution \(\displaystyle x=\sin y\, \). This makes that last integral\(\displaystyle \int_0^{z/\,\sqrt{1+z^2}}\frac{\log(1-y^2)}{\sqrt{1-y^2}}\,dy=\int_0^{\sin^{-1}({z/\,\sqrt{1+z^2}})}\log(1-\sin^2x)\,dx\equiv\)\(\displaystyle \int_0^{\tan^{-1}z}\log(1-\sin^2 x)\,dx=\int_0^{\tan^{-1}z}\log(\cos^2 x)\,dx=\)\(\displaystyle 2\int_0^{\tan^{-1}z}\log(\cos x)\,dx= 2\int_0^{\tan^{-1}z}\log(2\cos x)\,dx-2\log 2\int_0^{\tan^{-1}z}\,dx=\)\(\displaystyle 2\int_0^{\tan^{-1}z}\log(2\cos x)\,dx-2\log 2\, \tan^{-1}z\)Once final substitution, and then we're done (I promise!). Let \(\displaystyle x=\pi/2 - y/2\,\) to obtain\(\displaystyle \int_0^{\tan^{-1}z}\log(2\cos x)\,dx=\)\(\displaystyle -\int_{\pi}^{\pi-2\tan^{-1}z}\log\left[2\cos \left(\frac{\pi}{2}-\frac{y}{2}\right)\right]\,dy=\)\(\displaystyle -\int_{\pi}^{\pi-2\tan^{-1}z}\log\left[2\left(\cos\frac{\pi}{2}\cos\frac{y}{2}+\sin\frac{\pi}{2}\sin\frac{y}{2}\right)\right]\,dy=\)\(\displaystyle -\int_{\pi}^{\pi-2\tan^{-1}z}\log\left(2\sin\frac{y}{2}\right)\,dy=\text{Cl}_2(\pi-2\tan^{-1}z)-\text{Cl}_2(\pi)\)That last Clausen function is zero, however, since for \(\displaystyle k\in\mathbb{Z}\, ,\, \sin\pi k\equiv 0\,\Rightarrow\)

\(\displaystyle \text{Cl}_2(\pi)=\sum_{k=1}^{\infty}\frac{\sin \pi k}{k^2}=0\)So the final evaluation is\(\displaystyle \text{Ti}_2(z)=\)\(\displaystyle \text{Si}_2\left(\frac{z}{\sqrt{1+z^2}}\right)
+\frac{1}{2}\tan^{-1}z \log(1+z^2)-\frac{1}{2}\int_0^{z/\,\sqrt{1+z^2}}\frac{\log(1-y^2)}{\sqrt{1-y^2}}\,dy=\)\(\displaystyle \text{Si}_2\left(\frac{z}{\sqrt{1+z^2}}\right)
+\frac{1}{2}\tan^{-1}z \log(1+z^2)\)

\(\displaystyle -\frac{1}{2}\left[2\text{Cl}_2(\pi-2\tan^{-1}z)-2\log 2\, \tan^{-1}z\right]\)
----------------------------
And so, finally (!), we have the functional relation between the Inverse Tangent Integral and the Inverse Sine Integral:\(\displaystyle \text{Ti}_2(z)=\)\(\displaystyle \text{Si}_2\left(\frac{z}{\sqrt{1+z^2}}\right)+ \tan^{-1}z\log\left(2\sqrt{1+z^2}\right)-\text{Cl}_2(\pi-2\tan^{-1}z)\, . \,\Box\)

 

[DreamWeaver]

Here is the closed form evaluation of the Inverse Tangent Integral (basically I've just cut and pasted a post I made on another forum - bad mammal! (Rofl) )...-----------------------The Inverse Tangent Integral has the closed form:$$\text{Ti}_2(\tan \phi)=\phi\log(\tan\phi)+\tfrac{1}{2}\text{Cl}_2(2\phi)+\tfrac{1}{2}\text{Cl}_2(\pi-2\phi)$$

where

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

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

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

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

Next, we split the logtan integral as follows:

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

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

On the logsine integral we apply \(\displaystyle x\to y/2\,\), and on the logcosine integral we apply \(\displaystyle x\to \pi/2-y/2\,\), to obtain:

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

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

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

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

That last step/reduction is due to the fact that the (Fourier-type) series definition of $$\text{Cl}_2(\theta)$$ makes it clear that

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

[\(\displaystyle \sin \pi k=0\,\) for all integer k]

Hence,

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

 

[DreamWeaver]

Re: Commentary for &quot;Inverse Sine/Tangent Integrals and related functions&quot;

Hey Gethin , try proving

\(\displaystyle \int^1_0 \frac{1-a^2 t^2}{(1+a^2t^2)^2}\,\log^2(t) \, dt \,=\frac{2\text{Ti}_2(a)}{a}\)

Damn! And there was I just about to go to bed and all... You a bad mammal! (Rofl)

More seriously though, this looks an interesting problem. I'll get "on it like a car bonnet" tomorrow, and then - internet connection allowing - post a reply.

Cheers Z! (Beer)

- - - Updated - - -

Incidentally, it looks like a simple differentiation will get me most of the way there, since

\(\displaystyle \frac{d}{dz}\text{Ti}_{m+1}(z)=\frac{\text{Ti}_m(z)}{z}\)(Hug)(Hug)(Hug)

 

[DreamWeaver]

Hey Gethin , try proving

\(\displaystyle \int^1_0 \frac{1-a^2 t^2}{(1+a^2t^2)^2}\,\log^2(t) \, dt \,=\frac{2\text{Ti}_2(a)}{a}\)

Here's my solution... (Bandit)To start with, make the substitution \(\displaystyle y=at\,\) to obtain\(\displaystyle \frac{1}{a}\int_0^a\frac{(1-y^2)}{(1+y^2)^2}\,\log^2(y/a)\,dy=\)

\(\displaystyle \frac{1}{a}\int_0^a\frac{2-(1+y^2)}{(1+y^2)^2}\,\log^2(y/a)\,dy=\)

\(\displaystyle \frac{2}{a}\int_0^a\frac{\log^2(y/a)}{(1+y^2)^2}\,dy-

\frac{1}{a}\int_0^a\frac{\log^2(y/a)}{(1+y^2)}\,dy=\)\(\displaystyle \frac{2}{a}\int_0^a\frac{[(1+y^2)-y^2]\log^2(y/a)}{(1+y^2)^2}\,dy-

\frac{1}{a}\int_0^a\frac{\log^2(y/a)}{(1+y^2)}\,dy=\)
\(\displaystyle \left(\frac{2}{a}-\frac{1}{a}\right)\,\int_0^a\frac{\log^2(y/a)}{(1+y^2)}\,dy-\frac{2}{a}\int_0^a\frac{y^2\log^2(y/a)}{(1+y^2)^2}\,dy=\)\(\displaystyle \frac{1}{a}\,\int_0^a\frac{\log^2(y/a)}{(1+y^2)}\,dy-\frac{2}{a}\int_0^a\frac{y^2\log^2(y/a)}{(1+y^2)^2}\,dy=\)\(\displaystyle \frac{1}{a}\,\int_0^a\frac{\log^2(y/a)}{(1+y^2)}\,dy+\frac{1}{a}\int_0^a\left[\frac{-2y}{(1+y^2)^2}\right]y\,\log^2(y/a)\,dy\)The term in large square brackets in the last integral is the derivative:

\(\displaystyle \frac{d}{dy}\frac{1}{(1+y^2)}=\frac{-2y}{(1+y^2)^2}\)So we can perform an integration by parts on that last integral:\(\displaystyle \int_0^a\left[\frac{-2y}{(1+y^2)^2}\right]y\,\log^2(y/a)\,dy=\)\(\displaystyle \frac{y\,\log^2(y/a)}{(1+y^2)}\,\Biggr|_0^a-\int_0^a\frac{1}{(1+y^2)}\left[\log^2(y/a)+2\log(y/a)\right]\,dy=\)\(\displaystyle -\int_0^a\frac{\log^2(y/a)}{(1+y^2)}\,dy-2\,\int_0^a\frac{\log(y/a)}{(1+y^2)}\,dy\)Inserting this back into the previous partial evaluation we get\(\displaystyle \frac{1}{a}\,\int_0^a\frac{\log^2(y/a)}{(1+y^2)}\,dy+\frac{1}{a}\left[-\int_0^a\frac{\log^2(y/a)}{(1+y^2)}\,dy-2\,\int_0^a\frac{\log(y/a)}{(1+y^2)}\,dy\right]=\)\(\displaystyle -\frac{2}{a}\,\int_0^a\frac{\log(y/a)}{(1+y^2)}\,dy=\)\(\displaystyle -\frac{2}{a}\,\left[\tan^{-1}y\log(y/a)\,\Biggr|_0^a-\frac{1}{a}\,\int_0^a\frac{\tan^{-1}y}{(y/a)}\,dy\right]=\)

\(\displaystyle \frac{2}{a}\int_0^a\frac{\tan^{-1}y}{y}\,dy=\frac{2}{a}\,\text{Ti}_2(a)\, .\, \Box\)That was really good fun, Z! (Sun) If you have any others, I'll happily give them a go... Not saying I'll get them, but always up for a go...

Cheers!

 

[DreamWeaver]

Since the Inverse Tangent Integral is increasingly encountered in modern research papers - particularly in relation to various branches of quantum mechanics - and yet rarely expressed in closed form (as above), it seemed a good idea to give a few explicit evaluations (aside from the single standard case of \(\displaystyle \text{Ti}_2(\tan \pi/4)=\text{Ti}_2(1)=G\) (G is Catalan's constant), which seems to be about as ubiquitous as the other evaluations are decidely not).For certain lower order rational arguments, the closed form\(\displaystyle \text{Ti}_2(\tan \theta)=\theta\log(\tan\theta)+\frac{1}{2}\text{Cl}_2(2\theta)+\frac{1}{2}\text{Cl}_2(\pi-2\theta)\)can be expressed in terms of more elementary transcendental constants/functions and a single Clausen Function. To arrive at these evaluations, we will first need the duplication formula for the Clausen function (of second order) \(\displaystyle \text{Cl}_2(\theta)\). Starting from the integral form of the Clausen function\(\displaystyle \text{Cl}_2(\theta)=-\int_0^{\theta}\Bigg| 2\sin\frac{x}{2} \Bigg|\, dx=\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^2}\)we will drop the absolute value sign, on the understanding that we'll be working within the range \(\displaystyle 0< \theta < \pi\). This is purely to simplify notation, and for arguments outside of this range we can appeal to the series form given above, and then exploit the periodicity of the sine function to extend the results given below.

To find the duplication formula for the Clausen Function, we use elementary trigonometry to evaluate the following integral in two different ways, and then equate the results:\(\displaystyle \int_0^{\theta}\log(\sin x)\,dx\)For our first evaluation, we equate the above integral with a single Clausen Function, in part, by noting that\(\displaystyle \int_0^{\theta}\log(\sin x)\,dx=\int_0^{\theta}\log(2\sin x)\,dx-\log 2\, \int_0^{\theta}\,dx=\)\(\displaystyle \int_0^{\theta}\log(2\sin x)\,dx-\theta\log 2\)Next, apply the substitution \(\displaystyle x=y/2\) to obtain\(\displaystyle \frac{1}{2}\int_0^{2\theta}\log\left(2\sin \frac{y}{2}\right)\,dy-\theta\log 2=\)\(\displaystyle -\frac{1}{2}\text{Cl}_2(2\theta)-\theta\log 2\)Now we evaluate the same integral again, but this time by splitting the \(\displaystyle \sin x\) term according to the duplication formula for the sine function; \(\displaystyle \sin x=2\sin\frac{x}{2}\cos \frac{x}{2}\):\(\displaystyle \int_0^{\theta}\log(\sin x)\,dx=\int_0^{\theta}\log\left(2\sin\frac{x}{2} \cos \frac{x}{2}\right)\,dx\equiv\)\(\displaystyle \int_0^{\theta}\log\left(2\sin \frac{x}{2}\right)\,dx+\int_0^{\theta}\log\left(2\cos \frac{x}{2}\right)\,dx-\theta\log 2=\)\(\displaystyle -\text{Cl}_2(\theta)+\int_0^{\theta}\log\left(2\cos \frac{x}{2}\right)\,dx-\theta\log 2\)For the log-cosine integral, apply the substitution \(\displaystyle x=\pi-y\) to obtain\(\displaystyle \int_0^{\theta}\log\left(2\cos \frac{x}{2}\right)\,dx=-\int_{\pi}^{\pi-\theta}\log\left(2\sin \frac{y}{2}\right)\,dy=\)\(\displaystyle \text{Cl}_2(\pi-\theta)-\text{Cl}_2(\pi)=\text{Cl}_2(\pi-\theta)\)Hence\(\displaystyle \int_0^{\theta}\log(\sin x)\,dx=-\frac{1}{2}\text{Cl}_2(2\theta)-\theta\log 2=-\text{Cl}_2(\theta)+\text{Cl}_2(\pi-\theta)\)Or, equivalently\(\displaystyle \text{Cl}_2(2\theta)=2\,\text{Cl}_2(\theta)-2\,\text{Cl}_2(\pi-\theta)\, . \, \Box\)
-----------------------------------------
Setting \(\displaystyle \theta=\pi/3\) in the duplication formula gives:\(\displaystyle \text{Cl}_2\left(\frac{2\pi}{3}\right)=2\,\text{Cl}_2\left(\frac{\pi}{3}\right)-2\,\text{Cl}_2\left(\frac{2\pi}{3}\right)\, \Rightarrow\)\(\displaystyle (1)\quad \text{Cl}_2\left(\frac{\pi}{3}\right)=\frac{3}{2} \text{Cl}_2\left(\frac{2\pi}{3}\right)\)\(\displaystyle (2)\quad
\text{Cl}_2\left(\frac{2\pi}{3}\right)=\frac{2}{3} \text{Cl}_2\left(\frac{\pi}{3}\right)\)
Setting \(\displaystyle \theta=\pi/4\) in the duplication formula gives:\(\displaystyle \text{Cl}_2\left(\frac{\pi}{2}\right)=2\,\text{Cl}_2\left(\frac{\pi}{4}\right)-2\,\text{Cl}_2\left(\frac{3\pi}{4}\right)\, \Rightarrow\)\(\displaystyle \text{Cl}_2\left(\frac{\pi}{2}\right)=\sum_{k=1}^{\infty}\frac{\sin \pi k/2}{k^2}=1-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\cdots = G\)Hence\(\displaystyle \text{Cl}_2\left( \frac{\pi}{4}\right)=\text{Cl}_2\left( \frac{3\pi}{4}\right)-\frac{G}{2}\)\(\displaystyle \text{Cl}_2\left(\frac{3\pi}{4}\right)=\text{Cl}_2\left(\frac{ \pi}{4}\right)-\frac{G}{2}\)
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Applying the previous relations in conjunction with the closed form for the Inverse Tangent Integral gives the equivalent forms:\(\displaystyle (1)\quad \text{Ti}_2\left(\tan \frac{\pi}{6}\right)=\text{Ti}_2\left(\frac{1}{ \sqrt{3}}\right)=-\frac{\pi}{12}\log 3+\frac{5}{6}\text{Cl}_2\left(\frac{\pi}{3}\right)\)\(\displaystyle (2)\quad \text{Ti}_2\left(\tan \frac{\pi}{6}\right)=\text{Ti}_2\left(\frac{1}{ \sqrt{3}}\right)=-\frac{\pi}{12}\log 3+\frac{7}{8}\text{Cl}_2\left(\frac{2\pi}{3}\right)\)As well as\(\displaystyle (3) \quad \text{Ti}_2\left(\tan \frac{\pi}{8}\right)=\text{Ti}_2\left(\sqrt{2}-1\right)=\frac{\pi}{8}\log(\sqrt{2}-1)+\text{Cl}_2\left(\frac{\pi}{4}\right)-\frac{G}{4}\)\(\displaystyle (4) \quad \text{Ti}_2\left(\tan \frac{\pi}{8}\right)=\text{Ti}_2\left(\sqrt{2}-1\right)=\frac{\pi}{8}\log(\sqrt{2}-1)+\text{Cl}_2\left(\frac{3\pi}{4}\right)+\frac{G}{4}\)And\(\displaystyle (5) \quad \text{Ti}_2\left(\tan \frac{3 \pi}{8}\right)=\text{Ti}_2\left(\sqrt{2}+1\right)=\frac{3 \pi}{8}\log(\sqrt{2}-1)+\text{Cl}_2\left(\frac{\pi}{4}\right)-\frac{G}{4}\)\(\displaystyle (6) \quad \text{Ti}_2\left(\tan \frac{3 \pi}{8}\right)=\text{Ti}_2\left(\sqrt{2}+1\right)=\frac{3 \pi}{8}\log(\sqrt{2}-1)+\text{Cl}_2\left(\frac{3\pi}{4}\right)+\frac{G}{4}\)
More evaluations to follow shortly...

 

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Although further (basic) results can be obtained as above, there is an alternative, via the inversion formula. Noting that\(\displaystyle \tan^{-1}x+\cot^{-1}x=\text{sgn}(x)\frac{\pi}{2}\)and working with strictly positive x - hence dropping the \(\displaystyle sgn(x)\) notation from the above - differentiation of:\(\displaystyle \text{Ti}_2(x)-\text{Ti}_2\left(\frac{1}{x}\right)\)

gives:\(\displaystyle \frac{\tan^{-1}x}{x}-\left(-\frac{1}{x^2}\right)\, \frac{\tan^{-1}(1/x)}{(1/x)}=\)\(\displaystyle \frac{\tan^{-1}x}{x}+\frac{\cot^{-1}x}{x}=\frac{\pi}{2 x}\)So, by the fundamental theorem of calculus, re-integrating the RHS over the range \(\displaystyle 0\) to \(\displaystyle z\) gives:\(\displaystyle \text{Ti}_2(z)-\text{Ti}_2\left(\frac{1}{z}\right)=\int_0^z\frac{\pi}{2 x}\,dx=\frac{\pi}{2}\log z\)Hence the Inverse Tangent Integral has the inversion relation:\(\displaystyle \text{Ti}_2(z)-\text{Ti}_2\left(\frac{1}{z}\right)=\frac{\pi}{2} \log z\)
We can now apply inversion on some of our earlier results:

\(\displaystyle (1)\quad \text{Ti}_2\left(\tan \frac{\pi}{6}\right)=\text{Ti}_2\left(\frac{1}{ \sqrt{3}}\right)=-\frac{\pi}{12}\log 3+\frac{5}{6}\text{Cl}_2\left(\frac{\pi}{3}\right)\)\(\displaystyle (2)\quad \text{Ti}_2\left(\tan \frac{\pi}{6}\right)=\text{Ti}_2\left(\frac{1}{ \sqrt{3}}\right)=-\frac{\pi}{12}\log 3+\frac{7}{8}\text{Cl}_2\left(\frac{2\pi}{3}\right)\)
\(\displaystyle (3) \quad \text{Ti}_2\left(\tan \frac{\pi}{8}\right)=\text{Ti}_2\left(\sqrt{2}-1\right)=\frac{\pi}{8}\log(\sqrt{2}-1)+\text{Cl}_2\left(\frac{\pi}{4}\right)-\frac{G}{4}\)\(\displaystyle (4) \quad \text{Ti}_2\left(\tan \frac{\pi}{8}\right)=\text{Ti}_2\left(\sqrt{2}-1\right)=\frac{\pi}{8}\log(\sqrt{2}-1)+\text{Cl}_2\left(\frac{3\pi}{4}\right)+\frac{G}{4}\)
\(\displaystyle (5) \quad \text{Ti}_2\left(\tan \frac{3 \pi}{8}\right)=\text{Ti}_2\left(\sqrt{2}+1\right)=\frac{3 \pi}{8}\log(\sqrt{2}-1)+\text{Cl}_2\left(\frac{\pi}{4}\right)-\frac{G}{4}\)\(\displaystyle (6) \quad \text{Ti}_2\left(\tan \frac{3 \pi}{8}\right)=\text{Ti}_2\left(\sqrt{2}+1\right)=\frac{3 \pi}{8}\log(\sqrt{2}-1)+\text{Cl}_2\left(\frac{3\pi}{4}\right)+\frac{G}{4}\)

Applying inversion on (1) and (2) gives:\(\displaystyle (7)\quad \text{Ti}_2\left( \sqrt{3}\right)=-\frac{\pi}{6}\log 3+\frac{5}{6}\text{Cl}_2\left(\frac{\pi}{3}\right)\)\(\displaystyle (8)\quad \text{Ti}_2\left( \sqrt{3}\right)=-\frac{\pi}{6}\log 3+\frac{7}{8}\text{Cl}_2\left(\frac{2\pi}{3}\right)\)
Whereas applying inversion on (3) and (4) gives (5) and (6) respectively.
--------------------------------Here are a few more examples, firstly worked out via the Inverse Tangent Integral closed form, and then inverted:\(\displaystyle (9)\quad \text{Ti}_2\left(\tan\frac{\pi}{5}\right)=\text{Ti}_2\left(\sqrt{5-2\sqrt{5}}\right)=\)\(\displaystyle \frac{\pi}{10}\log(5-2\sqrt{5})+\frac{1}{2}\, \text{Cl}_2\left(\frac{2\pi}{5}\right)+\frac{1}{2}\, \text{Cl}_2\left(\frac{3\pi}{5}\right)\)

\(\displaystyle (10)\quad \text{Ti}_2\left(\tan\frac{3\pi}{10}\right)=\text{Ti}_2\left(\sqrt{1+\frac{2}{\sqrt{5}}}\right)=\)\(\displaystyle \frac{3\pi}{20} \log \left(1+\frac{2}{ \sqrt{5}}\right)+\frac{1}{2}\, \text{Cl}_2\left(\frac{3\pi}{5}\right)+\frac{1}{2}\, \text{Cl}_2\left(\frac{2\pi}{5}\right)\)
Inversion on (9) gives\(\displaystyle \text{Ti}_2\left(\frac{1}{\sqrt{5-2\sqrt{5}}}\right)=-\frac{3\pi}{20}\log(5-2\sqrt{5})+\frac{1}{2}\, \text{Cl}_2\left(\frac{2\pi}{5}\right)+\frac{1}{2}\, \text{Cl}_2\left(\frac{3\pi}{5}\right)\)
Whereas inversion on (10) gives\(\displaystyle \text{Ti}_2\left(\frac{1}{\sqrt{1+\frac{2}{\sqrt{5}}}}\right)=\)\(\displaystyle \frac{\pi}{20} \log \left(1+\frac{2}{ \sqrt{5}}\right)+\frac{1}{2}\, \text{Cl}_2\left(\frac{3\pi}{5}\right)+\frac{1}{2}\, \text{Cl}_2\left(\frac{2\pi}{5}\right)\)