MHB A generalized Arctangent integral/function

[DreamWeaver]

This is NOT a tutorial, so by all means, if you've a mind to, the please DO very much feel free to contribute...Preamble:As a consequence of various families of definite integrals I've been studying recently, I've been led to consider what I've come to call the q-shifted Inverse Tangent Integral (NB. I'm not sure about the notation, and I might well change it, but it'll do for now):$$\text{Etan}^{-1}(z,m,q)=\sum_{k=0}^{\infty}(-1)^k \frac{z^{2k+1}}{(2k+q+1)^m}$$This function has arisen quite naturally, as a generalization of a number of other functions. The following special cases will hopefully help illustrate the point:$$\lim_{z \to 1} \, \text{Etan}^{-1}(z,m,q)= \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+q+1)^m}$$

The sum on the RHS can variously be expressed as, or related to, polygamma and Hurwitz Zeta functions.Conversely, letting $$q$$ approach zero, we have:$$\lim_{q \to 0^{+}} \, \text{Etan}^{-1}(z,m,q)=
\sum_{k=0}^{\infty}(-1)^k \frac{z^{2k+1}}{(2k+1)^m}= \text{Ti}_m(z)$$Which is the order-$$m$$ generalization of the Inverse Tangent Integral $$\text{Ti}_2(z)$$:$$\text{Ti}_2(z)=\int_0^z\frac{\tan^{-1}x}{x}\, dx$$

I've a fair few results to follow, as and when I get time to post, but like I say, if any of you feel like joining in, then you are very much welcome to do so... (Heidy)

 

[DreamWeaver]

For illustrative purposes, here's an elementary integral that can be expressed in terms of $$\text{Etan}^{-1}(z,m,q)$$:$$T(z,m,q)=\int_0^zx^{q-1}\tan^{-1}x\,dx$$

An evaluation in terms of polygamma functions is not too hard to find, provided $$z=1$$. But why stop there...?

For $$0 < z \le 1$$, expand the arctangent into series form to obtain:

$$\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)}\int_0^zx^{q-1}x^{2k+1}\,dx=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)} \left[ \frac{x^{2k+q+1}}{(2k+q+1)} \right]_0^z=
$$

$$ z^q \sum_{k=0}^{\infty} \frac{(-1)^k z^{2k+1}}{(2k+q) (2k+q+1)}= $$$$ z^q \sum_{k=0}^{\infty} (-1)^k z^{2k+1} \frac{(2k+q+1)-(2k+1)}{q(2k+1) (2k+q+1)}=$$$$\frac{z^q}{q}\left[ \sum_{k=0}^{\infty} \frac{(-1)^k z^{2k+1}}{(2k+1) }-\sum_{k=0}^{\infty} \frac{(-1)^k z^{2k+1}}{ (2k+q+1)} \right]=$$$$\frac{z^q}{q}\left[ \text{Ti}_1(z)-\text{Etan}^{-1}(z,1,q) \right]=$$$$\frac{z^q}{q}\left[ \tan^{-1}z-\text{Etan}^{-1}(z,1,q) \right]$$(Heidy)
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Incidentally, note that:

$$\frac{d^n}{dq^n}T(z,m,q)=\int_0^zx^{q-1}(\log x)^n\tan^{-1}x\, dx$$

 

[alyafey22]

You can relate the q-shifted arctangent integral to polylogarithms by letting $$q \to 1$$ that reminds me of Q-series.

 

[DreamWeaver]

You can relate the q-shifted arctangent integral to polylogarithms by letting $$q \to 1$$ that reminds me of Q-series.

Well played, Sir! I've pages and pages of stuff worked out for this function, but I missed that one. Thanks! (Cool)Incidentally, it was the similarity with q-series that led to - at first - call this the Elliptic Tangent function, hence the "$$E$$" in $$\text{Etan}$$...

 

[alyafey22]

$$\text{Etan}^{-1}(z,m,1)=\sum_{k=0}^{\infty}(-1)^k \frac{z^{2k+1}}{(2k+2)^m}= \frac{z}{2^m}\sum_{k\geq 0}\frac{(-z^2)^{k}}{(k+1)^m}=\frac{-1}{2^m\, z}\sum_{k\geq 1}\frac{(-z^2)^{k}}{k^m}$$

So we have the following

$$\tag{1} \text{Etan}^{-1}(z,m,1) = -2^{-m} \frac{\text{Li}_m(-z^2)}{z}$$

Now you can use the established result

$$\text{Li}_m(z^2) = 2^{1-m} \left( \text{Li}_m(z)+\text{Li}_m(-z) \right)$$

Make good use of the complex numbers

$$\text{Li}_m(-z^2) = 2^{1-m} \left( \text{Li}_m(iz)+\text{Li}_m(-iz) \right)$$

Substituting in (1) then integrating with respect to z is surely interesting :) .

 

[DreamWeaver]

Here are a few definite integrals... $$p \ge 0, \, p \ne q$$

$$\int_0^1x^{p-1}\text{Etan}^{-1}(x,1,q)\,dx=\frac{1}{q-p}\left[ \text{Etan}^{-1}(1,1,p) - \text{Etan}^{-1}(1,1,q) \right]$$$$\int_0^1x^{p-1}\text{Etan}^{-1}(x,2,q)\,dx=\frac{1}{(q-p)^2}\left[ \text{Etan}^{-1}(1,1,p) - \text{Etan}^{-1}(1,1,q) \right]-\frac{1}{(q-p)}\text{Etan}^{-1}(1,2,q) $$

 

[DreamWeaver]

$$\text{Etan}^{-1}(z,m,1)=\sum_{k=0}^{\infty}(-1)^k \frac{z^{2k+1}}{(2k+2)^m}= \frac{z}{2^m}\sum_{k\geq 0}\frac{(-z^2)^{k}}{(k+1)^m}=\frac{-1}{2^m\, z}\sum_{k\geq 1}\frac{(-z^2)^{k}}{k^m}$$

So we have the following

$$\tag{1} \text{Etan}^{-1}(z,m,1) = -2^{-m} \frac{\text{Li}_m(-z^2)}{z}$$

Now you can use the established result

$$\text{Li}_m(z^2) = 2^{1-m} \left( \text{Li}_m(z)+\text{Li}_m(-z) \right)$$

Make good use of the complex numbers

$$\text{Li}_m(-z^2) = 2^{1-m} \left( \text{Li}_m(iz)+\text{Li}_m(-iz) \right)$$

Substituting in (1) then integrating with respect to z is surely interesting :) .

(Muscle) Nicely done...I was making use of the same relations, but in a different way, since:

$$\text{Ti}_2(x)=\frac{1}{2i}\left[ \text{Li}_2(ix)-\text{Li}_2(-ix)\right]$$

Or equivalently

$$\text{Li}_2(ix)=\frac{1}{4}\text{Li}_2(-x^2)+i\text{Ti}_2(x)$$EDIT: hence me considering $$\text{Etan}$$ in terms of my pet favourites, the regular Inverse Tangent Integrals...

Incidentally, and I'll add a bit more about this tomorrow if I get time, but the $$\text{Etan}$$ function gives a neat closed form to a particular Clausen Function generalization I've been considering recently (a different one to that I posted about in the tutorials board).

I'd be keen to get you take on feedback on that... But now, must sleep. Bed calling. NN :D