# A generalized Arctangent integral/function

#### DreamWeaver

##### Well-known member
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):

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

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:

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

The sum on the RHS can variously be expressed as, or related to, polygamma and Hurwitz Zeta functions.

Conversely, letting $$\displaystyle q$$ approach zero, we have:

$$\displaystyle \lim_{q \to 0^{+}} \, \text{Etan}^{-1}(z,m,q)= \sum_{k=0}^{\infty}(-1)^k \frac{z^{2k+1}}{(2k+1)^m}= \text{Ti}_m(z)$$

Which is the order-$$\displaystyle m$$ generalization of the Inverse Tangent Integral $$\displaystyle \text{Ti}_2(z)$$:

$$\displaystyle \text{Ti}_2(z)=\int_0^z\frac{\tan^{-1}x}{x}\, dx$$

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... #### DreamWeaver

##### Well-known member
For illustrative purposes, here's an elementary integral that can be expressed in terms of $$\displaystyle \text{Etan}^{-1}(z,m,q)$$:

$$\displaystyle T(z,m,q)=\int_0^zx^{q-1}\tan^{-1}x\,dx$$

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

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

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

$$\displaystyle z^q \sum_{k=0}^{\infty} \frac{(-1)^k z^{2k+1}}{(2k+q) (2k+q+1)}=$$

$$\displaystyle z^q \sum_{k=0}^{\infty} (-1)^k z^{2k+1} \frac{(2k+q+1)-(2k+1)}{q(2k+1) (2k+q+1)}=$$

$$\displaystyle \frac{z^q}{q}\left[ \sum_{k=0}^{\infty} \frac{(-1)^k z^{2k+1}}{(2k+1) }-\sum_{k=0}^{\infty} \frac{(-1)^k z^{2k+1}}{ (2k+q+1)} \right]=$$

$$\displaystyle \frac{z^q}{q}\left[ \text{Ti}_1(z)-\text{Etan}^{-1}(z,1,q) \right]=$$

$$\displaystyle \frac{z^q}{q}\left[ \tan^{-1}z-\text{Etan}^{-1}(z,1,q) \right]$$ - - - Updated - - -

Incidentally, note that:

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

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#### ZaidAlyafey

##### Well-known member
MHB Math Helper
You can relate the q-shifted arctangent integral to polylogarithms by letting $$\displaystyle q \to 1$$ that reminds me of Q-series.

#### DreamWeaver

##### Well-known member
You can relate the q-shifted arctangent integral to polylogarithms by letting $$\displaystyle q \to 1$$ 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! Incidentally, it was the similarity with q-series that led to - at first - call this the Elliptic Tangent function, hence the "$$\displaystyle E$$" in $$\displaystyle \text{Etan}$$...

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
$$\displaystyle \text{Etan}^{-1}(z,m,1)=\sum_{k=0}^{\infty}(-1)^k \frac{z^{2k+1}}{(2k+2)^m}= \frac{z}{2^m}\sum_{k\geq 0}\frac{(-z^2)^{k}}{(k+1)^m}=\frac{-1}{2^m\, z}\sum_{k\geq 1}\frac{(-z^2)^{k}}{k^m}$$

So we have the following

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

Now you can use the established result

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

Make good use of the complex numbers

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

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

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#### DreamWeaver

##### Well-known member
Here are a few definite integrals... $$\displaystyle p \ge 0, \, p \ne q$$

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

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

#### DreamWeaver

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

So we have the following

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

Now you can use the established result

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

Make good use of the complex numbers

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

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

I was making use of the same relations, but in a different way, since:

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

Or equivalently

$$\displaystyle \text{Li}_2(ix)=\frac{1}{4}\text{Li}_2(-x^2)+i\text{Ti}_2(x)$$

EDIT: hence me considering $$\displaystyle \text{Etan}$$ 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 $$\displaystyle \text{Etan}$$ 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 Last edited: