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#### DreamWeaver

##### Well-known member

- Sep 16, 2013

- 337

**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

**welcome to do so...**

*very much*