A generalized Arctangent integral/function

In summary, the conversation discussed a function called the q-shifted Inverse Tangent Integral and its various properties and special cases, including its relation to polygamma and Hurwitz Zeta functions. The conversation also mentioned its connection to q-series and its evaluation in terms of polylogarithms. The conversation also explored definite integrals involving the q-shifted Inverse Tangent Integral and its use in closed forms for other functions such as Clausen Functions.
  • #1
DreamWeaver
303
0
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... (Heidy)
 
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  • #2
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:

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

\(\displaystyle z^q \sum_{k=0}^{\infty} \frac{(-1)^k z^{2k+1}}{(2k+q) (2k+q+1)}= \)\(\displaystyle z^q \sum_{k=0}^{\infty} (-1)^k z^{2k+1} \frac{(2k+q+1)-(2k+1)}{q(2k+1) (2k+q+1)}=\)\(\displaystyle \frac{z^q}{q}\left[ \sum_{k=0}^{\infty} \frac{(-1)^k z^{2k+1}}{(2k+1) }-\sum_{k=0}^{\infty} \frac{(-1)^k z^{2k+1}}{ (2k+q+1)} \right]=\)\(\displaystyle \frac{z^q}{q}\left[ \text{Ti}_1(z)-\text{Etan}^{-1}(z,1,q) \right]=\)\(\displaystyle \frac{z^q}{q}\left[ \tan^{-1}z-\text{Etan}^{-1}(z,1,q) \right]\)(Heidy)
- - - 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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  • #3
You can relate the q-shifted arctangent integral to polylogarithms by letting \(\displaystyle q \to 1\) that reminds me of Q-series.
 
  • #4
ZaidAlyafey said:
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! (Cool)Incidentally, it was the similarity with q-series that led to - at first - call this the Elliptic Tangent function, hence the "\(\displaystyle E\)" in \(\displaystyle \text{Etan}\)...
 
  • #5
\(\displaystyle \text{Etan}^{-1}(z,m,1)=\sum_{k=0}^{\infty}(-1)^k \frac{z^{2k+1}}{(2k+2)^m}= \frac{z}{2^m}\sum_{k\geq 0}\frac{(-z^2)^{k}}{(k+1)^m}=\frac{-1}{2^m\, z}\sum_{k\geq 1}\frac{(-z^2)^{k}}{k^m}\)

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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  • #6
Here are a few definite integrals... \(\displaystyle p \ge 0, \, p \ne q\)

\(\displaystyle \int_0^1x^{p-1}\text{Etan}^{-1}(x,1,q)\,dx=\frac{1}{q-p}\left[ \text{Etan}^{-1}(1,1,p) - \text{Etan}^{-1}(1,1,q) \right]\)\(\displaystyle \int_0^1x^{p-1}\text{Etan}^{-1}(x,2,q)\,dx=\frac{1}{(q-p)^2}\left[ \text{Etan}^{-1}(1,1,p) - \text{Etan}^{-1}(1,1,q) \right]-\frac{1}{(q-p)}\text{Etan}^{-1}(1,2,q) \)
 
  • #7
ZaidAlyafey said:
\(\displaystyle \text{Etan}^{-1}(z,m,1)=\sum_{k=0}^{\infty}(-1)^k \frac{z^{2k+1}}{(2k+2)^m}= \frac{z}{2^m}\sum_{k\geq 0}\frac{(-z^2)^{k}}{(k+1)^m}=\frac{-1}{2^m\, z}\sum_{k\geq 1}\frac{(-z^2)^{k}}{k^m}\)

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 :) .
(Muscle) 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 :D
 
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Related to A generalized Arctangent integral/function

What is a generalized Arctangent integral/function?

A generalized Arctangent integral/function is a mathematical function that is defined as the inverse of the tangent function. It is commonly denoted as arctan(x) or tan^-1(x). It is a continuous, one-to-one function that maps real numbers to angles in the range of -π/2 to π/2 radians.

What is the domain and range of a generalized Arctangent function?

The domain of a generalized Arctangent function is all real numbers, while its range is restricted to the interval of -π/2 to π/2 radians. This means that the function can take any real number as an input, but its output will always be within the range of -π/2 to π/2 radians.

What is the derivative of a generalized Arctangent function?

The derivative of a generalized Arctangent function is 1/(1+x^2). This can be derived using the chain rule and the fact that the derivative of tan(x) is sec^2(x).

What are the important properties of a generalized Arctangent function?

Some important properties of a generalized Arctangent function include its periodicity, meaning that it repeats itself every π radians, and its symmetry, meaning that arctan(-x) = -arctan(x). It is also an odd function, meaning that arctan(-x) = -arctan(x), and its values are limited to the interval of -π/2 to π/2 radians.

How is a generalized Arctangent function used in real life?

A generalized Arctangent function has various applications in mathematics and engineering, such as in calculating angles and solving trigonometric equations. It is also used in fields like physics and astronomy to calculate the position and velocity of objects. In computer science, it is used in the development of algorithms and computer graphics.

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