Are there applications of $\mathrm{\int \frac {dx}{tan^{-1}(x)}}$?

Question

This will be a follow up question to my unexpectedly popular question:

Is there an exact solution for $\large\int \frac{dx}{\tan^{-1}(x)}$?

which is also nicely related to:

Indefinite Integral $\int\frac{1}{1+\tan^{-1}x}\,\text{d}x$

Note that @Yuri Negometyanov and @Nikos Bagis both gave 2 great sum representations:

The general result was the following. Here is a graph of what the function looks m

$$\mathrm{T(x)\mathop=^\text{def} \int\frac{dx}{\tan^{-1}(x)}=\int\frac{2i\,dx}{ln(1-ix)-ln(1+ix)}=\frac{x}{\tan^{-1}(x)}+ln\left(\tan^{-1}(x)\right)-\frac12 \sum_{n=2}^\infty\frac{(-4)^n\left(4^n-1\right)B_{2n} \left(\tan^{-1}(x)\right)^{2(n-1)}}{(n-1)(2n)!}+C=ln\left(tan^{-1}(x)\right)+\frac{x}{tan^{-1}(x)}-\frac{4}{\pi^2}\sum_{n\in \Bbb Z}\frac{ln\left|\pi(2n+1)-2tan^{-1}(x)\right|}{(2n+1)^2}+C}$$

This had the nice consequence of evaluating the hyperbolic version. Here is a visual representation:

$$\mathrm{Th(x)\mathop=^\text{def}\int\frac{dx}{tanh^{-1}(x)}=\int\frac{i\,dx}{tan^{-1}(ix)}=\int\frac{2dx}{ln(1+x)+ln(1-x)}=T(ix)=C+\frac{x}{tanh^{-1}(x)}+ln\left(tanh^{-1}(x)\right)+\frac12\sum_{n=2}^\infty \frac{4^n\left(4^n- 1\right) B_{2n}\left(tanh^{-1}(x)\right)^{2(n-1)}}{(n-1)(2n)!}\mathop=^?C+ \frac{x}{tanh^{-1}(x)}+ln\left(tanh^{-1}(x)\right)-\frac{4}{\pi^2}\sum_{n\in\Bbb Z}\frac{ln\left|\pi(2n+1)-2i\,tanh^{-1}(x)\right|}{(2n+1)^2}}$$

I would give other values, but they are already in the referenced question. I know it can be used to solve the following with a series expansion:

Does $\int_{-1}^1\frac{\arctan x}{\text{arctanh}\,x}\,\mathrm{d}x$ have a closed form?

It also poses a nice problem for finding $\mathrm{\int_0^1 \frac{1}{tan^{-1}(x)}-\frac{1}{tanh^{-1}(x)}dx=.359279…}$

I already can figure out these applications, but we need to come up with references of the integral in other works or actual applications. These references or applications can be for T(x) or Th(x). These are just names for fun. Please correct me and give me feedback!

Answer 1

There are 2 more applications I could think of with graphical proof. The Inverse Function Integral theorem is key here. $B_y$ the way, this notation is the y-th Bernoulli number:

What is the solution to this peculiar Integral: $\int \tan \left(\frac 1 x\right) \, dx$?

As a consequence of the inverse. Be careful when evaluating from a to b as to not put the $x^{2-2n}$ in the denominator. Simplified graph:

$$\mathrm{\int tan\left(\frac 1x\right)dx=\int\frac{dx}{e^\frac{i}{x}-i}-\int\frac{dx}{e^\frac{i}{x}+i}-ix=C+x\,tan\left(\frac1x\right)-T \left(tan\left(\frac1x\right)\right)}$$

$$\mathrm{t(x)\mathop=^{def}\int tan\left(\frac 1x\right)dx=C+x\, tan\left(\frac 1x\right)-\frac{tan\left(\frac 1x\right)}{tan^{-1}\left(tan\left(\frac 1x\right)\right)}-ln\left(tan^{-1}\left(tan\left(\frac 1x\right)\right)\right)+ \frac12 \sum_{n=2}^\infty\frac{(-4)^n\left(4^n-1\right)B_{2n} \left(\tan^{-1}(tan\left(\frac 1x\right))\right)^{2(n-1)}}{(n-1)(2n)!}= C+x\, tan\left(\frac 1x\right)-\frac{tan\left(\frac 1x\right)}{tan^{-1}\left(tan\left(\frac 1x\right)\right)}-ln\left(tan^{-1}\left(tan\left(\frac 1x\right)\right)\right)+ \frac{4}{\pi^2}\sum_{n\in \Bbb Z}\frac{ln\left|\pi(2n+1)-2tan^{-1}\left(tan\left(\frac 1x\right)\right)\right|}{(2n+1)^2}}$$

$$\mathrm{t(x)=\int tan\left(\frac1x\right)dx=C+ln(x)+\frac12\sum_{n=2}^\infty \frac{(-4)^n (4^n-1) B_{2n}x^{2-2n}}{(2n)!(n-1)}}$$

It was shown by @Jack D’Aurizio that:

$$ \int\tan\left(\frac{1}{x}\right)\,dx =C+\sum_{m\geq 0}\frac{4}{\pi^2(2m+1)^2}\log\left[(2m+1)^2\pi^2 x^2-4\right]$$

in this post:

Integral of $\tan(1/x)$

The hyperbolic version has a similar graph, hence the similar integral. Be careful when evaluating the integral to not put the $b^{-2(n-1)}-a^{-2(n-1)} $ part in the denominator as it is not the same as $\frac{1}{b^{2(n-1)}-a^{2(n-1)} }$

$$\mathrm{th(x)\mathop=^{def}\int tanh\left(\frac1x\right)dx=x+i\int\frac{dx}{e^\frac1 x-i}+\ int\frac{dx}{e^\frac1 x+i} =x\, tanh\left(\frac1x\right)-Th\left(tanh\left(\frac1x\right)\right)= C+ln(x)+\frac12\sum_{n=2}^\infty\frac{4^n\left(4^n-1\right)B_{2n} x^{-2(n-1)}}{(2n)!(n-1)}}$$

It can also be shown that t(x)=th(ix) and th(-ix)=t(x) as a consequence of being the same identity for their inverse integrand functions resulting in this graph:

$$\mathrm{th(x)= C+\frac{4}{\pi^2}\sum_{n=0}^\infty \frac{ln\left(\pi^2(2n+1)^2x^2+4\right)}{(2n+1)^2}}$$

More on t(x) and th(x): We can understand that the following holds because of the inverse function theorem and definitions of the four functions related to:

$$\mathrm{-i\,tan^{-1}(ix)=tanh^{-1}(x)\implies -i\,tanh^{-1}(x)=tan^{-1}(x)}$$

Using this knowledge, a bit of algebra, substitution, and already derived inverse integral function theorem definitions, here are the definitions of the functions in terms of one another:

$$\mathrm{T(ix)=Th(x)\implies Th(-ix)=T(x)}$$ $$\mathrm{t(ix)=th(x)\implies th(-ix)=t(x)}$$

$$\mathrm{t(x)= x\,tan\left(\frac1x\right)-T\left(tan\left(\frac1x\right)\right)= x\,tan\left(\frac1x\right)-Th\left(tanh\left(\frac1{-ix}\right)\right)}$$

$$\mathrm{th(x)=x\,tanh\left(\frac1x\right)-Th\left(tanh\left(\frac1x\right)\right)= x\,tanh\left(\frac1x\right)-T\left(tan\left(\frac1{ix}\right)\right)}$$

$$\mathrm{T(x)=\frac{x}{tan^{-1}(x)}-t\left(\frac{1}{tan^{-1}(x)}\right)= \frac{x}{tan^{-1}(x)}-th\left(\frac{1}{tanh^{-1}(-ix)}\right)}$$

$$\mathrm{Th(x)=\frac{x}{tanh^{-1}(x)}-th\left(\frac{1}{tanh^{-1}(x)}\right)= \frac{x}{tanh^{-1}(x)}-t\left(\frac{1}{tan^{-1}(ix)}\right)}$$

There is also a connection to $\mathrm{\int\frac{dx}{cot^{-1}(x)}\ and \ \int\frac{dx}{coth^{-1}(x)}}$, but it is cumbersome to get in terms of T(x). A similar technique to its derivation can be used to find the following result as seen in this graph demo. Note there are other ways to evaluate the integrals, but these are irrelevant for now. The new “name” should not be confused with the fresnel cosine integral. Also note we can easily find $\mathrm{\int cot\left(\frac1x\right)dx\ and\ \int coth\left(\frac1x\right)dx}$ using the aforementioned inverse function integral theorem:

$$\mathrm{C(x)\mathop=^{def} \int \frac{dx}{cot^{-1}(x)} =C+\frac{x}{cot^{-1}(x)}-\frac{1}{2\left(cot^{-1}(x)\right)^2}-\frac13 ln\left(cot^{-1}(x)\right)+\frac12\sum_{n=2}^\infty\frac{(-4)^n B_{2n} \left(cot^{-1}(x)\right)^{2(n-1)}}{(n-1)(2n)!}}$$

$$\mathrm{Ch(x)\mathop=^{def}\int \frac{dx}{coth^{-1}(x)}=C+\frac{x}{coth^{-1}(x)}-\frac{1}{2\left(coth^{-1}(x)\right)^2}-\frac13 ln\left(coth^{-1}(x)\right)+\frac12\sum_{n=2}^\infty\frac{(-4)^n B_{2n} \left(coth^{-1}(x)\right)^{2(n-1)}}{(n-1)(2n)!}}$$

$$\mathrm{C\left(\frac1x \right)+T(x)=C+\frac{1}{tan^{-1}(x)}\left(\frac1x+x\right)-\frac{1}{2\left(tan^{-1}(x)\right)^2}+\frac23 ln\left(tanh^{-1}(x)\right)+\frac12 \sum_{n=2}^\infty\frac{(-4)^{2n}B_{2n}\left(tan^{-1}(x)\right)^{2n-2}}{(n-1)(2n)!}}$$

$$\mathrm{Ch\left(\frac1x \right)+Th(x)=C+\frac{1}{tanh^{-1}(x)}\left(\frac1x+x\right)+\frac{1}{2\left(tanh^{-1}(x)\right)^2}+\frac23 ln\left(tanh^{-1}(x)\right)-\frac12 \sum_{n=2}^\infty\frac{4^{2n}B_{2n}\left(tanh^{-1}(x)\right)^{2n-2}}{(n-1)(2n)!}}$$

I wonder about the integral of the reciprocal argumented (co)secant and reciprocal inverse (co)secant formulas. What about the inverses, like $\mathrm {T^{-1}(x)\ and\ th^{-1}(x)}$? Please correct me and give me feedback!