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Do you know about the inverse tangent integral function? It is defined as: $$\mathrm{Ti_2(x)=Ti(x)=\int_0^x\frac{tan^{-1}(x)}{x}dx=-\frac1x\sum_{n\ge 1}\frac{(-1)^nx^{2n}}{(2n-1)^2}}$$

Expanding the denominator and then the sum gives many other forms of the function. Also, I wondered what other unsolved trigonometric integrals there are. You can click my profile questions to see similar inspiration questions. Here is the function I want to find. Here is an interactive graph. It is an odd injective function:

$$\mathrm{T(x)=\int \frac{dx}{tan^{-1}(x)} , \\ T(b)-T(a)=\int_a^b \frac{dx}{tan^{-1}(x)}=\int_{tan^{-1}(a)}^{tan^{-1}(b)} \frac{sec^2(x)}{x}dx\mathop=^{|x|<\frac\pi 2} \quad \int_{tan^{-1}(a)}^{tan^{-1}(b)}\sum_{N=0}^\infty\sum_{n=0}^\infty\frac{(-1)^{N+n}E_{2N}E_{2n}x^{2(N+n)-1}}{(2N)!(2n)!}dx= \quad \sum_{N=0}^\infty\sum_{n=0}^\infty\frac{(-1)^{N+n}E_{2N}E_{2n}(tan^{-1}(b))^{2(N+n)}-(tan^{-1}(a))^{2(N+n)}}{2(2N)!(2n)!(N+n)}}$$

This approach above uses the Euler numbers $\mathrm E_y$ and this series representation. This looks very complicated as I had to multiply the two series together to get a secant squared representation.

Here is a solution for T(x), but I do not know the Laurent series coefficient formula:

$$\mathrm{T(x)=\int \sum_{n=0}^\infty c_n x^{2n-1}dx=\int \frac1x +\frac x3-\frac{4x^3}{45}+\frac{44x^5}{945} -\frac{428x^7}{14175}+\frac{10196x^9}{467775}-\frac{10719068x^{11}}{638512875}+…dx=ln(x)+\sum_{n=1}^\infty \frac{c_n x^{2n}}{2n}=ln(x)+\frac {x^2}6-\frac{x^4}{45}+\frac{22x^6}{2835}-\frac{108x^8}{28350}+\frac{5098x^{10}}{2338875}-\frac{2679767x^{12}}{1915538625}+…,c_n=1,\frac13,-\frac4{45},\frac{44}{945},-\frac{428}{14175},\frac{10196}{467775},-\frac{10719068}{638512875},…}$$

Here is a graph of T(x). Notice the oblique asymptote which is a consequence of the fact that $\frac1{\tan^{-1}(\pm \infty)}=\frac2\pi$. The graph is for the area from x=1 to x=$\text x_0$. You can also see the vertical asymptote at x=0 implying infinite area over almost any interval containing x=0:

enter image description here

Here is motivation using trigonometric integral functions. I will assume a primitive here, no constant, for simplicity:

$$\mathrm{\int\frac{dx}{cos^{-1}(x)}=-Si(cos^{-1}(x))}$$

$$\mathrm{\int\frac{dx}{sin^{-1}(x)}=Ci(sin^{-1}(x)}$$

$$\mathrm{\int\frac{dx}{cosh^{-1}(x)}=Shi(cosh^{-1}(x))}$$

$$\mathrm{\int\frac{dx}{sinh^{-1}(x)}=Chi(sinh^{-1}(x)}$$

Just like the actual inverse tangent integral, I wonder if this T(x) function can also be expressed in exact form. If possible, please express in closed form, but an exact answer also works. I would be surprised if T(x) can even be expressed in terms of Ti(x), the inverse tangent integral.

Another answer is to find out if the coefficients I typed above have any pattern that can be written as an mathematical expression.

Please correct me and give me feedback!

Applications: Try the inverse integral theorem on $\tan\frac1x$ You can try this problem, find the integral $\mathrm{\int_{i}^{2i}\frac{dx}{tan^{-1}(x)}=\int_1^2\frac{i\,dx}{tanh^{-1}(x)}}$, and I will try to ask a question about the applications though .

Results from @Yuri Negometyanov and @Nikos Bagis show the following results. Graphical proof. Note that you can split the sum after distributing the factored numerator. Bernoulli Numbers. Notice the root of the desmos graph at different values. Be sure to set the slider to approach $\pm \infty$ for i by approximation of $\pm 10$ in the graph link:

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

I wonder about the integrals of the reciprocal of other inverse trigonometric functions…

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    $\begingroup$ You wrote "$T(x)=\int \frac{dx}{tan^{-1}(x)}=T(b)-T(a)=\int_a^b \frac{dx}{tan^{-1}(x)}$". Make up your mind, you are trying to calculate the indefinite integral or the definite integral. $\endgroup$ – jjagmath Jul 14 at 22:22
  • $\begingroup$ @jjagmath these are about the same integrals. If we add bounds to the indefinite integral, we get the definite integral. I only did the one with bounds so I did not have to substitute the bounds back when integrating, and desmos can plot the function in that case. The indefinite integral is my main goal. Do you have any ideas? $\endgroup$ – Tyma Gaidash Jul 14 at 22:29
  • $\begingroup$ Also the FTC, one of them, says that if F(x) is the integral of f(x), then the definite integral from a to b is F(b)-F(a). $\endgroup$ – Tyma Gaidash Jul 15 at 0:49
  • $\begingroup$ As you have noticed, $T(x)=\int{\sec^2x\over x}\,dx$. I think it very unlikely there's a closed form for this, any more than there is for $\int{\sin x\over x}\,dx$. But there is a theory of integration in finite terms which can probably answer the question for certain, and doing a websearch for the italicized phrase will probably get you started on learning about it. $\endgroup$ – Gerry Myerson Jul 15 at 3:47
  • $\begingroup$ @GerryMyerson In that case, I may change the question into finding an exact solution of the integral. I found one possible form of it. I was thinking more about the sine integral function which can be expressed in terms of the generalized exponential integral function. There should be a way to evaluate such a simple integrand using any functions in wolfram mathworld or the functions website. Please see equation (20) on the website. Thanks. $\endgroup$ – Tyma Gaidash Jul 15 at 11:40
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Let $$I(x)=\int\limits_1^x \dfrac{\text dt}{\arctan t},\tag1$$ then $$I(\tan y)=\int\limits_{\large^\pi/\mspace{1mu}_4}^y\dfrac1t\,\text d\tan t \;\overset{\text{IBP}}{=\!=}\;\dfrac{\tan t}t\bigg|_{\large^\pi/\mspace{1mu}_4}^y+\int\limits_{\large^\pi/\mspace{1mu}_4}^y\,\dfrac{\tan t}{t^2}\,\text dt = \dfrac{\tan y}y-\dfrac4\pi +\int\limits_{\large^\pi/\mspace{1mu}_4}^y\dfrac{\tan t}{t^2}\,\text dt,\tag2$$ or, applying the known Maclaurin series of the tangent function in the form of $$\tan t =\sum\limits_{n=1}^\infty (-1)^{n-1}\dfrac{4^n(4^n-1)\text B_{2n}}{(2n)!}t^{2n-1}= t+\dfrac13\,t^3+\dfrac2{15}\,t^5+\dfrac{17}{315}\,t^7+\dfrac{62}{2835}\,t^9+\dots,\tag3$$

$$\color{green}{\mathbf{I(\tan y)= \dfrac{\tan y}y-\dfrac4\pi+\ln\dfrac{4y}\pi + \sum\limits_{n=2}^\infty c_n \left(y^{2n-2}-\left(\dfrac\pi4\right)^{2n-2}\right),}}\tag4$$ where $$\color{green}{\mathbf{c_n = (-1)^{n-1}\dfrac{4^n(4^n-1)\text B_{2n}}{(2n-2)(2n)!}.}}\tag5$$

I.e., we have got $\;1D\;$ series.

Alternative form of the solution $(4)$ is $$\color{green}{\mathbf{I(x)= \dfrac x{\arctan x}-\dfrac4\pi+\ln\dfrac{4\arctan x}\pi + \sum\limits_{n=2}^\infty c_n\left(\arctan^{2n-2} x -\left(\dfrac\pi4\right)^{2n-2}\right).}}\tag6$$

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  • $\begingroup$ [+1] This is a great answer, and I will most likely choose it as the answer to the problem. However, in case another person chooses to answer it, I will leave it here for the next few days. Can you write the final answer in terms of x only? As a final note, please simplify as much as possible. I will check this. Thanks! $\endgroup$ – Tyma Gaidash Jul 22 at 18:48
  • $\begingroup$ @TymaGaidash Thank you for the comments. Done. $\endgroup$ – Yuri Negometyanov Jul 22 at 19:49
  • $\begingroup$ Thanks again. Now we have a solution for this type of inverse tangent integral. I am not too sure where this may show up though. $\endgroup$ – Tyma Gaidash Jul 22 at 19:58
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    $\begingroup$ @TymaGaidash I have a lot of materials and have not articles, and I'm interested too. Try to ask in Academy SE. Will be glad to know your results. $\endgroup$ – Yuri Negometyanov Jul 22 at 20:15
  • $\begingroup$ You got the accepted answer as you method is simplest and more graphically verified. There were 2 answers, but NikosBagis put more effort even without the extra edits. $\endgroup$ – Tyma Gaidash Jul 24 at 11:57
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It is known that $$ \csc^2 z=\sum^{\infty}_{k=-\infty}\frac{1}{(z-k\pi)^2}\textrm{, }z\neq 0,\pm\pi,\pm 2\pi,\ldots\tag 1 $$ Hence with $z\rightarrow z-\frac{\pi}{2}$, we have $$ \sec^2 z=\sum^{\infty}_{n=-\infty}\frac{1}{(z-\pi/2-k\pi)^2}\tag 2 $$ Also $$ \frac{d}{dt}\int^{\tan t}_{c}\frac{1}{\arctan x}dx=\frac{\sec^2 t}{t} $$ Hence using (1): $$ I=\int^{\tan t}_{c}\frac{1}{\arctan x}dx= $$ $$ =\sum _{k=-\infty }^{\infty } [\frac{4 \pi }{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}+\frac{8 k \pi }{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}- $$ $$ -\frac{4 \pi \log(\pi +2 k \pi -2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}-\frac{8 k \pi \log(\pi +2 k \pi -2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}+ $$ $$ +\frac{8 t \log(\pi +2 k \pi -2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}+\frac{4 \pi \log(2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}+ $$ $$ +\frac{8 k \pi \log(2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}-\frac{8 t \log(2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}]+C. $$ But $$ S_1=\sum^{\infty}_{k=-\infty}\frac{4\pi}{(\pi+2k\pi)^2(\pi+2 k\pi-2t)}=\frac{\pi (-t+\tan t)}{2 t^2} $$ $$ S_2=\sum _{k=-\infty }^{\infty } \frac{8 k \pi }{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}=\frac{\pi t-\pi \tan t+2 t \tan t}{2 t^2} $$ $$ S_3=\sum _{k=-\infty }^{\infty } \frac{4 \pi \log (2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}=\frac{\pi \log(2 t) (-t+\tan t)}{2 t^2} $$ $$ S_4=\sum _{k=-\infty }^{\infty } \frac{8 k \pi \log(2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}=\frac{\log(2 t) (\pi t-(\pi -2 t) \tan t)}{2 t^2} $$ $$ S_5=-\sum _{k=-\infty }^{\infty } \frac{8 t \log(2 t)}{(\pi +2 k \pi )^2 (\pi +2 k \pi -2 t)}=\frac{\log(2 t) (t-\tan t)}{t} $$ Hence $$ I=S_1+S_2+S_3+S_4+S_5-\sum^{\infty}_{k=-\infty}\frac{4\log(\pi+2k\pi-2t)}{\pi^2(2k+1)^2}+C $$ Hence $$ \int^{\tan t}_{c}\frac{dx}{\arctan x}=\log(2t)+\frac{\tan t}{t}-\sum_{k\in\textbf{Z},k-odd}\frac{4\log(\pi k-2t)}{\pi^2 k^2}+C\textrm{, }0<t<\frac{\pi}{2}.\tag 3 $$ The logarithm is defined as $\log x:=\{\log x$, $x>0$ and $i\pi+\log(-x)$, when $x<0\}$.
$$ \int\frac{dx}{\arctan x}=\log\left(2\arctan x\right)+\frac{x}{\arctan x}- $$ $$ -4\sum^{+\infty}_{k=-\infty}\frac{\log\left(\pi (2k+1)-2\arctan x\right)}{\pi^2 (2k+1)^2}+C= $$ $$ =\log\left(2\arctan x\right)+\frac{x}{\arctan x} -4\sum^{+\infty}_{k=-\infty}\frac{\log(\left|\pi (2k+1)-2\arctan x\right|)}{\pi^2 (2k+1)^2}+C,\tag 4 $$ where $x>0$.

Notes.

This is the graph of $\int^{\tan t}_{\tan 1}\frac{1}{\arctan x}dx$ and the evaluation I found where the sum is trancated at $k=5$

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At $t=e/2$ cutting the sum in $k=100$, we get

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At $t=e/2$ cutting the sum in $k=1000$, we get

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This is the graph's of $\int^{t}_{1}\frac{1}{\arctan(x)}dx$ and the evaluations and the evaluation (4):

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    $\begingroup$ @TymaGaidash. I have add some numerical results (graphical proof and approximations). $\endgroup$ – Nikos Bagis Jul 23 at 13:13
  • $\begingroup$ I see the result working, but this graph shows it slightly off. What does “$k\in \pmb Z,k-\text{odd}$” mean? Please write $k=…-5,-3,-1,1,3,5…$ or $k\in 2\Bbb Z+1$ or $k\in 2z+1, z\in \Bbb Z$. If it does go over the negatives, then complex logarithms are involved in the sum. You answer looks the simplest which is why I like it. As your answer is a definite integral, there should be no +C, unless it is the lower bound for tan(c). Finally, please write the final answer in terms of x only like Yuri. Thanks again! $\endgroup$ – Tyma Gaidash Jul 23 at 13:42
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    $\begingroup$ @TymaGaidash. Ok. I have made the changes. Thank you for your good words. $\endgroup$ – Nikos Bagis Jul 23 at 15:02
  • $\begingroup$ Here is your desmos graph. You can see your result is the most elementary one so far, so you may get the bounty. You put in great effort, and you result works. Good job. I will put your answer in the question. $\endgroup$ – Tyma Gaidash Jul 23 at 15:45
  • $\begingroup$ @TymaGaidash. I add two new graph's. Please check them. $\endgroup$ – Nikos Bagis Jul 24 at 3:59
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Long Comment

Similar idea to @YuriNegometyanov:

$$I(x)=\int\limits_1^x \dfrac{\text dt}{\arctan t}=\int_{\tan ^{-1}(1)}^{\tan ^{-1}(x)} \frac{\tan ^2(u)+1}{u} \, du,\tag1$$

$$I(x)=\int_{\tan ^{-1}(1)}^{\tan ^{-1}(x)} \frac{1}{u} \, du+\int_{\tan ^{-1}(1)}^{\tan ^{-1}(x)} \frac{\tan ^2(u)}{u} \, du,\tag2$$

$$I(x)=\log \left(\frac{16 \tan ^{-1}(x)}{\pi^2 }\right)+x^2 \log \left(\tan ^{-1}(x)\right)-2\int_{\tan ^{-1}(1)}^{\tan ^{-1}(x)} \log (u) \tan (u) \sec ^2(u) \, du,\tag3$$

The integral in (3) can also be expressed in terms of the infinite series for $\tan u$, since

$$\frac{d^2 \tan (u)}{d u^2}=2\tan (u) \sec ^2(u)=2 \sum _{k=1}^{\infty } \frac{(2 k-2) (2 k-1) \left(\left(2^{2 k}-1\right) \zeta (2 k)\right) x^{2 k-3}}{\pi ^{2 k}}$$

and

$$\int x^{2 k-3} \log (x) \, dx=\frac{ (2 (k-1) \log (x)-1)}{4 (k-1)^2}x^{2 k-2}$$

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  • $\begingroup$ @JamesArathroon Here is my graph on what seems to be the final answer, but it seems to be a bit off. Either I made a mistake or something else. Notice the summation starting at n=2 for Yuri and that my typed result is at the very bottom of the desmos graph. $\endgroup$ – Tyma Gaidash Jul 23 at 2:27
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This is not a complete answer either, but more observations. Consider

$$\int \frac{1}{\tan^{-1}(x)} dx$$

and substitute $x = \tan u$, so $dx = \sec^2 u\ du$. Then

$$\int \frac{1}{\tan^{-1}(x)} dx = \int \frac{\sec^2 u}{u}\ du$$

Now note that $\sec^2 u = 1 + \tan^2 u$ so we can write this as

$$\int \frac{\sec^2 u}{u} du = \int \frac{1}{u}\ du + \int \frac{\tan^2 u}{u}\ du = \ln u + \int \frac{\tan^2 u}{u}\ du$$

Hence, to have a form similar to the ones you've given for $\frac{1}{\sin^{-1}(x)}$ and the like, you'd need to have a new special function for either

$$\int \frac{\sec^2 u}{u}\ du$$

or

$$\int \frac{\tan^2 u}{u}\ du$$

I don't know if these can be solved in terms of

$$\int \frac{\sec u}{u}\ du$$

or

$$\int \frac{\tan u}{u}\ du$$

.

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  • $\begingroup$ It would be redundant to define a new function unless academically recognized. Even my “T(x)” function was for fun. Wolfram Alpha refuses to even try to integrate $\frac{\tan^2(x)}x$. Maybe you can find a series expansion for $\sec^2(x)$ or $\tan^2(x)$? Only one series. I used two series as seen above. These are good observations of yours. $\endgroup$ – Tyma Gaidash Jul 23 at 3:21
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This addendum will be for extra ideas found. All answers are pretty much similar and are for fun:

Note that C can be a complex number because of $\mathrm{ln(tan^{-1}(ix))=\frac{\pi i}{2}+ln(tanh^{-1}(x))}:$

  1. $$\mathrm{Th(x)=\int\frac{dx}{tanh^{-1}(x)}=\int\frac{i\,dx}{tan^{-1}(ix)}=\int\frac{2dx}{ln(1+x)+ln(1-x)}=T(ix)=\frac{x}{tanh^{-1}(x)}+ln\left(tanh^{-1}(x)\right)+\frac12\sum_{n=2}^\infty \frac{4^n\left(4^n- 1\right) B_{2n}\left(tanh^{-1}(x)\right)^{2(n-1)}}{(n-1)(2n)!}+C}$$

Specific values with expanded logarithms. Result verification and graph of Th(x) with root at x=$.376…$. For the graph, note the “i” parameter has to be infinity. Then, one takes a limit as x approaches 1 of the evaluation limits for the sum. This has to be done if an integral bound is 1 as the inverse hyperbolic tangent of 1 is 0 and approaches infinity as a result of an improper integral.

2.$$\mathrm{\int_\frac1{\sqrt{3}}^1\frac{dx}{tan^{-1}(x)}= T\left(\frac1{\sqrt3}\right)-T(1)=\frac{4-2\sqrt3}\pi+ln(3)-\ln(2)-\frac1{2\pi^2}\sum_{n=2}^\infty \frac{B_{2n}(-4)^n \pi^{2n}\left(4^n-1\right)\left(4^{2-2n}-6^{2-2n}\right)}{(n-1)(2n)!}=.6457455…}$$

3.$$\mathrm{\int_{1-\frac{2}{e+1}}^\frac12 \frac{dx}{tanh^{-1}(x)}=Th\left(1-\frac{2}{e+1}\right)-Th\left(\frac12\right)=\frac{4}{e+1}-\frac1{ln(3)}+ln(ln(3))-2+2\sum_{n=2}^\infty\frac{B_{2n}4^n\left(4^n-1\right)\left(ln^{2n-2}(3)-1\right)}{2^{2n}(n-1)(2n)!}=.0722855…}$$

Please correct me and give me feedback!

I wonder what would happen if we expanded and separated the sums…

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