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I tried to integrate:

$$\int \tan \left(\frac 1 x\right) \, dx$$

using integration by parts, and also by universal substitution but to no avail. WolframAlpha reports that "no result found in terms of standard mathematical functions."

What could it be?

Тyma Gaidash
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Bak1139
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    Sometimes there's just no nice answer. Do you have a reason to expect that there is a closed form (or do you actually have need of one)? –  Mar 20 '18 at 17:34
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    If you write down an integral at random there's typically no way to "do" it. See https://en.wikipedia.org/wiki/Risch_algorithm – David C. Ullrich Mar 20 '18 at 17:47

1 Answers1

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I will note you never stated you wanted a closed form. I will solve the integral using a summation expansion. I will also use this reference on properties of the tangent function. $B_y$ are Bernoulli numbers:

$$\mathrm{\int tan\left(\frac1x\right)dx=\int \frac{dx}{e^\frac ix-1}-\int\frac{dx}{e^\frac ix+1}-ix=\int i+2i\sum_{n=0}^\infty(-1)^n e^\frac{2in}{x}dx=ix+ \sum_{n=0}^\infty(-1)^n 2i\int e^\frac{2in}{x}dx}$$

There is a way to integrate for $|x|\ge\frac2\pi$:

It is seen in this

reference

that:

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

graphical proof. Using the Inverse Function Integral theorem:

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

For a simplified integral of $|x|\ge\frac2\pi$,evaluate using $x^{2-2n}\to b^{2-2n}-a^{2-2n}$ and not $x^{2-2n}\to \frac{1}{b^{2n-2}-a^{2n-2}}$. This integral needs no constant for the definite integral in the graph. Unfortunately, I cannot simply using the sum of logarithm expression. Let’s name the integral t(x) for fun:

$$\mathrm{t(x)=\int tan\left(\frac1x\right)dx=C+ln(x)+\frac12\sum_{n=2}^\infty \frac{(-4)^n (4^n-1) B_{2n}x^{2-2n}}{(2n)!(n-1)}}$$ There should be others ways to integrate it though.

Also see this great answer for a modified simplified version of the other sum expression:

Integral of $\tan(1/x)$

Тyma Gaidash
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