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As per title, I am looking for the solution of the integral involving the trigonometric cot function.

The problem comes from a differential equation that I derived for a time-dependent variable $x\equiv x(t)$, \begin{equation} x' = 2 B_1 x - \sum_{k=1}^n (-1)^k \frac{2^{2k}B_{2k}}{(2k)!} x^{2k} = x - x \, \text{cot}(x) + 1, \end{equation} where $B_j$ are Bernoulli numbers, $B_1=1/2$ and we used the connection to the series expansion of cot. Proceeding with the separation of variables we obtain \begin{equation} \int \frac{1}{x-x \, \text{cot}(x)+1} \, dx = \int dt= t+ C, \end{equation}

Similarly, if we consider $B_1=-1/2$ (the first Bernoulli number can be either positive or negative) we obtain (up to a global minus sign) the slightly different variant of the LHS, namely, \begin{equation} \int \frac{1}{x+x \, \text{cot}(x)-1} \, dx . \end{equation} Online calculators have not been helpful and no valid approach to tackle the integrals comes to my mind. Is at least one of these two integrals solvable?

edit: what I find surprising is that substituting cot with coth Wolfram outputs an explicit solution: \begin{equation} \int \frac{1}{x-x \, \text{coth}(x)+1} \, dx = 1/2 [x + \log(x \, \text{cosh}(x) - (1 + x)\, \text{sinh}(x))]+C \end{equation}

Can this somehow help? (considering $\text{coth}(x) = i \text{cot}(ix)$)

Тyma Gaidash
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EmFed
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    What's the source of this integral? – Gwen Oct 08 '24 at 18:07
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    The integrands can be written in terms of one another. Reciprocals of a mix of trigonometric and non-trigonometric functions seem to have no closed form, but maybe substituting $y=x+x\cot(x)-1$ and solving for $x$ as a series/integral would help/ – Тyma Gaidash Oct 08 '24 at 18:14
  • Adding 3 elements in the denominator can be tricky for an integrand. The integral might not even be expressible by standard functions. See : https://math.stackexchange.com/questions/612879/integral-int-1-infty-fracdx12x3x

    and that is not even an indefinite integral.

    Nevertheless I will try to solve this one ...

     – mick Oct 08 '24 at 18:33
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    I see two close votes. perhaps add some source or motivation. I like it and upvoted btw. – mick Oct 08 '24 at 18:35
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    Not an answer but the similar looking $\int (1 + \cot(x))^{-1} dx = 1/2(x- \ln(\sin(x) + \cos(x)))$ might help. Maybe the functional inverse of $x + x\cot(x)$ or $x + x\cot(x) -1$ could be part of the solution. Just thinking out loud – mick Oct 08 '24 at 19:07
  • If I am not mistaken Ramanujan considered the functional inverse Tyma and I have mentioned. But I was not able to find a reference. – mick Oct 08 '24 at 19:10
  • @mick I have edited my question with some more context. Although, the problem is about the integral... I am not very familiar with integrals and "functional inverse", could you please elaborate a bit? – EmFed Oct 08 '24 at 19:55
  • @Gwen I added some more context, in short it comes from a differential equation involving aBernoulli numbers series that imply the cot function. – EmFed Oct 08 '24 at 19:58
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    @EmFed I think your last integral is indeed the key to the solution ! – mick Oct 08 '24 at 20:32
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    Im not sure if this is solvable but the edit shows hope , effort , context and motivation. I already upvoted and now I voted to reopen and I suggest others to do the same ! – mick Oct 08 '24 at 20:46
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    @mick thanks a lot for championing the reopening of my question :) So you think that the integral with the hyperbolic function coth can shed light? It is true that coth(x)= i cot(ix), but I cannot see a clear way through... – EmFed Oct 08 '24 at 20:49
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    I don't understand the idea of writing $\coth(x)$ as $i \cot (ix)$, I would rather go for the $e^x$-related definition and get an $e^x$ behind the $d$ ($d e^x$). – Dominique Oct 09 '24 at 12:10
  • This might help? $$\frac12\int\frac{(x\sin x-x\cos x+\sin x)+(x\cos x+x\sin x+\sin x)-2x\sin x}{x\sin x-x\cos x+\sin x}\mathrm dx$$ – Integreek Dec 13 '24 at 16:50
  • @MathGuy I imagine that the integral you wrote is related to my problem, but I cannot clearly identify the direct connection? Also, I tried to feed your input into some online integral calculators: apparently this doesn't have a closed form solution either... – EmFed Dec 13 '24 at 21:53
  • Using this relation and multivalued product log function $W_k(z)$ with $x=\operatorname{Im}(W(t))$ gives the integral as $\frac{i-1}2\int \frac{d(W_{-1}(t)-W(t))}{W(t)+iW_{-1}(t)+1+i},t<-\frac1e$ – Тyma Gaidash Dec 16 '24 at 02:11
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    The integral can be written in terms of $\int\frac {dx}{1-\cot(x)+\frac1x}$ and there is a fairly complex expansion for the inverse of $\frac1x-\cot(x)$, but it would give a triple sum solution to the integral. Is this answer wanted? – Тyma Gaidash Dec 19 '24 at 01:45
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    nice question ........... +1 – TShiong Dec 20 '24 at 02:04
  • @ТymaGaidash indeed, I currently Taylor-expand the inverse of the denominator and then integrate the terms individually. This gives a good approximation, but I was hoping that we can find an exact, closed form solution... – EmFed Dec 20 '24 at 09:43
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    @EmFed I just multiplied the numerator and the denominator by $\sin x$, then tried to create the derivative of the denominator in the numerator. Then, all you remains is to handle the third term in the numerator, i.e., $x\sin x$, i.e., we need to evaluate $\displaystyle\int\frac{x,\mathrm dx}{x-x\cot x+1}$. – Integreek Feb 14 '25 at 15:44
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    I still believe it can be done ... – mick Mar 30 '25 at 22:47

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