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I need to develop $\mathrm{ln}(x)$ into series, where $x \geq 1$, and I don`t know how? In literature I only found series of $\mathrm{ln}(x)$, where:

  1. $|x-1| \leq 1 \land x \neq 0$, $ \,\,\,\,\, \mathrm{ln}(x) = x - 1 - \dfrac{(x-1)^2}{2} + ...$

  2. $|x| \leq 1 \land x \neq -1$, $ \,\,\,\,\, \mathrm{ln}(x+1) = x - \dfrac{x^2}{2}+ ...$

My problem is problem in area of fluid dynamics, and $x$ is non-dimensional coordinate and it signifies radial coordinate of annular tube (it starts in the center of the tube). At the wall of inner tube $x=1$, and at the wall of outer tube it only can be larger (and values ​​are not limited), because of that I need to fulfill a condition $x \geq 1$, for developing $\mathrm{ln}(x)$ into series.

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Additional explanation:

I am looking for series of $\mathrm{ln}(x)$, where $x \geq 1$, because I have two integrals that I can't solve. That is, I can solve it in Wolfram Mathematica (I don't know how to solve it analyticaly), but in obtained solution appear Gamma and ExpIntegralEi function. These functions are formulated over an imaginary number $z=x+iy$, and the solutions of my integrals represents the mass flow and the pressure field in the pipe. So, in my solution I got Gamma function, formulated over $z=x+iy$, on the other side I am sure that solution of mass flow does not contain any imaginary number. Maybe the right question is: what is the behavior of Gamma and ExpIntegralEi function? Could I look at these functions as black boxes? And if input is real number ($0i$) will the output be a real number?

Integrals:

$$\int \dfrac{ r^3 + r \mathrm{ln}r}{\mathrm{ln}r}\mathrm{d}r$$ $$\int \dfrac{r}{(\mathrm{ln}r)^{0.5} }\mathrm{d}r$$

nick_name
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    There is no power series which only converges for $x \ge 1$. A power series which converges for arbitrarily large positive $x$ must also converge for arbitrarily large negative $x$ - and there is no such series for $\ln$. What do you need the series for? We may be able to help solve your real problem. – preferred_anon Nov 15 '21 at 15:01
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    I concur with preferred_anon here. The domain of $\ln$ is ${x\in\mathbb{R},:,x\gt0}$. Due to the natural boundary at 0, any series expansion of $\ln$ can only have a finite radius of convergence. If the center of expansion is $p$, then the radius of convergence is also $p$, meaning that the series converges for $0\lt{x}\leq2p$. But as such, it is not possible to give an expansion for $\ln$ that converges for every $x\geq1$. There is also no non-trivial asymptotic expansion for $\ln$. So you will need to provide more context here. – Angel Nov 15 '21 at 15:27
  • Are you restricting yourself to Laurent series? – A rural reader Nov 15 '21 at 15:58
  • I added explanation why I am looking for series of ln(x), shortly: because I have two integrals with ln(x) that I cannot solve. – nick_name Nov 15 '21 at 17:35

2 Answers2

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If I understand correctly, I am allowed to assume that for all $y$ such that $1 \leq y < 2$, you are able to satisfactorily express $\ln(y)$ as a series.

Given $x$ such that $2 \leq x$, let $r$ denote the largest positive integer such that $2^r \leq x$.

Define $\displaystyle y = \frac{x}{2^r} \implies 1 \leq y < 2.$

Further, $\ln(y) = \ln(x) - r\ln(2)$.

Therefore, since $\ln(2)$ is a constant, as long as it is acceptable to be able to assume that $r$ may be computed in terms of $x$, then you are done.

One approach that you may or may not consider acceptable is to define $r$ as $\displaystyle \left\lfloor \frac{\ln(x)}{\ln(2)}\right\rfloor,$ where $\lfloor a\rfloor$ denotes the largest integer $\leq a.$

user2661923
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Render $y=(x-1)/(x+1), \therefore x=(1+y)/(1-y)$. With the aid of the familiar Maclaurin series render

$\ln x =\ln(1+y)-\ln(1-y)=[y-\dfrac{y^2}2+\dfrac{y^3}3-...]-[-y-\dfrac{y^2}2-\dfrac{y^3}3-...]$

$=2y+\dfrac{2y^3}3+\dfrac{2y^5}5+...$

$=\dfrac{2(x-1)}{x+1}+\dfrac{2(x-1)^3}{3(x+1)^3}+\dfrac{2(x-1)^5}{5(x+1)^5}+...$

which by the ratio test converges for all positive $x$. Also for all positive $\Re(x)$ in the complex domain.

Oscar Lanzi
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