Is there a relationship between $\arctan(x)$ and $\ln(x)$ involving only real numbers?
I am aware that the equation $$\arctan x=\frac12i[\ln(1-ix)-\ln(1+ix)]$$ holds, but it uses the imaginary number “$i$” and its derivation involves the Euler equation which “defines” a relation between the exponential function and trigonometric functions. But without assuming Euler’s formula, is there a relation between $\arctan(x)$ and $\ln(x)$ over the real numbers only?
The question is rather vague (which, I suspect, may be one of the reasons why it was closed on MO). However, one nontrivial way of making it precise is to ask whether $\arctan$ and $\log$ are first-order definable (with parameters) from each other over the ambient structure of the real field $(\mathbb R,0,1,+,\cdot,<)$: that is, if one can express the relation $\log x=y$ by means of a formula built from equalities and inequalities between terms composed of variables, $+$, $\cdot$, $\arctan$, and real constants (and possibly also $-$ and $/$), using propositional connectives $\land,\lor,\neg$, and quantification over reals $\exists z\in\mathbb R$, $\forall z\in\mathbb R$; and vice versa. (This is a rather generous reading of “definable”.)
The answer to one half of the question is that log is not first-order definable from arctan. This can be shown by considering the structure $\mathbb R_{\mathrm{an}}$ of restricted analytic functions: this is the expansion of the real field with all functions $f\colon[0,1]^n\to[0,1]$ that are real-analytic on an open neighbourhood of $[0,1]^n$. (For details, see e.g. van den Dries, Macintyre, Marker: The elementary theory of restricted analytic fields with exponentiation, Annals of Mathematics 140 (1994), 183–205, JSTOR doi 10.2307/2118545.)
On the one hand, $\arctan$ is definable in $\mathbb R_{\mathrm{an}}$: specifically, it is easy to define from the restricted analytic function $\arctan\restriction[0,1]$ using $\arctan x=\frac\pi2-\arctan x^{-1}$ for $x\ge1$ and $\arctan x=-\arctan(-x)$ for $x<0$.
On the other hand, the structure $\mathbb R_{\mathrm{an}}$ is known to be polynomially bounded (every function $f\colon\mathbb R\to\mathbb R$ definable in $\mathbb R_{\mathrm{an}}$ satisfies $|f(x)|\le p(x)$ for large enough $x$, where $p$ is a polynomial). Thus, $\exp$ is not definable in $\mathbb R_{\mathrm{an}}$, hence $\log$ is not definable in $\mathbb R_{\mathrm{an}}$ either, and a fortiori it is not definable in $(\mathbb R,0,1,+,\cdot,<,\arctan)$.
I fully expect that conversely, $\arctan$ is not first-order definable from $\log$, but I do not know if there is a simple argument (modulo known results) as above.