Is it possible to express $(\sin(x))^x$ as a complex number, if it is, then how is it done?

Question

I have looked at the graph and I know it is real for all $x$ that satisfy $2\pi \mathbb{Z} <x<2 \pi (\mathbb{Z}+1)$ and imaginary or complex for all other $x$. I want to know if it is possible to express this equation as $f(x)+g(x) i$ where both $f(x)$ and $g(x)$ are real for all real domain.

Answer 1

The trouble you run into is that $b^a$ generally needs to be handled as a multi-valued expression in contexts like this where $a$ is not an integer and $b$ is not a positive real.

What this means is that for most values of $x$ there are many different complex numbers that all have about equally good claims to be $\sin(x)^x$. When $\sin(x)\ge 0$ one of those choices will be real, and it's useful to declare that to be the value of $\sin(x)^x$, but we're not so lucky for the $\sin(x)<0$.

If we really have to select one of the choices, there's one of them called the "principal value" of the expression -- but note that this is still a quite arbitrary choice (and especially so when the base of the exponent is a negative real).

With all these caveats, deciding to use the principal value leads us to (for real $x$ only): $$ \sin(x)^x = |\sin(x)|^x\cos(\theta(x)) + i |\sin(x)|^x\sin(\theta(x)) $$ where the helper function $\theta$ can be defined as $$ \theta(x) = \begin{cases} 0 & \text{when } \sin(x) \ge 0 \\ \pi x & \text{when } \sin(x) < 0 \end{cases} $$

(The other possible values arise by adding some multiple of $2\pi x$ to $\theta(x)$).

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How does this work? We start by defining $e^z$ for all complex numbers $z$, giving the quite robustly defined complex exponential function $$ e^{x+iy} = e^x\cos(y) + i e^x\sin(y) $$ Then we can define $b^a$ to mean $e^{ca}$ where $c$ is some complex number such that $e^c=b$. However, there are always many choices for $c$ that satisfy this -- and that is where the multi-valuedness $c$ comes in. By convention the "principal" $c$ means the one whose imaginary part is in $(-\pi,\pi]$.

When $\sin x=0$ this definition won't work. We get at least a continuous function here if we follow common convention and set $\sin(0)^0=0^0=1$ but $\sin(n\pi)^{n\pi} = 0^{n\pi}=0$ for every positive integer $n$.

Answer 2

We generalize the logarithm to be $$ \log x=\log|x|+i\theta, $$ where $\theta=0$ if $x>0$ and $\theta=\pi$, when $x<0$.

It is $\log(\sin x)=\log|\sin(x)|+i\theta$, where $\theta=0$ if $x\in(2k\pi,2k\pi+\pi)=D_1(k)$, $k\in\textbf{Z}$ and $\theta=\pi$ if $x\in(2k\pi+\pi,2k\pi+2\pi)=D_2(k)$.

Hence when $k\neq 0$, we have $$ \sin(x)^x=e^{x\log(\sin x)}=|\sin x|^x\in\textbf{R}\textrm{, if }x\in D_1(k) $$ and $$ \sin(x)^x=e^{x\log(\sin x)}=e^{x\log|\sin x|+ix\pi}=|\sin x|^{x}e^{i\pi x}\textrm{, if }x\in D_2(k) $$

For the case $x\rightarrow 0$, we have

1. When $x\rightarrow 0^{+}$: $$ \frac{2 x}{\pi}<\sin x\leq x\textrm{, }\forall x\in \left(0,\frac{\pi}{4}\right).\tag 1 $$ Hence $$ x\log\left(\frac{2x}{\pi}\right)<x\log\sin x<x \log x\textrm{, }\forall x\in\left(0,\frac{\pi}{4}\right) $$ Hence knowing that $\lim_{x\rightarrow{0}^+}x\log x=0$, we get $$ \lim_{x\rightarrow 0^{+}}x\log\sin x=0.\tag 2 $$ and consequently $$ \lim_{x\rightarrow 0^+}(\sin x)^x=1 $$ 2. When $x\rightarrow 0^{-}$: We have $(\sin x)^x=e^{x\log|\sin x|+ix\pi}=e^{x\log\sin(-x)+ix\pi}\textrm{, }\forall x\in\left(-\frac{\pi}{4},0\right) $ and if we set $h=-x>0$, then $(\sin x)^x=e^{-h\log \sin h-ih\pi}$, $\forall h\in\left(0,\frac{\pi}{4}\right)$. Hence in the same way we get $$ \lim_{x\rightarrow 0^{-}}x\log(|\sin x|)=\lim_{x\rightarrow 0^{-}}x\log(\sin(-x))=-\lim_{h\rightarrow 0^{+}}h\log\sin h=0. $$ and consequently $$ \lim_{x\rightarrow 0^{-}}(\sin x)^x=1 $$ 3. When $x\rightarrow \pi^{+}$, we get that $\lim_{x\rightarrow \pi^{+}}\log\sin x=-\infty$ and consequently $$ \lim_{x\rightarrow \pi^{+}}(\sin x)^x=0. $$ and when $x\rightarrow \pi^{-}$, we get that $\lim_{x\rightarrow \pi^{-}}\log\sin x=-\infty$ and consequently $$ \lim_{x\rightarrow \pi^{-}}(\sin x)^x=0. $$

4. The cases $x\rightarrow 2\pi^{-}$ and $x\rightarrow 2\pi^{+}$ are the same with 3. $$ \lim_{x\rightarrow 2\pi^{+}}(\sin x)^x=\lim_{x\rightarrow 2\pi^{-}}(\sin x)^x=0. $$