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I've recently been searching around the web for an inverse of the zeta function, $ \zeta^{-1}(s) $, somewhat unsuccessfully. I then came across $$ \zeta(s) = \frac{\eta(s)}{1-2^{1-s}}, $$ where $$\eta(s) = \sum^\infty_{k=1} \frac{-1^{k-1}}{k^s}$$ I then manipulated it like so:

$$\begin{align*} \zeta(s) &= \frac{\eta(s)}{1-2^{1-s}} \\[0.1in] \frac{1}{\zeta(s)} &= \frac{1-2^{1-s}}{\eta(s)} \\[0.1in] \frac{\eta(s)}{\zeta(s)} &= 1-2^{1-s} \\[0.1in] \frac{\eta(s)}{\zeta(s)} - 1 &= -2^{1-s} \\[0.1in] -\frac{\eta(s)}{\zeta(s)} - 1 &= 2^{1-s} \\[0.1in] \log_2\Bigl(-\frac{\eta(s)}{\zeta(s)} - 1\Bigr) &= 1-s \\[0.1in] -\log_2\Bigl(-\frac{\eta(s)}{\zeta(s)} - 1\Bigr) -1 &= s \end{align*}$$

So there you go? An inverse of the zeta function. However, when I input it into Wolfram Alpha, it doesn't work, as in:

$$ \zeta^{-1}(2) \approx -1.58496-4.53236i $$ $$ \zeta(-1.58496 + -4.53236i) = 0.282432... - 0.377775... i $$

Any help?

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    One could attempt Lagrange Inversion Theorem. – 2017-02-27

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You made an arithmetic mistake in multiplying both sides by $-1$:

$$\frac{\eta(s)}{\zeta(s)} - 1 = -2^{1-s} \implies -\frac{\eta(s)}{\zeta(s)} + 1 = 2^{1-s} $$

(By the way, you haven't really produced an inverse of the Riemann zeta function, since your formula depends on the value of $s$ itself, in order to plug into $\eta$.)

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    Surely then I could rewrite it as – 2017-02-27
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    $$ \eta(s) = (1-2^{1-s})\zeta(s) $$ – 2017-02-27
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    @OllyBritton: That still depends on $s$. – 2017-02-27
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    But surely to evaluate $\zeta^{-1}(s)$ all you need to do is plug it into the formula and you can calculate the values of $\zeta(s)$ yourself? – 2017-02-27
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    @OllyBritton: Plug **what** into the formula? Remember, the whole point of the inverse of a function is that I can tell you a number, $x$, and you can tell me which number $s$ has the property that $\zeta(s)=x$. – 2017-02-27
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    Note that using $\eta(s)=(1-2^{1-s})\zeta(s)$ also makes your formula a complete tautology, because the $\zeta(s)$ cancels and has no impact on anything: $$-\log_2\Bigl(-\frac{(1-2^{1-s})\zeta(s)}{\zeta(s)} + 1\Bigr) -1 = s$$ $$-\log_2\bigl(-(1-2^{1-s}) + 1\bigr) -1 = s$$ $$-\log_2(2^{1-s}) -1 = s$$ $$-(1-s)-1=s$$ $$s=s$$ Your formula takes in a value of $\zeta(s)$ and then never uses it for anything. – 2017-02-27
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    Also, maybe the point I should have started with is that the Riemann zeta function is not injective, and therefore does not have an inverse everywhere. For example, there are many known values of $s$ for which $\zeta(s)=0$, so there is no way to define $\zeta^{-1}(0)$. – 2017-02-27