# Solve for $a$ as a function of $\beta$?

I am trying to solve this equation:

$$\beta^{-a} \Gamma(a) \sin(a \pi) + e^\beta \beta^{2 a - 1} \Gamma(1 - a) \sin(a \pi) = 0$$

I tried the following:

Solve[β^-a Gamma[a] Sin[a π] + E^β β^(2 a - 1) Gamma[1 - a] Sin[a π] == 0, {a, β}]


However this is not working. Can you help me?

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First, this is a single equation in two unknowns so we'd expect infinitely many solutions. So, let's try to solve for one variable in terms of the other:

Solve[\[Beta]^-a Gamma[a] Sin[a \[Pi]] +
E^\[Beta] \[Beta]^(2 a - 1) Gamma[1 - a] Sin[a \[Pi]] ==
0, \[Beta]]


That message and the appearance of the ProductLog indicates that we simply might not be able to do much better. If you try to solve for $a$, as you asked for, the empty list is returned; not surprising, as you've got an $a$ in an exponent, a $\sin$, and a $\Gamma$ function.

There certainly are solutions which we can see:

ContourPlot[E^\[Beta] \[Beta]^(-1 + 2 a)
Gamma[1 - a] Sin[a \[Pi]] + \[Beta]^-a Gamma[a] Sin[a \[Pi]] == 0,
{a, -3, 3}, {\[Beta], 0, 2}, AspectRatio -> Automatic]


Using that image, we could certainly find good starting points for FindRoot to get good numerical values, if you like.

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And a == 0 OR b == 0 – belisarius has settled Sep 19 '12 at 6:45
@belisarius The l.h.s. is undefined at all integral values of $a$, including $0$ (although it does have removable singularities there). It is not defined for $\beta=0$ because one or the other of the exponents $-a$ and $2a-1$ will be negative. – whuber Sep 19 '12 at 18:34
@whuber Yep. Thanks! I shouldn't try to do maths when I'm supposed to sleep :( – belisarius has settled Sep 19 '12 at 18:44

A quick annotation:

I'm always sceptic when a weird function is spitted out by any mathematics program. With this in mind, I took the liberty to do the following analysis:

Suppose $a \not \in \mathbb{Z}$, and $\beta \neq 0$. Using Euler's reflection formula

$$\Gamma(a) \Gamma(1-a) = \frac{\pi}{\sin \pi a}$$

is easy to see that

$$\beta^{-a} \Gamma(a) \sin(a \pi) + e^\beta \beta^{2 a - 1} \Gamma(1 - a) \sin(a \pi) = 0$$

transforms into

$$1 + \frac{\pi}{\Gamma^2(a)\sin (\pi a)}e^\beta \beta^{3a-1} = 0.$$

Then, for $\beta > 0$, there are solutions when $(2n - 1) < a < 2n$, for $n \in \mathbb{Z}$. Now, what happens when $a \rightarrow n$?

If $-1 < n$ then

$$\lim_{a \rightarrow n} \bigl(\beta^{-a} \Gamma(a) \sin(a \pi) + e^\beta \beta^{2 a - 1} \Gamma(1 - a) \sin(a \pi)\bigl) = \frac{\pi}{(n-1)!} \beta^{2n-1}e^\beta,$$ so $\beta$ should be equal to zero, but then, in order to apply the implicit function theorem, the limits in the reverse order should exist, which is not the case.

If $n < -1$ the analog argument applies.

Now, if $\beta < 0$, the condition that $a \in \mathbb{R}$ implies that there is no solution at all.

Whit all this in mind, the implicit function theorem will guaranty that $a =a(\beta)$ as long as

$$\partial_a \left\{\frac{\pi}{\Gamma^2(a)\sin (\pi a)}e^\beta \beta^{3a-1} \right\}\neq 0$$

and $\beta = \beta(a)$ if we take $\partial_\beta\{\}$ instead. With a bit more of effort, one can show that the ContourPlotshown by Mark is qualitatively correct (not that it wasn't in the first place, but since the original expression is so obscure, it's hard to see -at least for me- if there aren't any artifacts behind the plot), and continue to work with confidence.

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