# Can the gamma function be generalized to quaternions and how? [duplicate]

The gamma function is a generalization of the operator !n. The question is: Can the concept of the gamma function be generalized to quaternion analysis and the use of quaternions, and how?

• math.stackexchange.com/a/3290693/269764. You generally don't end up with anything new when extending holomorphic functions with real power series to the quaternions. (Note Gamma is holomorphic away from the poles) Commented Jul 17 at 7:56
• Following up on the connection with zeta functions, you can think of Gamma as a special function arising as an integral over $\mathbb R$ related to the zeta function. There is also an analogue for $\mathbb C$, and $\mathbb H$, but here $s$ is still complex, and you just get some elementary expression in terms of the usual $\Gamma$-function. Commented Jul 18 at 4:50

The quaternions (1,i,k,l) form an abstract representation of the Lie algebra of the orthogonal group. Its lowest dimensional linear representation are the imaginary multiples of the Pauli matrices. Any function of the Pauli algebra is defined by the power series with matrix powers. Since any function of linear combinations of the Pauli matrice, that can be diagonalized, reduces to a function of $$(\mathbb 1_2m \sigma_3)$$ by a rotation matrix in $$\mathit {SU}_2(\mathbb C)$$

So calculate

$$\int_0^\infty e^{-t} \left(t(a \ \mathbb 1 + i b \sigma_3\right))^{\alpha-1} dt$$

$$= \int_0^\infty \ \left( \begin{array}{cc} e^{-t} \ (t(a+ i b))^{\alpha-1} & 0\\ 0 & e^{-t}\ (t(a- i b))^{\alpha-1} \end{array}\right) \ dt = \left( \begin{array}{cc} \Gamma (\alpha ) (a+i b)^{\alpha -1} & 0 \\ 0 & \Gamma (\alpha ) (a-i b)^{\alpha -1} \\ \end{array} \right)$$

$$\mathbf \Gamma_{a +i b \mathbf k}(n) = \Gamma(n) \ (a + b \mathbf k)^{n-1}$$

So for a domain of linear combinations of the quaternions where power series converge, one has an algebraic extension from the positive real line to matrices with positive real parts of the eigenvalues. This algebra of extensions obeys the functional equation in the given direction.

$$\mathbf \Gamma_{q}(n) = \Gamma(n) \ q^{n-1}$$