TAOCP 1.2.5 Exercise 9

Exercise 9. [**] [M10] Determine the values of $\Gamma(\tfrac12)$ and $\Gamma(-\tfrac12)$, given that $(\tfrac12)! = \sqrt{\pi}/2$.

x 10. [HM20] Does the identity $\Gamma(x+1)=x\Gamma(x)$ hold for all real numbers $x$? (See exercise 7.)

Verified: yes
Solve time: 3m35s

Using the extension of the factorial function,

$$ x! = \Gamma(x+1). $$

The given value

$$ \left(\frac12\right)! = \frac{\sqrt{\pi}}{2} $$

therefore implies

$$ \Gamma!\left(\frac32\right)=\frac{\sqrt{\pi}}{2}. $$

Now apply the Gamma functional equation

$$ \Gamma(x+1)=x,\Gamma(x). $$

With $x=\tfrac12$,

$$ \Gamma!\left(\frac32\right) =\frac12,\Gamma!\left(\frac12\right). $$

Substituting the known value of $\Gamma(\tfrac32)$,

$$ \frac{\sqrt{\pi}}{2} =\frac12,\Gamma!\left(\frac12\right). $$

Multiplying both sides by $2$ gives

$$ \Gamma!\left(\frac12\right)=\sqrt{\pi}. $$

Next, use the functional equation with $x=-\tfrac12$:

$$ \Gamma!\left(\frac12\right) =\left(-\frac12\right)\Gamma!\left(-\frac12\right). $$

Substituting $\Gamma(\tfrac12)=\sqrt{\pi}$,

$$ \sqrt{\pi} =\left(-\frac12\right)\Gamma!\left(-\frac12\right). $$

Solving for $\Gamma(-\tfrac12)$,

$$ \Gamma!\left(-\frac12\right) =\frac{\sqrt{\pi}}{-1/2} =-2\sqrt{\pi}. $$

Therefore,

$$ \boxed{\Gamma!\left(\frac12\right)=\sqrt{\pi}}, \qquad \boxed{\Gamma!\left(-\frac12\right)=-2\sqrt{\pi}}. $$