Question

Use (5.4) to show that

(a) $\mathbf{\Gamma }\mathbf{(}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{-}\mathbf{n}\mathbf{)}\mathbf{\Gamma }\mathbf{(}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{+}\mathbf{n}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{n}}\mathit{\pi }$if $\mathit{n}\mathbf{=}\mathit{a}$ positive integer

(b)$\mathbf{(}\mathit{z}\mathbf{!}\mathbf{)}\mathbf{(}\mathbf{-}\mathit{z}\mathbf{)}\mathbf{!}\mathbf{=}\mathit{\pi }\mathit{z}\mathbf{/}\mathit{s}\mathit{i}\mathit{n}\mathit{\pi }\mathit{z}$, where zis not necessarily an integer; see comment after equation (3.3).

Short answer

The following are proved.

(a)$\Gamma \left(p\right)\Gamma (1-p)=(-1{)}^n\pi$$\Gamma \left(p\right)\Gamma (1-p)=(-1{)}^n\pi$

(b)$(z!)(-z)!=\frac{\pi z}{\sin \left(\pi z\right)}$

Step-by-step solution

d

d

Given Information

Here, we need to prove that and $\Gamma \left(\frac{1}{2}-n\right)\Gamma \left(\frac{1}{2}+n\right)=(-1{)}^n\pi$and$(\mathrm{z}!)(-\mathrm{z})!=\frac{\pi \mathrm{z}}{\sin \left(\pi \mathrm{z}\right)}$.

Begin by using known identities of Gamma functions.

Use the formula to prove the two questions.

$\Gamma \left(p\right)\Gamma (1-p)=\frac{\pi }{\sin \left(\pi p\right)}$

Use$p=\frac{1}{2}-n$

$\Gamma \left(p\right)\Gamma (1-p)=\frac{\pi }{\sin \left(\frac{\pi }{2}-\pi n\right)}$

Use the trigonometric product identity to simplify (a)

Use the sine formula to expand the denominator.

$\begin{array}{c}\sin (A\pm B)=\sin A\cos B\pm \cos A\sin B\\ \sin \left(\frac{\pi }{2}-n\pi \right)=\sin \frac{\pi }{2}\cos n\pi -\cos \frac{\pi }{2}\sin n\pi \end{array}$

Apply this identity to simplify further.

$\Gamma \left(p\right)\Gamma (1-p)=\frac{\pi }{\sin \frac{\pi }{2}\cos \left(n\pi \right)-\cos \frac{\pi }{2}\sin \left(n\pi \right)}$

From trigonometric tables, use the following formulae.

$\begin{array}{c}\sin \frac{\pi }{2}=1\\ \cos \frac{\pi }{2}=0\\ \cos \left(n\pi \right)=(-1{)}^n\end{array}$

The equation can be simplified further.

$\Gamma \left(p\right)\Gamma (1-p)=(-1{)}^n\pi$

Use the relation between Gamma function and factorials (b)

The following relations between the Gamma function and factorials are known.

$\begin{array}{c}\Gamma (p+1)=p!\\ \Gamma (p+1)=p\Gamma \left(p\right)\end{array}$

Use these formulae to simplify the left hand side of the equation in the question.

$\begin{array}{c}(z!)(-z)!=\Gamma (z+1)\Gamma (-z+1)\\ (z!)(-z)!=z\Gamma \left(z\right)\Gamma (1-z)\end{array}$

Simplify using established identities.

$\Gamma \left(p\right)\Gamma (1-p)=\frac{\pi }{\sin \left(\pi p\right)}$

Replace $p=z$.

$\begin{array}{l}\Gamma \left(z\right)\Gamma (1-z)=\frac{\pi }{\sin \left(\pi z\right)}\\ (z!)(-z)!=\frac{\pi z}{\sin \left(\pi z\right)}\end{array}$

Therefore, the equation is proved.

$(\mathrm{z}!)(-\mathrm{z})!=\frac{\pi \mathrm{z}}{\sin \left(\pi \mathrm{z}\right)}$