Difference between revisions of "Gamma(z+1)=zGamma(z)"

From specialfunctionswiki
Jump to: navigation, search
m (Tom moved page Factorial property of gamma to Gamma(z+1)=zGamma(z): Gamma(z+1)=zGamma(z))
Line 1: Line 1:
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
+
==Theorem==
<strong>[[Factorial property of gamma|Theorem]]:</strong>
+
The following formula holds:
$$\Gamma(x+1)=x\Gamma(x), \quad x>0,$$
+
$$\Gamma(z+1)=z\Gamma(z),$$
where $\Gamma$ denotes the [[gamma]] function.
+
where $\Gamma$ denotes [[gamma]].
<div class="mw-collapsible-content">
+
 
<strong>Proof:</strong> Use [[integration by parts]] to compute
+
==Proof==
 +
Use [[integration by parts]] to compute
 
$$\begin{array}{ll}
 
$$\begin{array}{ll}
\Gamma(x+1) &= \displaystyle\int_0^{\infty} \xi^x e^{-\xi} \mathrm{d}\xi \\
+
\Gamma(z+1) &= \displaystyle\int_0^{\infty} \xi^z e^{-\xi} \mathrm{d}\xi \\
&= -\xi^x e^{-\xi}\Bigg|_0^{\infty}- \displaystyle\int_0^{\infty}  x \xi^{x-1} e^{-\xi} \mathrm{d}\xi \\
+
&= -\xi^z e^{-\xi}\Bigg|_0^{\infty}- \displaystyle\int_0^{\infty}  z \xi^{z-1} e^{-\xi} \mathrm{d}\xi \\
&= x\Gamma(x),
+
&= z\Gamma(z),
 
\end{array}$$
 
\end{array}$$
 
as was to be shown. █
 
as was to be shown. █
</div>
+
 
</div>
+
==References==
 +
* {{BookReference|Special Functions for Scientists and Engineers|1968|W.W. Bell|prev=Gamma(1)=1|next=Gamma(n+1)=n!}}: $(2.2)$
 +
 
 +
[[Category:Theorem]]
 +
[[Category:Proven]]
 +
[[Category:Justify]]

Revision as of 19:46, 15 March 2018

Theorem

The following formula holds: $$\Gamma(z+1)=z\Gamma(z),$$ where $\Gamma$ denotes gamma.

Proof

Use integration by parts to compute $$\begin{array}{ll} \Gamma(z+1) &= \displaystyle\int_0^{\infty} \xi^z e^{-\xi} \mathrm{d}\xi \\ &= -\xi^z e^{-\xi}\Bigg|_0^{\infty}- \displaystyle\int_0^{\infty} z \xi^{z-1} e^{-\xi} \mathrm{d}\xi \\ &= z\Gamma(z), \end{array}$$ as was to be shown. █

References