Difference between revisions of "Gamma(1)=1"

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where $\Gamma$ denotes the [[gamma]] function.
 
where $\Gamma$ denotes the [[gamma]] function.
 
==Proof==
 
==Proof==
Compute
+
Compute using the [[fundamental theorem of calculus]],
 
$$\begin{array}{ll}
 
$$\begin{array}{ll}
 
\Gamma(1) &= \displaystyle\int_0^{\infty} \xi^{0} e^{-\xi} \mathrm{d}\xi \\
 
\Gamma(1) &= \displaystyle\int_0^{\infty} \xi^{0} e^{-\xi} \mathrm{d}\xi \\
 
&= \displaystyle\int_0^{\infty} e^{-\xi} \mathrm{d}\xi \\
 
&= \displaystyle\int_0^{\infty} e^{-\xi} \mathrm{d}\xi \\
&= \left[ -e^{-\xi} \right]_{0}^{\infty} \\
+
&= \left[ -e^{-\xi} \right.\Bigg|_{0}^{\infty} \\
 
&= 1,
 
&= 1,
 
\end{array}$$
 
\end{array}$$
 
as was to be shown. █
 
as was to be shown. █
 +
 +
==References==
 +
* {{BookReference|Special Functions for Scientists and Engineers|1968|W.W. Bell|prev=Beta|next=Gamma(z+1)=zGamma(z)}}: Theorem 2.1
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 +
[[Category:Theorem]]
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[[Category:Proven]]

Latest revision as of 19:47, 15 March 2018

Theorem

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

Proof

Compute using the fundamental theorem of calculus, $$\begin{array}{ll} \Gamma(1) &= \displaystyle\int_0^{\infty} \xi^{0} e^{-\xi} \mathrm{d}\xi \\ &= \displaystyle\int_0^{\infty} e^{-\xi} \mathrm{d}\xi \\ &= \left[ -e^{-\xi} \right.\Bigg|_{0}^{\infty} \\ &= 1, \end{array}$$ as was to be shown. █

References