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

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==Theorem==
<strong>[[Value of Gamma(1)|Theorem]]:</strong> The following formula holds:
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The following formula holds:
 
$$\Gamma(1)=1,$$
 
$$\Gamma(1)=1,$$
 
where $\Gamma$ denotes the [[gamma]] function.
 
where $\Gamma$ denotes the [[gamma]] function.
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==Proof==
<strong>Proof:</strong> Compute
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Compute using the [[fundamental theorem of calculus]],
 
$$\begin{array}{ll}
 
$$\begin{array}{ll}
\Gamma(1) &= \displaystyle\int_0^{\infty} \xi^{0} e^{-\xi} d\xi \\
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\Gamma(1) &= \displaystyle\int_0^{\infty} \xi^{0} e^{-\xi} \mathrm{d}\xi \\
&= \displaystyle\int_0^{\infty} e^{-\xi} d\xi \\
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&= \displaystyle\int_0^{\infty} e^{-\xi} \mathrm{d}\xi \\
&= \left[ -e^{-\xi} \right]_{0}^{\infty} \\
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&= \left[ -e^{-\xi} \right.\Bigg|_{0}^{\infty} \\
&= 1.
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&= 1,
 
\end{array}$$
 
\end{array}$$
 
as was to be shown. █
 
as was to be shown. █
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==References==
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* {{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