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 | + | &= \left[ -e^{-\xi} \right.\Bigg|_{0}^{\infty} \\ |
&= 1, | &= 1, | ||
\end{array}$$ | \end{array}$$ | ||
as was to be shown. █ | as was to be shown. █ | ||
Revision as of 09:32, 4 June 2016
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. █
