Difference between revisions of "Gamma(1)=1"
From specialfunctionswiki
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$\Gamma(1)=1,$$ | $$\Gamma(1)=1,$$ | ||
where $\Gamma$ denotes the [[gamma]] function. | where $\Gamma$ denotes the [[gamma]] function. | ||
| − | + | ==Proof== | |
| − | + | Compute | |
$$\begin{array}{ll} | $$\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 \\ | ||
| Line 12: | Line 12: | ||
\end{array}$$ | \end{array}$$ | ||
as was to be shown. █ | as was to be shown. █ | ||
| − | |||
| − | |||
Revision as of 09:31, 4 June 2016
Theorem
The following formula holds: $$\Gamma(1)=1,$$ where $\Gamma$ denotes the gamma function.
Proof
Compute $$\begin{array}{ll} \Gamma(1) &= \displaystyle\int_0^{\infty} \xi^{0} e^{-\xi} \mathrm{d}\xi \\ &= \displaystyle\int_0^{\infty} e^{-\xi} \mathrm{d}\xi \\ &= \left[ -e^{-\xi} \right]_{0}^{\infty} \\ &= 1, \end{array}$$ as was to be shown. █
