Difference between revisions of "Gamma(1)=1"
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&= \displaystyle\int_0^{\infty} e^{-\xi} d\xi \\ | &= \displaystyle\int_0^{\infty} e^{-\xi} d\xi \\ | ||
&= \left[ -e^{-\xi} \right]_{0}^{\infty} \\ | &= \left[ -e^{-\xi} \right]_{0}^{\infty} \\ | ||
| − | &= 1 | + | &= 1, |
\end{array}$$ | \end{array}$$ | ||
as was to be shown. █ | as was to be shown. █ | ||
</div> | </div> | ||
</div> | </div> | ||
Revision as of 05:41, 16 May 2016
Theorem: The following formula holds: $$\Gamma(1)=1,$$ where $\Gamma$ denotes the gamma function.
Proof: Compute $$\begin{array}{ll} \Gamma(1) &= \displaystyle\int_0^{\infty} \xi^{0} e^{-\xi} d\xi \\ &= \displaystyle\int_0^{\infty} e^{-\xi} d\xi \\ &= \left[ -e^{-\xi} \right]_{0}^{\infty} \\ &= 1, \end{array}$$ as was to be shown. █
