Euler reflection formula
Theorem 1
Γ(x)Γ(1-x)=πsin(πx) |
Proof: We have
1Γ(x)=xeγx∞∏n=1((1+xn)e-x/n) |
and thus
1Γ(x)1Γ(-x)=-x2eγxe-γx∞∏n=1((1+xn)e-x/n)((1-xn)ex/n)=-x2∞∏n=1(1-x2n2) |
But Γ(1-x)=-xΓ(-x) and thus
1Γ(x)1Γ(1-x)=x∞∏n=1(1-x2n2) |
Now, using the formula (http://planetmath.org/ExamplesOfInfiniteProducts) for sinx/x, we have
sin(πx)=πx∞∏n=1(1-x2n2) |
so that
1Γ(x)1Γ(1-x)=sin(πx)π |
and the result follows.
Title | Euler reflection formula |
---|---|
Canonical name | EulerReflectionFormula |
Date of creation | 2013-03-22 16:23:37 |
Last modified on | 2013-03-22 16:23:37 |
Owner | rm50 (10146) |
Last modified by | rm50 (10146) |
Numerical id | 5 |
Author | rm50 (10146) |
Entry type | Theorem |
Classification | msc 30D30 |
Classification | msc 33B15 |