Gamma Function evaluation problem?

Question

So we know that Γ(1/2) = √pi, 
and that Γ(x+1) = xΓ(x)

Therefore 
Γ(3/2) = Γ(1+1/2)
          = 1/2Γ(1/2)

but I'm stuck when I'm faced with Γ(-3/2) and Γ(-5/2) and it becomes much more complicated. Can someone help me out ?

? · Answer

Γ(1/2) = √π

Γ(x+1) = xΓ(x)
Γ(x) = Γ(x+1)/x

Γ(−3/2)
= Γ(−1/2)/(−3/2)
= Γ(1/2)/((−1/2)(−3/2))
= √π / (3/4)
= (4√π)/3

Γ(−5/2)
= Γ(−3/2)/(−5/2)
= (−2/5) Γ(−3/2)
= (−2/5) (4√π)/3
= −(8√π)/15

cidyah · Answer

https://gyazo.com/b9376df52c237ad6f731013b957e3b5f
Source: 
https://www.quora.com/What-is-the-gamma-function-of-1-2

Γ(x+1) = xΓ(x)
Γ(x) = Γ(x+1) /x
Γ(-3/2) = Γ(-3/2 +1) /(-3/2) = Γ(-1/2) (-2/3) = (-2/3)(-2)√π  = (4/3)√π <------

Γ(-5/2) = Γ(-5/2+1) /(-5/2) = (-2/5)Γ(-3/2) = (-2/5)(4/3)√π = (-8/15)√π <------

? · Answer

Hint: From  Γ(x+1) = xΓ(x)  -->  Γ(-1/2+1) = (-1/2)Γ(-1/2) . 
Now that you have Γ(-1/2), reiterate to get your answers...

Done!

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