Γ(z)=(z−1)!=∫∞0tz−1e−tdt⟹Γ(32)=12!=∫∞0t12e−tdt
t12=x⟹2∫∞0x2e−x2dx (2 carpanini unutmayalin en sonda)
∫∞0x2e−x2dx⟹x=u,∫xe−x2dx=∫dv⟹−12e−x2=v
∫∞0udv=uv|∞0−∫∞0vdu⟹
∫∞0x2e−x2dx=−x2e−x2|∞0+12∫∞0e−x2dx=0+12√π2
Γ(32)=12!=∫∞0t12e−tdt=2√π4=√π2
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Hata fonksiyonun tam degeri, bazi ozel sinirlar disinda, bulunamaz.
Erf(x)=2√π∫x0e−t2dt⟹
Erf(∞)=2√π∫∞0e−t2dt=1⟹∫∞0e−t2dt=√π2
∫∞0e−t2dt=√π2 oldugunu gostermek cok kolay degil ve surada gosterilmis.