On the Properties of Beta Function

I have been working on the iterated properties of especially Beta Function and hope you enjoy it too. We have,

\(\int_0^1 t^{x-1}(\,1-t)\,^{y-1}dt\) \(=\)\(\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\) \(\implies\) \(\int_0^1t^{x-2}(\,1-t)\,^y\cdot\frac{t}{1-t}dt\) ,then by using the geometric series we have ,


    \implies Γ(x)Γ(y)Γ(x+y)Γ(y+1)\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)\Gamma(y+1)}= n=1Γ(x+n1)Γ(x+y+n)\sum_{n=1}^{\infty}\frac{\Gamma(x+n-1)}{\Gamma(x+y+n)} .Then taking natural logarithm of both sides yields,

lnΓ(x)\ln \Gamma(x) +lnΓ(y)\ln \Gamma(y)-lnΓ(x+y)\ln \Gamma(x+y)-lnΓ(y+1)\ln \Gamma(y+1)= lnn=1Γ(x+n1)Γ(x+y+n)\ln \sum_{n=1}^{\infty}\frac{\Gamma(x+n-1)}{\Gamma(x+y+n)} .Then taking d.w.r.x,

ψ(x)ψ(x+y)=Γ(x+y)Γ(y+1)Γ(x)Γ(y)n=1Γ(x+n1)Γ(x+y+n)(ψ(x+n1)ψ(x+y+n))\boxed{\psi(x)-\psi(x+y)=\frac{\Gamma(x+y)\Gamma(y+1)}{\Gamma(x)\Gamma(y)}\sum_{n=1}^{\infty}\frac{\Gamma(x+n-1)} {\Gamma(x+y+n)}(\psi(x+n-1)-\psi(x+y+n))} as the result. Any mistake spotted then please inform or any unknown step that you don't understand.(I wasn't able to find the code for infinity 'cause this is my first latex!!!)

Note by Aruna Yumlembam
1 year, 1 month ago

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For infinty, use: \infty. - It looks like this: \infty @Aruna Yumlembam

A Former Brilliant Member - 1 year, 1 month ago

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Aruna Yumlembam - 1 year, 1 month ago

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