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 ,

Γ(x)Γ(y)Γ(x+y)\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}==∑n=1∞Γ(x+n−1)Γ(y+1)Γ(x+y+n)\sum_{n=1}^{\infty}\frac{\Gamma(x+n-1)\Gamma(y+1)}{\Gamma(x+y+n)}

  ⟹  \implies Γ(x)Γ(y)Γ(x+y)Γ(y+1)\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)\Gamma(y+1)}= ∑n=1∞Γ(x+n−1)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,

ln⁡Γ(x)\ln \Gamma(x) +ln⁡Γ(y)\ln \Gamma(y)-ln⁡Γ(x+y)\ln \Gamma(x+y)-ln⁡Γ(y+1)\ln \Gamma(y+1)= ln⁡∑n=1∞Γ(x+n−1)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=1∞Γ(x+n−1)Γ(x+y+n)(ψ(x+n−1)−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!!!)

#Calculus

Note by Aruna Yumlembam
1 year ago

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Comments

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

A Former Brilliant Member - 1 year ago

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Thanks!

Aruna Yumlembam - 1 year ago
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