Integral #1

Prove that $\displaystyle\int_{0}^{1} \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-logx)} \mathrm{d}x\tag*{}$
Equal to
$\displaystyle log \left[\frac{\Gamma (b+c+1)\Gamma(c+a+1)\Gamma(a+b+1)}{\Gamma(a+1)\Gamma(b+1)\Gamma(c+1)\Gamma(a+b+c+1)}\right]$

Misal $\displaystyle F(c) = \int_{0}^{1} \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-logx)} \mathrm{d}x\tag*{}$

Turunkan $\displaystyle F(c)$ terhadap $\displaystyle c$ , didapat

$\displaystyle F'(c) = \int_{0}^{1} \frac{(1-x^a)(1-x^b)x^c}{(1-x)} \mathrm{d}x\tag*{}$

Senilai dengan

$\displaystyle F'(c) =  \int_{0}^{1} \frac{(1-x^a-x^b+x^{a+b})x^c}{(1-x)} \mathrm{d}x\tag*{}$

$\displaystyle F'(c) =  \int_{0}^{1} \frac{(x^c-x^{a+c}-x^{b+c}+x^{a+b+c}}{(1-x)} \mathrm{d}x$

$\displaystyle F'(c) = -\int_{0}^{1}\frac{1-x^c}{1-x} \mathrm{d}x + \int_{0}^{1} \frac{1-x^{a+c}}{1-x} \mathrm{d}x$ $\displaystyle + \int_{0}^{1}\frac{1-x^{b+c}}{1-x}\mathrm{d}x + \int_{0}^{1} \frac{1-x^{a+b+c}}{1-x} \mathrm{d}x $

Kita tahu bahwa

$\displaystyle \psi(s+1)= -\gamma + \int_{0}^{1} \frac{1-x^s}{1-x} \mathrm{d}x\tag*{}$

Sehingga bentuk diatas dapat disederhanakan menjadi

$\displaystyle F'(c) = -\psi(c+1)+\psi(a+c+1)$$\displaystyle +\psi(b+c+1)-\psi(a+b+c+1)\tag*{}$

Integral kan terhadap $\displaystyle c$ didapat :

$\displaystyle F(c) = -log[\Gamma(c+1)] + log[\Gamma(b+c+1)]$$ -log[\Gamma(a+b+c+1)] + e$

Dapat di sederhanakan menjadi

$\displaystyle log\left[\frac{\Gamma(a+c+1)\Gamma(b+c+1)}{\Gamma(c+1)\Gamma(a+b+c+1)}\right] +e \tag*{}$

Misal c = 0 , maka

$\displaystyle 0 = \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)} + e\tag*{}$

Dengan mudah didapat nilai dari e , yaitu

$\displaystyle e = -log\left[\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)}\right] \tag*{}$

Dengan begitu, kita punya

$\displaystyle F(c) = log\left[\frac{\Gamma(a+c+1)\Gamma(b+c+1)}{\Gamma(c+1)\Gamma(a+b+c+1)}\right]$
$\displaystyle -log\left[\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)}\right]$

Dengan begitu di dapat hasil dari

$\displaystyle\int_{0}^{1} \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-logx)} \mathrm{d}x\tag*{}$ adalah

$\displaystyle log \left[\frac{\Gamma (b+c+1)\Gamma(c+a+1)\Gamma(a+b+1)}{\Gamma(a+1)\Gamma(b+1)\Gamma(c+1)\Gamma(a+b+c+1)}\right]$
$\blacksquare$

Reference:advanced integration technique By Zaid Alyafeai

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Ditulis Mohammad_fatshira — Sunday, September 1, 2019 — Add Comment — Mathematics