The most ugly looking integral

Prove the following

$$I= \log \left\{\frac{\Gamma(b+c+1) \Gamma(c+a+1)\Gamma(a+b+1)}{\Gamma(a+1) \Gamma(b+1) \Gamma(c+1) \Gamma(a+b+c+1)} \right\}$$

 

where

 

$$I = \int_0^1 \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-\log x)}dx$$

$$\textit{proof}$$

First note that since there is a log in the denominator that gives as an idea to use differentiation under the integral sign.

Let

$$F(c)=\int_0^1 \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-\log x)}dx$$

Differentiate with respect to c

$$F'(c)=\int_0^1 \frac{(1-x^a)(1-x^b)x^c}{(1-x)}dx$$

By expanding

$$F'(c)=\int_0^1 \frac{x^c\,-\,x^{a+c}\,-\,x^{b+c}\,+\,x^{a+b+c}}{(1-x)}dx$$

We can add and subtract one to use the second integral representation

$$F'(c)=\int_0^1 \frac{(x^c\,-1)\,+\,(1-\,x^{a+c})\,+\,(1-\,x^{b+c})\,+\,(x^{a+b+c}-1)}{(1-x)}dx$$

Distribute the integral over the terms

$$F'(c)=-\int_0^1 \frac{1-x^c\,}{1-x}\,dx+\int_0^1\frac{1-x^{a+c}}{1-x}\,dx+\\\int_0^1\frac{1-x^{b+c}}{1-x}\,dx-\int_0^1\frac{1-x^{a+b+c}}{1-x}dx$$

Which simplifies to

$$F'(c)=-\psi(c+1)+\psi(a+c+1)+\psi(b+c+1)-\psi(a+b+c+1)$$

Integrate with respect to $c$

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

Which reduces to

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

Now put $c= 0$ we have

$$0=\log \left[ \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)}\right]+e$$

The constant

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

So we have the following

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

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