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Question Number 116796 by mnjuly1970 last updated on 08/Oct/20

$$...calculus...$$$$$$$$provethat::$$$$$$$${\int }_0^1\frac{(1-x^p)(1-x^q)x^{r-1}}{log\left(x\right)}dx\stackrel{???}{=}log\left(\frac{(p+q+r+1)r}{(p+r)(q+r)}\right)$$$$$$$$m.n.1970$$$$$$

Answered by mindispower last updated on 08/Oct/20

$$letf\left(a\right)={\int }_0^1\frac{(1-x^p)(1-x^q)x^{r-1+a}}{log\left(x\right)}dx$$$$f^{\prime }\left(a\right)={\int }_0^1\frac{(1-x^p)(1-x^q)x^{r-1+a}}{log\left(x\right)}.log\left(x\right)dx$$$$={\int }_0^1(x^{r+a-1}+x^{p+q+r-1+a}-x^{p+r-1+a}-x^{q+r-1+a})dx$$$$=\frac{1}{r+a}+\frac{1}{r+q+a+1}-\frac{1}{p+r+a}-\frac{1}{q+r+a}$$$$wewantf\left(0\right)$$$$f\left(a\right)=\int (\frac{1}{r+a}+\frac{1}{r+q+a+1+p}-\frac{1}{p+r+a}-\frac{1}{q+r+a})da$$$$=ln\frac{(r+a)(r+q+p+a+1)}{(p+r+a)(q+r+a)}+c$$$$\underset{a\to \mathrm{\infty }}{\lim }f\left(a\right)=0\Rightarrow c=0$$$$f\left(0\right)=ln\left(\frac{r(p+q+r+1)}{(p+r)(q+r)}\right)$$$${\int }_0^1\frac{(1-x^p)(1-x^q)x^{r-1}}{log\left(x\right)}dx=ln\left(\frac{r(p+q+r+1)}{(p+r)(q+r)}\right)$$$$$$

Commented by mnjuly1970 last updated on 08/Oct/20

$$tayyeballahbravothankyou..$$

Commented by mindispower last updated on 08/Oct/20

$$withePleasur$$

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