calculus-prove-that-0-1-1-x-p-1-x-q-x-r-1-log-x-dx-log-p-q-r-1-r-p-r-q-r-m-n-1970-

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

\dots c a l c u l u s \dots

p r o v e t h a t ::

\int_{0}^{1} \frac{(1 - x^{p}) (1 - x^{q}) x^{r - 1}}{l o g (x)} d x \overset{? ? ?}{=} l o g (\frac{(p + q + r + 1) r}{(p + r) (q + r)})

m . n . 1970

Answered by mindispower last updated on 08/Oct/20

l e t f (a) = \int_{0}^{1} \frac{(1 - x^{p}) (1 - x^{q}) x^{r - 1 + a}}{l o g (x)} d x

f^{'} (a) = \int_{0}^{1} \frac{(1 - x^{p}) (1 - x^{q}) x^{r - 1 + a}}{l o g (x)} . l o g (x) d x

= \int_{0}^{1} (x^{r + a - 1} + x^{p + q + r - 1 + a} - x^{p + r - 1 + a} - x^{q + r - 1 + a}) d x

= \frac{1}{r + a} + \frac{1}{r + q + a + 1} - \frac{1}{p + r + a} - \frac{1}{q + r + a}

w e w a n t f (0)

f (a) = \int (\frac{1}{r + a} + \frac{1}{r + q + a + 1 + p} - \frac{1}{p + r + a} - \frac{1}{q + r + a}) d a

= l n \frac{(r + a) (r + q + p + a + 1)}{(p + r + a) (q + r + a)} + c

\lim_{a \to \infty} f (a) = 0 \Rightarrow c = 0

f (0) = l n (\frac{r (p + q + r + 1)}{(p + r) (q + r)})

\int_{0}^{1} \frac{(1 - x^{p}) (1 - x^{q}) x^{r - 1}}{l o g (x)} d x = l n (\frac{r (p + q + r + 1)}{(p + r) (q + r)})

Commented by mnjuly1970 last updated on 08/Oct/20

t a y y e b a l l a h b r a v o t h a n k y o u . .

Commented by mindispower last updated on 08/Oct/20

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