Ω = ∫_0 ^( (1/3)) x^(2n) ln(1−x)dx


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Question Number 128633 by Lordose last updated on 09/Jan/21

$Ω = \int_{0}^{\frac{1}{3}} x^{2 n} \ln (1 - x) dx$
	Answered by mathmax by abdo last updated on 09/Jan/21
	$let try another way x=(t/3) ⇒ Ω=∫_0 ^1 ((t/3))^(2n) ln(1−(t/3))(dt/3) =(1/3^(2n+1) )∫_0 ^1 t^(2n) ln(((3−t)/3))dt =(1/3^(2n+1) ) ∫_0 ^1 t^(2n) ln(3−t)dt−(1/3^(2n+1) )ln(3)[(t^(2n+1) /(2n+1))]_0 ^1 =(1/3^(2n+1) )∫_0 ^1 t^(2n) ln(3−t)dt −((ln3)/((2n+1)3^(2n+1) )) we have by parts ∫_0 ^1 t^(2n) ln(3−t)dt =[(t^(2n+1) /(2n+1))ln(3−t)]_0 ^1 −∫_0 ^1 (t^(2n+1) /(2n+1))×((−1)/(3−t))dt =(1/(2n+1))ln(2)+(1/(2n+1))∫_0 ^1 (t^(2n+1) /(3−t))dt and ∫_0 ^1 (t^(2n+1) /(3−t)) dt =_(3−t=u) −∫_2 ^3 (((3−u)^(2n+1) )/u)(−du) =−∫_2 ^3 (((u−3)^(2n+1) )/u) du =−∫_2 ^3 ((Σ_(k=0) ^(2n+1) C_(2n+1) ^k u^k (−3)^(2n+1−k) )/u)du =3^(2n+1) Σ_(k=1) ^(2n+1) (−3)^(−k) C_(2n+1) ^k ∫_2 ^3 u^(k−1) du −3^(2n+1) ln((3/2)) =3^(2n+1) Σ_(k=0) ^(2n+1) (−3)^(−k) C_(2n+1) ^k (1/k){ 3^(k−1) −2^(k−1) }−3^(2n+1) ln((3/2))...$
	$let try another way x = \frac{t}{3} \Rightarrow$ $Ω = \int_{0}^{1} {(\frac{t}{3})}^{2 n} \ln (1 - \frac{t}{3}) \frac{dt}{3} = \frac{1}{3^{2 n + 1}} \int_{0}^{1} t^{2 n} \ln (\frac{3 - t}{3}) dt$ $= \frac{1}{3^{2 n + 1}} \int_{0}^{1} t^{2 n} \ln (3 - t) dt - \frac{1}{3^{2 n + 1}} \ln (3) {[\frac{t^{2 n + 1}}{2 n + 1}]}_{0}^{1}$ $= \frac{1}{3^{2 n + 1}} \int_{0}^{1} t^{2 n} \ln (3 - t) dt - \frac{\ln 3}{(2 n + 1) 3^{2 n + 1}} we have by parts$ $\int_{0}^{1} t^{2 n} \ln (3 - t) dt = {[\frac{t^{2 n + 1}}{2 n + 1} \ln (3 - t)]}_{0}^{1} - \int_{0}^{1} \frac{t^{2 n + 1}}{2 n + 1} \times \frac{- 1}{3 - t} dt$ $= \frac{1}{2 n + 1} \ln (2) + \frac{1}{2 n + 1} \int_{0}^{1} \frac{t^{2 n + 1}}{3 - t} dt and$ $\int_{0}^{1} \frac{t^{2 n + 1}}{3 - t} dt =_{3 - t = u} - \int_{2}^{3} \frac{{(3 - u)}^{2 n + 1}}{u} (- du)$ $= - \int_{2}^{3} \frac{{(u - 3)}^{2 n + 1}}{u} du = - \int_{2}^{3} \frac{\sum_{k = 0}^{2 n + 1} C_{2 n + 1}^{k} u^{k} {(- 3)}^{2 n + 1 - k}}{u} du$ $= 3^{2 n + 1} \sum_{k = 1}^{2 n + 1} {(- 3)}^{- k} C_{2 n + 1}^{k} \int_{2}^{3} u^{k - 1} du - 3^{2 n + 1} \ln (\frac{3}{2})$ $= 3^{2 n + 1} \sum_{k = 0}^{2 n + 1} {(- 3)}^{- k} C_{2 n + 1}^{k} \frac{1}{k} {3^{k - 1} - 2^{k - 1}} - 3^{2 n + 1} \ln (\frac{3}{2}) . . .$
	Answered by mathmax by abdo last updated on 09/Jan/21

	$Ω = \int_{0}^{\frac{1}{3}} x^{2 n} \ln (1 - x) dx by parts u^{^{'}} = x^{2 n} [and v = \ln (1 - x)$ $Ω = {[\frac{x^{2 n + 1}}{2 n + 1} \ln (1 - x)]}_{0}^{\frac{1}{3}} - \int_{0}^{\frac{1}{3}} \frac{x^{2 n + 1}}{2 n + 1} \times \frac{- 1}{1 - x} dx$ $= \frac{1}{2 n + 1 {(3)}^{2 n + 1}} \ln (\frac{2}{3}) + \frac{1}{2 n + 1} \int_{0}^{\frac{1}{3}} \frac{x^{2 n + 1}}{1 - x} dx but$ $\int_{0}^{\frac{1}{3}} \frac{x^{2 n + 1}}{1 - x} dx = \int_{0}^{\frac{1}{3}} \frac{x^{2 n + 1} - 1 + 1}{1 - x} dx = \int_{0}^{\frac{1}{3}} \frac{(x - 1) (1 + x + x^{2} + . . . + x^{2 n})}{1 - x} dx$ $+ {[- \ln (1 - x)]}_{0}^{\frac{1}{3}} = - \int_{0}^{\frac{1}{3}} (\sum_{k = 0}^{2 n} x^{k}) dx - \ln (\frac{2}{3})$ $= - \sum_{k = 0}^{2 n} {[\frac{x^{k + 1}}{k + 1}]}_{0}^{\frac{1}{3}} - \ln 2 + \ln 3$ $= - \sum_{k = 0}^{2 n} \frac{1}{(k + 1) 3^{k + 1}} + \ln 3 - \ln 2 rsst to find the value of this serie . . .$
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