Can anyone do this? ∫^( +∞) _( 1) ((t−1)/((1+t)^3 lnt))dt


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Question Number 197772 by Erico last updated on 28/Sep/23

$Can anyone do this ?$ ${\int_{1}}^{+ \infty} \frac{t - 1}{{(1 + t)}^{3} lnt} dt$
Commented by witcher3 last updated on 28/Sep/23

$we can Give$ $close forme of$ $\frac{{(t - 1)}^{m}}{\ln^{n} (t)} . \frac{dt}{{(1 + t)}^{p}}$ $m ⩾ n$ $p ⩾ m + 2$ $m, n, p \in N$
	Answered by witcher3 last updated on 03/Oct/23
	$(1/t)=y I=∫_0 ^1 ((y−1)/((y+1)^3 ln(y))) y→0 ((y−1)/((y+1)ln(y)))→0 y→1,((y−1)/(ln(y))).(1/((y+1)^3 ))→(1/8) (1/(ln(y)))=−∫_0 ^∞ y^a dy y∈]0,1[ ln(t)<0 I=−∫_0 ^∞ ∫_0 ^1 ((y−1)/((y+1)^3 ))y^a dyda= ((y−1)/((y+1)^3 ))=(y−1).(−(1/2)Σ_(n≥2) (−1)^n n(n−1)y^(n−2) ) =−(1/2)Σ_(n≥2) (−1)^n n(n−1)(y^(n−1) −y^(n−2) ) I=(1/2)∫_0 ^∞ ∫_0 ^1 Σ_(n≥2) (−1)^n n(n−1)(y^(n+a−1) −y^(n+a−2) )dy =(1/2)Σ_(n≥2) (−1)^n n(n−1)∫_0 ^∞ (1/((n+a)))−(1/((n+a−1)))da =(1/2)Σ_(n≥2) (−1)^n n(n−1)ln(((n−1)/n)) =(1/2)Σ_(n≥1) 2n(2n−1)ln(((2n−1)/(2n)))−(2n+1)(2n)ln(((2n)/(2n+1))) =−Σ_(n≥1) [2n(2n)^2 ln(2n)+2(2n+1)(2n+1)^2 ln(2n+1)] −Σ_(n≥1) n^2 ln(n)=ζ^∗ ′(−2) ζ^∗ prlangement of Zeta function Over C−plane−{1}$
	$\frac{1}{t} = y$ $I = \int_{0}^{1} \frac{y - 1}{{(y + 1)}^{3} \ln (y)}$ $y \to 0 \frac{y - 1}{(y + 1) \ln (y)} \to 0$ $y \to 1, \frac{y - 1}{\ln (y)} . \frac{1}{{(y + 1)}^{3}} \to \frac{1}{8}$ $\frac{1}{\ln (y)} = - \int_{0}^{\infty} y^{a} dy y \in] 0, 1 [\ln (t) < 0$ $I = - \int_{0}^{\infty} \int_{0}^{1} \frac{y - 1}{{(y + 1)}^{3}} y^{a} dyda =$ $\frac{y - 1}{{(y + 1)}^{3}} = (y - 1) . (- \frac{1}{2} \sum_{n ⩾ 2} {(- 1)}^{n} n (n - 1) y^{n - 2})$ $= - \frac{1}{2} \sum_{n ⩾ 2} {(- 1)}^{n} n (n - 1) (y^{n - 1} - y^{n - 2})$ $I = \frac{1}{2} \int_{0}^{\infty} \int_{0}^{1} \sum_{n ⩾ 2} {(- 1)}^{n} n (n - 1) (y^{n + a - 1} - y^{n + a - 2}) dy$ $= \frac{1}{2} \sum_{n ⩾ 2} {(- 1)}^{n} n (n - 1) \int_{0}^{\infty} \frac{1}{(n + a)} - \frac{1}{(n + a - 1)} da$ $= \frac{1}{2} \sum_{n ⩾ 2} {(- 1)}^{n} n (n - 1) \ln (\frac{n - 1}{n})$ $= \frac{1}{2} \sum_{n ⩾ 1} 2 n (2 n - 1) \ln (\frac{2 n - 1}{2 n}) - (2 n + 1) (2 n) \ln (\frac{2 n}{2 n + 1})$ $= - \sum_{n ⩾ 1} [2 n {(2 n)}^{2} \ln (2 n) + 2 (2 n + 1) {(2 n + 1)}^{2} \ln (2 n + 1)]$ $- \sum_{n ⩾ 1} n^{2} \ln (n) = {\overset{}{ζ}}^{'} (- 2)$ $\overset{}{ζ} prlangement of Zeta function Over C - plane - {1}$
	Answered by MathematicalUser2357 last updated on 15/Jan/24

	$0.213011170749$
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