∫_0 ^( ∞) ((xln(1+x))/(1+x^4 ))dx


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Question Number 125911 by Lordose last updated on 15/Dec/20

$\int_{0}^{\infty} \frac{xln (1 + x)}{1 + x^{4}} dx$
	Answered by mathmax by abdo last updated on 15/Dec/20
	$A =∫_0 ^∞ ((xln(1+x))/(1+x^4 ))dx =∫_0 ^1 ((xln(1+x))/(1+x^4 ))dx+∫_1 ^∞ ((xln(x+1))/(1+x^4 ))dx ∫_1 ^∞ ((xln(x+1))/(1+x^4 ))dx =_(x=(1/t)) −∫_0 ^1 ((ln(1+(1/t)))/(t(1+(1/t^4 ))))(−(dt/t^2 )) =∫_0 ^1 ((ln(t+1)−ln(t)/(t^3 +(1/t)))dt =∫_0 ^1 ((tln(t+1)−tlnt)/(1+t^4 ))dt ⇒A =2∫_0 ^1 ((xln(x+1))/(1+x^4 ))dx−∫_0 ^1 ((xlnx)/(1+x^4 ))dx we have ∫_0 ^1 ((xlnx)/(1+x^4 ))dx =∫_0 ^1 xln(x)Σ_(n=0) ^∞ (−1)^n x^(4n) dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(4n+1) ln(x)dx =Σ_(n=0) ^∞ (−1)^n u_n u_n =∫_0 ^1 x^(4n+1) ln(x)dx =[(1/(4n+2))x^(4n+2) ln(x)]_0 ^1 −∫_0 ^1 (x^(4n+1) /(4n+2))dx =−(1/((4n+2)^2 )) ⇒∫_0 ^1 ((xlnx)/(1+x^4 ))dx =−Σ_(n=0) ^∞ (((−1)^n )/(4(2n+1)^2 )) =−(k/4)( katalan constant) ∫_0 ^1 ((xln(x+1))/(1+x^4 ))dx =∫_0 ^1 xln(x+1)Σ_(n=0) ^∞ (−1)^n x^(4n) dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(4n+1) ln(x+1)dx =Σ_(n=0) ^∞ (−1)^n v_n v_n =∫_0 ^1 x^(4n+1) ln(x+1)dx =_(x+1=t) ∫_1 ^2 (t−1)^(4n+1) ln(t)dt =∫_1 ^2 ln(t)Σ_(k=0) ^(4n+1) C_(4n+1) ^k t^k (−1)^(4n+1−k) dt =−Σ_(k=0) ^(4n+1) (−1)^k C_(4n+1) ^k ∫_1 ^2 t^(k ) ln(t)dt and by psrts ∫_1 ^2 t^k ln(t)dt =[(1/(k+1))t^(k+1) ln(t)]_1 ^2 −∫_1 ^2 (t^(k+1) /(k+1))(dt/t) =((2^(k+1) ln(2))/(k+1))−(1/(k+1))∫_1 ^2 t^k dt =((2^(k+1) ln(2))/(k+1))−(1/((k+1)^2 ))[t^(k+1) ]_1 ^2 =((2^(k+1) ln(2))/(k+1))−(1/((k+1)^2 ))(2^(k+1) −1) ⇒ v_n =−Σ_(k=0) ^(4n+1) (−1)^k C_(4n+1) ^k {((2^(k+1) ln(2))/(k+1))−((2^(k+1) −1)/((k+1)^2 ))} ....be continued...$
	$A = \int_{0}^{\infty} \frac{xln (1 + x)}{1 + x^{4}} dx = \int_{0}^{1} \frac{xln (1 + x)}{1 + x^{4}} dx + \int_{1}^{\infty} \frac{xln (x + 1)}{1 + x^{4}} dx$ $\int_{1}^{\infty} \frac{xln (x + 1)}{1 + x^{4}} dx =_{x = \frac{1}{t}} - \int_{0}^{1} \frac{\ln (1 + \frac{1}{t})}{t (1 + \frac{1}{t^{4}})} (- \frac{dt}{t^{2}})$ $= \int_{0}^{1} \frac{\ln (t + 1) - \ln (t}{t^{3} + \frac{1}{t}} dt = \int_{0}^{1} \frac{tln (t + 1) - tlnt}{1 + t^{4}} dt$ $\Rightarrow A = 2 \int_{0}^{1} \frac{xln (x + 1)}{1 + x^{4}} dx - \int_{0}^{1} \frac{xlnx}{1 + x^{4}} dx we have$ $\int_{0}^{1} \frac{xlnx}{1 + x^{4}} dx = \int_{0}^{1} xln (x) \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{4 n} dx$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} x^{4 n + 1} \ln (x) dx = \sum_{n = 0}^{\infty} {(- 1)}^{n} u_{n}$ $u_{n} = \int_{0}^{1} x^{4 n + 1} \ln (x) dx = {[\frac{1}{4 n + 2} x^{4 n + 2} \ln (x)]}_{0}^{1} - \int_{0}^{1} \frac{x^{4 n + 1}}{4 n + 2} dx$ $= - \frac{1}{{(4 n + 2)}^{2}} \Rightarrow \int_{0}^{1} \frac{xlnx}{1 + x^{4}} dx = - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 {(2 n + 1)}^{2}}$ $= - \frac{k}{4} (katalan constant)$ $\int_{0}^{1} \frac{xln (x + 1)}{1 + x^{4}} dx = \int_{0}^{1} xln (x + 1) \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{4 n} dx$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} x^{4 n + 1} \ln (x + 1) dx = \sum_{n = 0}^{\infty} {(- 1)}^{n} v_{n}$ $v_{n} = \int_{0}^{1} x^{4 n + 1} \ln (x + 1) dx =_{x + 1 = t} \int_{1}^{2} {(t - 1)}^{4 n + 1} \ln (t) dt$ $= \int_{1}^{2} \ln (t) \sum_{k = 0}^{4 n + 1} C_{4 n + 1}^{k} t^{k} {(- 1)}^{4 n + 1 - k} dt$ $= - \sum_{k = 0}^{4 n + 1} {(- 1)}^{k} C_{4 n + 1}^{k} \int_{1}^{2} t^{k} \ln (t) dt and by psrts$ $\int_{1}^{2} t^{k} \ln (t) dt = {[\frac{1}{k + 1} t^{k + 1} \ln (t)]}_{1}^{2} - \int_{1}^{2} \frac{t^{k + 1}}{k + 1} \frac{dt}{t}$ $= \frac{2^{k + 1} \ln (2)}{k + 1} - \frac{1}{k + 1} \int_{1}^{2} t^{k} dt = \frac{2^{k + 1} \ln (2)}{k + 1} - \frac{1}{{(k + 1)}^{2}} {[t^{k + 1}]}_{1}^{2}$ $= \frac{2^{k + 1} \ln (2)}{k + 1} - \frac{1}{{(k + 1)}^{2}} (2^{k + 1} - 1) \Rightarrow$ $v_{n} = - \sum_{k = 0}^{4 n + 1} {(- 1)}^{k} C_{4 n + 1}^{k} {\frac{2^{k + 1} \ln (2)}{k + 1} - \frac{2^{k + 1} - 1}{{(k + 1)}^{2}}}$ $. . . . be continued . . .$
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