∫_0 ^∞ ((log(1+z^4 ))/( (√z)(1+z)))dz


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Question Number 140601 by mathdanisur last updated on 10/May/21

$\underset{0}{\int^{\infty}} \frac{l o g (1 + z^{4})}{\sqrt{z} (1 + z)} d z$
	Answered by mathmax by abdo last updated on 10/May/21

	$Φ = \int_{0}^{\infty} \frac{\log (1 + z^{4})}{\sqrt{z} (1 + z)} dz \Rightarrow Φ =_{\sqrt{z} = x} \int_{0}^{\infty} \frac{\log (1 + x^{8})}{x (1 + x^{2})} (2 x) dx$ $= 2 \int_{0}^{\infty} \frac{\log (1 + x^{8})}{x^{2} + 1} dx = \int_{- \infty}^{+ \infty} \frac{\log (x^{8} + 1)}{x^{2} + 1} dx let$ $φ (z) = \frac{\log (z^{8} + 1)}{z^{2} + 1} \Rightarrow φ (z) = \frac{\log (z^{8} + 1)}{(z - i) (z + i)}$ $\int_{R} φ (z) dz = 2 i π Res (φ, i) = 2 i π \times \frac{\log (i^{8} + 1)}{2 i} = π \log (2)$ $this method by residus is not sure . . .!$
	Commented by mathmax by abdo last updated on 10/May/21
	$let try another way Φ=2 ∫_0 ^∞ ((log(x^8 +1))/(x^2 +1))dx =∫_(−∞) ^(+∞) ((log(x^8 +1))/(x^2 +1))dx let f(a) =∫_(−∞) ^(+∞) ((log(x^8 +a^8 ))/(x^2 +1))dx we have f(1)=Φ f^′ (a)=∫_(−∞) ^(+∞) ((8a^7 )/((x^8 +a^8 )(x^2 +1)))dx =8a^7 ∫_(−∞) ^(+∞) (dx/((x^8 +a^8 )(x^2 +1))) x^8 +a^8 =0 ⇒((x/a))^8 =−1 =e^((2k+1)iπ) ⇒x^8 =a^8 e^((2k+1)iπ) ⇒ x_k =a e^(i(2k+1)(π/8)) ⇒x^8 +a^8 =Π_(k=0) ^7 (x−a e^(i(2k+1)(π/8)) ) let ϕ(z)=(1/((z^8 +a^8 )(z^2 +1))) ⇒ϕ(z)=(1/((z−i)(z+i)Π_(k=0) ^7 (z−ae^(i(2k+1)(π/8)) ))) z_k =ae^((i(2k+1)π)/8) ⇒z_0 =a e^((iπ)/8) ,z_1 =ae^(i((3π)/8)) ,z_2 =ae^((i5π)/8) z_3 =a e^((i7π)/8) ,z_4 =ae^((i9π)/8) ,z_5 =ae^((i11π)/8) ,z_6 =ae^((i13π)/8) ,z_7 =ae^((i15π)/8) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,i)+Res(ϕ ,ae^((iπ)/8) )+Res(ϕ,ae^((i3π)/8) ) +Res(ϕ,a e^((i5π)/8) )} Res(ϕ,i)=(1/(2iΠ_(k=0) ^7 (i−ae^(i(2k+1)(π/8)) ))) ϕ(z)=(1/(z^(10) +z^8 +a^8 z^2 +a^8 ))=(1/(p(z))) ⇒Res(ϕ ,z_i ) =(1/(p^′ (z_i ))) =(1/(10z_i ^9 +8 z_i ^7 +2a^8 z_i )) ⇒ Res(ϕ,ae^((iπ)/8) ) =(1/(10(ae^((iπ)/8) )^9 +8(ae^((iπ)/8) )^7 +2a^8 (ae^((iπ)/8) ))) ....be continued...$
	$let try another way Φ = 2 \int_{0}^{\infty} \frac{\log (x^{8} + 1)}{x^{2} + 1} dx = \int_{- \infty}^{+ \infty} \frac{\log (x^{8} + 1)}{x^{2} + 1} dx$ $let f (a) = \int_{- \infty}^{+ \infty} \frac{\log (x^{8} + a^{8})}{x^{2} + 1} dx we have f (1) = Φ$ $f^{^{'}} (a) = \int_{- \infty}^{+ \infty} \frac{{8 a}^{7}}{(x^{8} + a^{8}) (x^{2} + 1)} dx = {8 a}^{7} \int_{- \infty}^{+ \infty} \frac{dx}{(x^{8} + a^{8}) (x^{2} + 1)}$ $x^{8} + a^{8} = 0 \Rightarrow {(\frac{x}{a})}^{8} = - 1 = e^{(2 k + 1) i π} \Rightarrow x^{8} = a^{8} e^{(2 k + 1) i π} \Rightarrow$ $x_{k} = a e^{i (2 k + 1) \frac{π}{8}} \Rightarrow x^{8} + a^{8} = \prod_{k = 0}^{7} (x - a e^{i (2 k + 1) \frac{π}{8}})$ $let φ (z) = \frac{1}{(z^{8} + a^{8}) (z^{2} + 1)} \Rightarrow φ (z) = \frac{1}{(z - i) (z + i) \prod_{k = 0}^{7} (z - {ae}^{i (2 k + 1) \frac{π}{8}})}$ $z_{k} = {ae}^{\frac{i (2 k + 1) π}{8}} \Rightarrow z_{0} = a e^{\frac{i π}{8}}, z_{1} = {ae}^{i \frac{3 π}{8}}, z_{2} = {ae}^{\frac{i 5 π}{8}}$ $z_{3} = a e^{\frac{i 7 π}{8}}, z_{4} = {ae}^{\frac{i 9 π}{8}}, z_{5} = {ae}^{\frac{i 11 π}{8}}, z_{6} = {ae}^{\frac{i 13 π}{8}}, z_{7} = {ae}^{\frac{i 15 π}{8}}$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, i) + Res (φ, {ae}^{\frac{i π}{8}}) + Res (φ, {ae}^{\frac{i 3 π}{8}})$ $+ Res (φ, a e^{\frac{i 5 π}{8}})}$ $Res (φ, i) = \frac{1}{2 i \prod_{k = 0}^{7} (i - {ae}^{i (2 k + 1) \frac{π}{8}})}$ $φ (z) = \frac{1}{z^{10} + z^{8} + a^{8} z^{2} + a^{8}} = \frac{1}{p (z)} \Rightarrow Res (φ, z_{i}) = \frac{1}{p^{^{'}} (z_{i})}$ $= \frac{1}{{10 z}_{i}^{9} + 8 z_{i}^{7} + {2 a}^{8} z_{i}} \Rightarrow$ $Res (φ, {ae}^{\frac{i π}{8}}) = \frac{1}{10 {({ae}^{\frac{i π}{8}})}^{9} + 8 {({ae}^{\frac{i π}{8}})}^{7} + {2 a}^{8} ({ae}^{\frac{i π}{8}})}$ $. . . . be continued . . .$
	Answered by Dwaipayan Shikari last updated on 10/May/21

	$z = u^{2}$ $2 \int_{0}^{\infty} \frac{l o g (1 + u^{8})}{1 + u^{2}}$ $= 2 \int_{1}^{\infty} \frac{l o g (1 + u^{8})}{1 + u^{2}} d u + 2 \int_{0}^{1} \frac{l o g (1 + u^{8})}{1 + u^{2}} d u u = \frac{1}{j}$ $= 2 \int_{0}^{1} \frac{l o g (1 + j^{8}) - l o g (j^{8})}{1 + j^{2}} d j + 2 \int_{0}^{1} \frac{l o g (1 + u^{8})}{1 + u^{2}} d u$ $= 4 \int_{0}^{1} \frac{l o g (1 + j^{8})}{1 + j^{2}} - 16 \int_{0}^{1} \frac{l o g (j)}{1 + j^{2}} d j$ $= 4 \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{1} j^{2 n} l o g (1 + j^{8}) d j - \int_{0}^{1} \frac{k^{- \frac{3}{4}} l o g (k)}{1 - k} d k$ $= 4 \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} (\frac{l o g (2)}{2 n + 1} - 8 \int_{0}^{1} \frac{j^{2 n + 8}}{1 + j^{8}} d j) + ψ^{'} (\frac{1}{4})$ $= π l o g (2) - 32 \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{1} \frac{j^{2 n + 8} - j^{2 n + 16}}{1 - j^{16}} d j + ψ^{'} (\frac{1}{4})$ $= π l o g (2) - 2 \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{1} \frac{k^{\frac{n}{8} - \frac{7}{8}} - k^{\frac{n}{8} + \frac{1}{8}}}{1 - k} d k + ψ^{'} (\frac{1}{4})$ $= π l o g (2) - 2 \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} (ψ (\frac{n}{8} + \frac{9}{8}) - ψ (\frac{n}{8} + \frac{1}{8})) + ψ^{'} (\frac{1}{4})$ $= π l o g (2) - 2 \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} (\frac{8}{n + 1}) + ψ^{'} (\frac{1}{4})$ $= π l o g (2) - 16 l o g (2) + ψ^{'} (\frac{1}{4})$
	Commented by mathdanisur last updated on 11/May/21

	$S i r, a n s w e r 4 π l n (\sqrt{2} + \sqrt{(2 + \sqrt{2}}))$ $s i m p l i f y, p s i (1 / 4) p l i z$
	Commented by mathdanisur last updated on 12/May/21

	$S i r, D w a i p a y a n S h i k a r i p l i z . . .$
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