Ω=∫_( 1) ^( ∞) (((√x) ln x)/(x^2 + 1)) dx = ?


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Question Number 148887 by mathdanisur last updated on 01/Aug/21

$Ω = \underset{1}{\int^{\infty}} \frac{\sqrt{x} l n x}{x^{2} + 1} d x = ?$
	Answered by Kamel last updated on 01/Aug/21

	$Ω = \underset{1}{\int^{\infty}} \frac{\sqrt{x} l n x}{x^{2} + 1} d x \overset{t = \frac{1}{x}}{=} - \int_{0}^{1} \frac{L n (t)}{\sqrt{t} (1 + t^{2})} d t$ $= - \underset{n = 0}{\sum^{+ \infty}} {(- 1)}^{n} \int_{0}^{1} t^{2 n - \frac{1}{2}} L n (t) d t = \underset{n = 0}{\sum^{+ \infty}} \frac{{(- 1)}^{n}}{{(2 n + \frac{1}{2})}^{2}}$ $= \underset{n = 0}{\sum^{+ \infty}} \frac{1}{{(4 n + \frac{1}{2})}^{2}} - \underset{n = 0}{\sum^{+ \infty}} \frac{1}{{(4 n + \frac{5}{2})}^{2}}$ $= \frac{1}{16} (Ψ^{(1)} (\frac{1}{8}) - Ψ^{(1)} (\frac{5}{8}))$
	Commented by mathdanisur last updated on 01/Aug/21

	$T h a n k y o u S e r$
	Answered by Ank0369 last updated on 01/Aug/21

	$S o l u t i o n : Ω \overset{x = 1 / y}{=} \int_{0}^{1} - \frac{\log_{e} (y)}{\sqrt{y} (1 + y^{2})} d y \overset{y = x^{2}}{=} \int_{0}^{1} - 4 \frac{\log_{e} (x)}{1 + x^{4}} d x$ $- \frac{Ω}{4} = \underset{n = 0}{\sum^{n = \infty}} {(- 1)}^{n} \int_{0}^{1} x^{4 n} \cdot \log_{e} (x) d x \Rightarrow \underset{n = 0}{\sum^{n = \infty}} \frac{{(- 1)}^{n + 1}}{{(4 n + 1)}^{2}}$ $- 4 Ω = \underset{n = 0}{\sum^{n = \infty}} \frac{{(- 1)}^{n + 1}}{{(n + \frac{1}{4})}^{2}} \Rightarrow \frac{1}{4} [ζ (2, \frac{5}{8}) - ζ (2, \frac{1}{8})]$ $Ω = \int_{1}^{\infty} \frac{\sqrt{x} \cdot \log_{e} (x)}{x^{2} + 1} d x = \frac{1}{16} [ψ^{(1)} (\frac{1}{8}) - ψ^{(1)} (\frac{5}{8})]$
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