∫_0 ^( 1) ∫_0 ^( 1) (1/(1−xy^2 )) dx dy =?


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Question Number 128057 by bramlexs22 last updated on 04/Jan/21

$\int_{0}^{1} \int_{0}^{1} \frac{1}{1 - {xy}^{2}} dx dy = ?$
	Answered by liberty last updated on 04/Jan/21

	$\frac{1}{1 - {xy}^{2}} = \underset{m = 0}{\sum^{\infty}} {({xy}^{2})}^{m}$ $I = \int_{0}^{1} \int_{0}^{1} \frac{1}{1 - {xy}^{2}} dx dy = \underset{m = 0}{\sum^{\infty}} \int_{0}^{1} x^{m} dx \int_{0}^{1} y^{2 m} dy$ ${{= \underset{m = 0}{\sum^{\infty}} \frac{x^{m + 1}}{m + 1}]}_{0}^{1} . \frac{y^{2 m + 1}}{2 m + 1}]}_{0}^{1}$ $= \underset{m = 0}{\sum^{\infty}} \frac{1}{(m + 1) (2 m + 1)} = \underset{k = 1}{\sum^{\infty}} \frac{1}{k (2 k - 1)}; k = m + 1$ $= 2 \underset{k = 1}{\sum^{\infty}} (\frac{1}{2 k - 1} - \frac{1}{2 k}) = 2 \underset{i = 1}{\sum^{\infty}} \frac{{(- 1)}^{i - 1}}{i}$ $= 2 \ln (2)$
	Answered by Olaf last updated on 04/Jan/21

	$Ω = \int_{0}^{1} \int_{0}^{1} \frac{d x d y}{1 - {x y}^{2}}$ $Ω = \int_{0}^{1} \frac{1}{2 \sqrt{x}} \int_{0}^{1} [\frac{\sqrt{x}}{1 + \sqrt{x} y} + \frac{\sqrt{x}}{1 - \sqrt{x} y}] d y d x$ $Ω = \int_{0}^{1} \frac{1}{2 \sqrt{x}} {[\ln (\frac{1 + \sqrt{x} y}{1 - \sqrt{x} y})]}_{0}^{1} d x$ $Ω = \int_{0}^{1} \frac{\ln (\frac{1 + \sqrt{x}}{1 - \sqrt{x}})}{2 \sqrt{x}} d x$ $Ω = \int_{0}^{1} \ln (\frac{1 + u}{1 - u}) d u$ $Ω = {[(1 + u) \ln (1 + u) + (1 - u) \ln (1 - u)]}_{0}^{1}$ $Ω = 2 \ln 2$
	Answered by mathmax by abdo last updated on 04/Jan/21

	$\int_{0}^{1} \int_{0}^{1} \frac{1}{1 - {xy}^{2}} dxdy = \int_{0}^{1} (\int_{0}^{1} \frac{dy}{1 - {xy}^{2}}) dx we have$ $\int_{0}^{1} \frac{dy}{1 - {xy}^{2}} = \int_{0}^{1} \frac{xy}{(1 - \sqrt{x} y) (1 + \sqrt{x} y)} = \frac{1}{2} \int_{0}^{1} (\frac{1}{1 - \sqrt{x} y} + \frac{1}{1 + \sqrt{x} y}) dy$ $= - \frac{1}{2 \sqrt{x}} {[\ln ∣ 1 - \sqrt{x} y ∣]}_{y = 0}^{1} + \frac{1}{2 \sqrt{x}} {[\ln ∣ 1 + \sqrt{x} y ∣]}_{y = 0}^{1}$ $= \frac{1}{2 \sqrt{x}} \ln (1 + \sqrt{x}) - \frac{1}{2 \sqrt{x}} \ln (1 - \sqrt{x}) = \frac{1}{2 \sqrt{x}} \ln (\frac{1 + \sqrt{x}}{1 - \sqrt{x}}) \Rightarrow$ $I = \frac{1}{2} \int_{0}^{1} \frac{1}{\sqrt{x}} \ln (\frac{1 + \sqrt{x}}{1 - \sqrt{x}}) dx =_{\sqrt{x} = t} \frac{1}{2} \int_{0}^{1} \frac{1}{t} \ln (\frac{1 + t}{1 - t}) (2 t) dt$ $= \int_{0}^{1} \ln (1 + t) dt (\to 1 + t = u) - \int_{0}^{1} \ln (1 - t) dt (\to 1 - t = v)$ $= \int_{1}^{2} lnu du - \int_{1}^{0} lnv (- dv) = \int_{1}^{2} lnu du - \int_{0}^{1} lnv dv$ $= {[ulnu - u]}_{1}^{2} - {[vlnv - v]}_{0}^{1} = 2 \ln 2 - 2 + 1 - (- 1) = 2 \ln 2 - 2 + 2$ $\Rightarrow I = 2 \ln (2)$
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