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Question Number 200417 by hardmath last updated on 18/Nov/23
find:   Ω = ∫_1 ^( ∞)  ((√x)/((1 + x^2 ))) dx = ?
$$\mathrm{find}:\:\:\:\Omega\:=\:\int_{\mathrm{1}} ^{\:\infty} \:\frac{\sqrt{\mathrm{x}}}{\left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} \right)}\:\mathrm{dx}\:=\:? \\ $$
Answered by witcher3 last updated on 18/Nov/23
(√x)=y  =∫_1 ^∞ ((2y^2 )/((1+y^4 )))dy  =∫_1 ^∞ ((2y^2 )/((y^2 +y(√2)+1)(y^2 −y(√2)+1)))  =(1/( (√2)))∫_1 ^∞ −(y/(y^2 +y(√2)+1))+(y/(y^2 −y(√2)+1))dy    =(1/( 2(√2)))∫_1 ^∞ ((2y−(√2)+(√2))/(y^2 −y(√2)+1))−((2y+(√2)−(√2))/(y^2 +y(√2)+1))dy  =(1/(2(√2)))ln(((2−(√2))/(2+(√2))))+(1/2)∫_1 ^∞ (dy/((y−(1/( (√2))))^2 +(1/2)))+(1/2)∫_1 ^∞ (dy/((y+(1/( (√2))))^2 +(1/2)))  =((ln(((2+(√2))/(2−(√2)))))/(2(√2)))+(1/( (√2)))[tan^(−1) (y(√2)−1)+tan^(−1) (y(√2)+1)]_1 ^∞   =((ln(((2+(√2))/(2−(√2)))))/(2(√2)))+(1/( (√2)))[tan^(−1) ((√2)−1)+tan^(−1) ((√2)+1)]  tan^(−1) ((√2)+1)+tan^(−1) ((√2)−1)=(π/2)  we get   (π/( 2(√( 2))))+((ln(1+(√(2))))/( (√2)))
$$\sqrt{\mathrm{x}}=\mathrm{y} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{2y}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{y}^{\mathrm{4}} \right)}\mathrm{dy} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{2y}^{\mathrm{2}} }{\left(\mathrm{y}^{\mathrm{2}} +\mathrm{y}\sqrt{\mathrm{2}}+\mathrm{1}\right)\left(\mathrm{y}^{\mathrm{2}} −\mathrm{y}\sqrt{\mathrm{2}}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{1}} ^{\infty} −\frac{\mathrm{y}}{\mathrm{y}^{\mathrm{2}} +\mathrm{y}\sqrt{\mathrm{2}}+\mathrm{1}}+\frac{\mathrm{y}}{\mathrm{y}^{\mathrm{2}} −\mathrm{y}\sqrt{\mathrm{2}}+\mathrm{1}}\mathrm{dy} \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{2}}}\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{2y}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}}{\mathrm{y}^{\mathrm{2}} −\mathrm{y}\sqrt{\mathrm{2}}+\mathrm{1}}−\frac{\mathrm{2y}+\sqrt{\mathrm{2}}−\sqrt{\mathrm{2}}}{\mathrm{y}^{\mathrm{2}} +\mathrm{y}\sqrt{\mathrm{2}}+\mathrm{1}}\mathrm{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\left(\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{dy}}{\left(\mathrm{y}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{dy}}{\left(\mathrm{y}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\right)}{\mathrm{2}\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{y}\sqrt{\mathrm{2}}−\mathrm{1}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{y}\sqrt{\mathrm{2}}+\mathrm{1}\right)\right]_{\mathrm{1}} ^{\infty} \\ $$$$=\frac{\mathrm{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\right)}{\mathrm{2}\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\right] \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}+\mathrm{1}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{we}\:\mathrm{get}\: \\ $$$$\frac{\pi}{\:\mathrm{2}\sqrt{\:\mathrm{2}}}+\frac{\mathrm{ln}\left(\mathrm{1}+\sqrt{\left.\mathrm{2}\right)}\right.}{\:\sqrt{\mathrm{2}}} \\ $$$$ \\ $$
Commented by hardmath last updated on 19/Nov/23
thank you dear professor
$${thank}\:{you}\:{dear}\:{professor} \\ $$

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