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Question Number 209521 by mokys last updated on 12/Jul/24
find the integral ∫ (dx/(x^4 +a^4 )) by complex number ?
$${find}\:{the}\:{integral}\:\int\:\frac{{dx}}{{x}^{\mathrm{4}} +{a}^{\mathrm{4}} }\:{by}\:{complex}\:{number}\:?\: \\ $$
Answered by mathmax last updated on 14/Jul/24
I=∫ (dx/(x^4 −i^2 a^4 ))=∫(dx/((x^2 −ia^2 )(x^2 +ia^2 ))) =(1/(2ia^2 ))∫((1/(x^2 −ia^2 ))−(1/(x^2 +ia^2 )))dx =(1/(2ia^2 )){∫(dx/(x^2 −ia^2 ))−∫(dx/(x^2 +ia^2 ))}dx I_1 =∫ (dx/((x−(√i)a)(x+(√i)a))) =(1/(2(√i)a))∫((1/(x−(√i)a))−(1/(x+(√i)a))))dx =(1/(2(√i)a))ln(((x−(√i)a)/(x+(√i)a))) +c_1 I_2 =∫ (dx/(x^2 +ia^2 ))=∫(dx/(x^2 −(−i)a^2 )) =(1/(2(√(−i))a))∫((1/(x−(√(−i))a))−(1/(x+(√(−i))a)))dx =(1/(2(√(−i))a))ln(((x−(√(−i))a)/(x+(√(−i))a)))+c_2 I =(1/(2ia^2 ))×(1/(2(√i)a))ln(((x−(√i)a)/(x+(√i)a))) −(1/(2ia^2 ))×(1/(2(√(−i))a))ln(((x−(√(−i))a)/(x+(√(−i))a)))+C =(1/(4a^3 i(√i)))ln(((x−(√i)a)/(x+(√i)a)))−(1/(4a^3 i(√(−i))))ln(((x−(√(−i))a)/(x+(√(−i))a)))+C
$${I}=\int\:\frac{{dx}}{{x}^{\mathrm{4}} −{i}^{\mathrm{2}} {a}^{\mathrm{4}} }=\int\frac{{dx}}{\left({x}^{\mathrm{2}} −{ia}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{ia}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ia}^{\mathrm{2}} }\int\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{ia}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{ia}^{\mathrm{2}} }\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ia}^{\mathrm{2}} }\left\{\int\frac{{dx}}{{x}^{\mathrm{2}} −{ia}^{\mathrm{2}} }−\int\frac{{dx}}{{x}^{\mathrm{2}} +{ia}^{\mathrm{2}} }\right\}{dx} \\ $$$${I}_{\mathrm{1}} =\int\:\:\frac{{dx}}{\left({x}−\sqrt{{i}}{a}\right)\left({x}+\sqrt{{i}}{a}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{i}}{a}}\int\left(\frac{\mathrm{1}}{{x}−\sqrt{{i}}{a}}−\frac{\mathrm{1}}{\left.{x}+\sqrt{{i}}{a}\right)}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{i}}{a}}{ln}\left(\frac{{x}−\sqrt{{i}}{a}}{{x}+\sqrt{{i}}{a}}\right)\:+{c}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\int\:\frac{{dx}}{{x}^{\mathrm{2}} +{ia}^{\mathrm{2}} }=\int\frac{{dx}}{{x}^{\mathrm{2}} −\left(−{i}\right){a}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{i}}{a}}\int\left(\frac{\mathrm{1}}{{x}−\sqrt{−{i}}{a}}−\frac{\mathrm{1}}{{x}+\sqrt{−{i}}{a}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{i}}{a}}{ln}\left(\frac{{x}−\sqrt{−{i}}{a}}{{x}+\sqrt{−{i}}{a}}\right)+{c}_{\mathrm{2}} \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{2}{ia}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{2}\sqrt{{i}}{a}}{ln}\left(\frac{{x}−\sqrt{{i}}{a}}{{x}+\sqrt{{i}}{a}}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}{ia}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{i}}{a}}{ln}\left(\frac{{x}−\sqrt{−{i}}{a}}{{x}+\sqrt{−{i}}{a}}\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{a}^{\mathrm{3}} {i}\sqrt{{i}}}{ln}\left(\frac{{x}−\sqrt{{i}}{a}}{{x}+\sqrt{{i}}{a}}\right)−\frac{\mathrm{1}}{\mathrm{4}{a}^{\mathrm{3}} {i}\sqrt{−{i}}}{ln}\left(\frac{{x}−\sqrt{−{i}}{a}}{{x}+\sqrt{−{i}}{a}}\right)+{C} \\ $$

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