1. find ∫_1 ^(+∞) (dx/(x^2 −i)) and ∫_1 ^(+∞) (dx/(x^2 +i)) (i=(√(−1))) 2. find the value of ∫


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Question Number 96373 by mathmax by abdo last updated on 01/Jun/20

$1 . find \int_{1}^{+ \infty} \frac{dx}{x^{2} - i} and \int_{1}^{+ \infty} \frac{dx}{x^{2} + i} (i = \sqrt{- 1})$ $2 . find the value of \int_{1}^{+ \infty} \frac{dx}{x^{4} + 1}$
	Answered by Sourav mridha last updated on 01/Jun/20

	$l e t I = \int_{1}^{+ \infty} \frac{d x}{x^{2} - i} a n d J = \int_{1}^{+ \infty} \frac{d x}{x^{2} + i}$ $n o w, I + J = \int_{1}^{+ \infty} \frac{2 x^{2}}{x^{4} + 1} d x . . . . . (i)$ $= \int_{1}^{+ \infty} \frac{x^{2} + 1}{x^{4} + 1} d x + \int_{1}^{\infty} \frac{x^{2} - 1}{x^{4} + 1} d x$ $. . . . . . . . (I_{1}) . . . . . . . . . . (I_{2})$ $= \int_{1}^{+ \infty} \frac{d (x - \frac{1}{x})}{{(x - \frac{1}{x})}^{2} + {(\sqrt{2})}^{2}} + \int_{1}^{+ \infty} \frac{d (x + \frac{1}{x})}{{(x + \frac{1}{x})}^{2} - {(\sqrt{2})}^{2}}$ $= \frac{1}{\sqrt{2}} {[\tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{2}})]}_{1}^{+ \infty} + \frac{1}{2 \sqrt{2}} {[l n (\frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}})]}_{1}^{+ \infty}$ $= \frac{π}{2 \sqrt{2}} - \frac{1}{2 \sqrt{2}} l n (3 - \sqrt{2})$ $n o w I - J = \frac{1}{2 i} \int_{1}^{+ \infty} \frac{d x}{x^{4} + 1} . . . . . . (i i)$ $= \frac{1}{4 i} [\int_{1}^{+ \infty} \frac{x^{2} + 1}{x^{4} + 1} d x - \int_{1}^{+ \infty} \frac{x^{2} - 1}{x^{4} + 1} d x]$ $= \frac{1}{4 i} [I_{1} - I_{2}] = - \frac{i}{4} [\frac{π}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}} l n (3 - \sqrt{2})]$ $n o w (i) + (i i)$ $I = [\frac{π}{4 \sqrt{2}} - \frac{1}{4 \sqrt{2}} l n (3 - \sqrt{2})] - i [\frac{π}{16 \sqrt{2}} + \frac{1}{16 \sqrt{2}} l n (3 - \sqrt{2})]$ $a n d (i) - (i i)$ $J = I^{*} = c o m p l e x c o n j u g a t e o f I .$ $n o w f r o m (i i) . . .$ $\int_{1}^{+ \infty} \frac{d x}{x^{4} + 1} = 2 i (I - J)$ $= [\frac{π}{4 \sqrt{2}} + \frac{1}{4 \sqrt{2}} l n (3 - \sqrt{2})] .$ $v e r y n i c e p r o b l e m . . . {i t}^{'} s r e a l l y$ $i n t e r e s t i n g .$
	Commented by abdomathmax last updated on 01/Jun/20

	$thanks sir$
	Answered by abdomathmax last updated on 01/Jun/20
	$1. ∫_1 ^(+∞) (dx/(x^2 −i)) =∫_1 ^(+∞) (dx/((x−(√i))(x+(√i)))) =(1/(2(√i)))∫_1 ^(+∞) ((1/(x−(√i)))−(1/(x+(√i))))dx =(1/(2(√i)))[ln(((x−(√i))/(x+(√i))))]_1 ^(+∞) =(1/(2(√i)))(−ln(((1−(√i))/(1+(√i))))) =(1/(2(√i)))( ln(1+(√i))−ln(1−(√i))) but 1+(√i)=1+e^((iπ)/4) =1+(1/(√2))+(i/(√2)) =(√((1+(1/(√2)))^2 +(1/2)))× e^(iarctan((1/((√2)(1+(1/(√2)))))) =(√((3/2)+(√2)+(1/2)))e^(iarctan((1/(1+(√2))))) =(√(2+(√2))) e^(iarctan((1/(1+(√2))))) ⇒ ln(1+(√i))=(1/2)ln(2+(√2))+iarctan((1/(1+(√2)))) ln(1−(√i)) =conj(..)=(1/2)ln(2+(√2))−iarctan((1/(1+(√2)))) ⇒(1/(2(√i)))(ln(1+(√i))−ln(1−(√i))) =(1/(2(√i))){2i arctan((1/(1+(√2))))} =(√i)arctan((√2)−1) =(√i)((π/8)) =(π/8)e^((iπ)/4) ∫_1 ^(+∞) (dx/(x^2 +i)) =conj(∫_1 ^(+∞) (dx/(x^2 −i))) =(π/8)e^(−((iπ)/4)) 2.∫_1 ^(+∞) (dx/(x^4 +1)) =∫_1 ^(+∞) (dx/((x^2 −i)(x^2 +i))) =(1/(2i))∫_1 ^(+∞) ((1/(x^2 −i))−(1/(x^2 +i)))dx =(1/(2i))((π/8)e^((iπ)/4) −(π/8)e^(−((iπ)/4)) ) =(π/8)sin((π/4)) =(π/8)×((√2)/2) =((π(√2))/(16))$
	$1 . \int_{1}^{+ \infty} \frac{dx}{x^{2} - i} = \int_{1}^{+ \infty} \frac{dx}{(x - \sqrt{i}) (x + \sqrt{i})}$ $= \frac{1}{2 \sqrt{i}} \int_{1}^{+ \infty} (\frac{1}{x - \sqrt{i}} - \frac{1}{x + \sqrt{i}}) dx$ $= \frac{1}{2 \sqrt{i}} {[\ln (\frac{x - \sqrt{i}}{x + \sqrt{i}})]}_{1}^{+ \infty} = \frac{1}{2 \sqrt{i}} (- \ln (\frac{1 - \sqrt{i}}{1 + \sqrt{i}}))$ $= \frac{1}{2 \sqrt{i}} (\ln (1 + \sqrt{i}) - \ln (1 - \sqrt{i})) but$ $1 + \sqrt{i} = 1 + e^{\frac{i π}{4}} = 1 + \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} = \sqrt{{(1 + \frac{1}{\sqrt{2}})}^{2} + \frac{1}{2}} \times$ $e^{iarctan (\frac{1}{\sqrt{2} (1 + \frac{1}{\sqrt{2}}})} = \sqrt{\frac{3}{2} + \sqrt{2} + \frac{1}{2}} e^{iarctan (\frac{1}{1 + \sqrt{2}})}$ $= \sqrt{2 + \sqrt{2}} e^{iarctan (\frac{1}{1 + \sqrt{2}})} \Rightarrow$ $\ln (1 + \sqrt{i}) = \frac{1}{2} \ln (2 + \sqrt{2}) + iarctan (\frac{1}{1 + \sqrt{2}})$ $\ln (1 - \sqrt{i}) = conj (. .) = \frac{1}{2} \ln (2 + \sqrt{2}) - iarctan (\frac{1}{1 + \sqrt{2}})$ $\Rightarrow \frac{1}{2 \sqrt{i}} (\ln (1 + \sqrt{i}) - \ln (1 - \sqrt{i}))$ $= \frac{1}{2 \sqrt{i}} {2 i \arctan (\frac{1}{1 + \sqrt{2}})} = \sqrt{i} \arctan (\sqrt{2} - 1)$ $= \sqrt{i} (\frac{π}{8}) = \frac{π}{8} e^{\frac{i π}{4}}$ $\int_{1}^{+ \infty} \frac{dx}{x^{2} + i} = conj (\int_{1}^{+ \infty} \frac{dx}{x^{2} - i}) = \frac{π}{8} e^{- \frac{i π}{4}}$ $2 . \int_{1}^{+ \infty} \frac{dx}{x^{4} + 1} = \int_{1}^{+ \infty} \frac{dx}{(x^{2} - i) (x^{2} + i)}$ $= \frac{1}{2 i} \int_{1}^{+ \infty} (\frac{1}{x^{2} - i} - \frac{1}{x^{2} + i}) dx$ $= \frac{1}{2 i} (\frac{π}{8} e^{\frac{i π}{4}} - \frac{π}{8} e^{- \frac{i π}{4}})$ $= \frac{π}{8} \sin (\frac{π}{4}) = \frac{π}{8} \times \frac{\sqrt{2}}{2} = \frac{π \sqrt{2}}{16}$
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