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Question Number 67850 by mathmax by abdo last updated on 01/Sep/19
calculate ∫_(−∞) ^(+∞) (dx/(x^2 −z)) with z from C
$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} −{z}}\:\:{with}\:{z}\:{from}\:{C} \\ $$
Commented by mathmax by abdo last updated on 01/Sep/19
let I =∫_(−∞) ^(+∞) (dx/(x^2 −z)) let z =re^(iθ) ⇒I =∫_(−∞) ^(+∞) (dx/(x^2 −((√r)e^((iθ)/2) )^2 )) =∫_(−∞) ^(+∞) (dx/((x−(√r)e^((iθ)/2) )(x+(√r)e^((iθ)/2) ))) =(1/(2(√r)e^((iθ)/2) ))∫_(−∞) ^(+∞) {(1/(x−(√r)e^((iθ)/2) ))−(1/(x+(√r)e^((iθ)/2) ))}dx =(1/(2(√r)e^((iθ)/2) )){ ∫_(−∞) ^(+∞) (dx/(x−(√r)e^((iθ)/2) )) −∫_(−∞) ^(+∞) (dx/(x−(−(√r)e^((iθ)/2) )))} we have proved that for a ∈C ∫_(−∞) ^(+∞) (dx/(x−a)) =iπ if Im(a)>0 and −iπ if Im(a)<0 so if θ>0 ⇒ I =(1/(2(√r)e^((iθ)/2) )){iπ−(−iπ)} =((2iπ)/(2(√r)e^((iθ)/2) )) =((iπ)/( (√r)e^((iθ)/2) )) if θ<0 weget I =(1/(2(√r)e^((iθ)/2) )){−iπ−(iπ)} =−((iπ)/( (√r)e^((iθ)/2) ))
$${let}\:{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{2}} −{z}}\:\:{let}\:{z}\:={re}^{{i}\theta} \:\Rightarrow{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{2}} −\left(\sqrt{{r}}{e}^{\frac{{i}\theta}{\mathrm{2}}} \right)^{\mathrm{2}} } \\ $$$$=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}−\sqrt{{r}}{e}^{\frac{{i}\theta}{\mathrm{2}}} \right)\left({x}+\sqrt{{r}}{e}^{\frac{{i}\theta}{\mathrm{2}}} \right)}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{r}}{e}^{\frac{{i}\theta}{\mathrm{2}}} }\int_{−\infty} ^{+\infty} \left\{\frac{\mathrm{1}}{{x}−\sqrt{{r}}{e}^{\frac{{i}\theta}{\mathrm{2}}} }−\frac{\mathrm{1}}{{x}+\sqrt{{r}}{e}^{\frac{{i}\theta}{\mathrm{2}}} }\right\}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{r}}{e}^{\frac{{i}\theta}{\mathrm{2}}} }\left\{\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}−\sqrt{{r}}{e}^{\frac{{i}\theta}{\mathrm{2}}} }\:−\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}−\left(−\sqrt{{r}}{e}^{\frac{{i}\theta}{\mathrm{2}}} \right)}\right\} \\ $$$${we}\:{have}\:{proved}\:{that}\:{for}\:{a}\:\in{C}\:\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}−{a}}\:={i}\pi\:{if}\:{Im}\left({a}\right)>\mathrm{0}\:{and} \\ $$$$−{i}\pi\:{if}\:{Im}\left({a}\right)<\mathrm{0}\:{so}\:\:{if}\:\theta>\mathrm{0}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{r}}{e}^{\frac{{i}\theta}{\mathrm{2}}} }\left\{{i}\pi−\left(−{i}\pi\right)\right\}\:=\frac{\mathrm{2}{i}\pi}{\mathrm{2}\sqrt{{r}}{e}^{\frac{{i}\theta}{\mathrm{2}}} }\:=\frac{{i}\pi}{\:\sqrt{{r}}{e}^{\frac{{i}\theta}{\mathrm{2}}} } \\ $$$${if}\:\theta<\mathrm{0}\:\:{weget}\:\:{I}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{r}}{e}^{\frac{{i}\theta}{\mathrm{2}}} }\left\{−{i}\pi−\left({i}\pi\right)\right\}\:=−\frac{{i}\pi}{\:\sqrt{{r}}{e}^{\frac{{i}\theta}{\mathrm{2}}} } \\ $$

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