let-f-a-dx-x-2-1-a-e-ix-with-a-gt-0-1-find-a-explicit-form-of-f-a-2-determine-also-g-a-dx-x-2-1-a-e-ix-2-3-let-I-Re-dx-

Question Number 67542 by mathmax by abdo last updated on 28/Aug/19

$${let}\:{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({a}\:+{e}^{{ix}} \right)}\:\:\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right){find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{also}\:{g}\left({a}\right)=\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({a}+{e}^{{ix}} \right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right){let}\:{I}\:={Re}\left(\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}+{e}^{{ix}} \right)}\right)\:{and}\:{J}={Im}\left(\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}+{e}^{{x}} \right)}\right) \\ $$$$\:{determine}\:{I}\:{and}\:{J}\:\:{and}\:\:{its}\:{values}. \\ $$

Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19

$${let}\:{change}\:{x}={tanu}\: \\ $$$${f}\left({a}\right)=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{du}}{{h}\left({u}\right)}\:\:\:\:\:\:\:\:{with}\:\:\:{h}\left({u}\right)={a}+{e}^{{itanu}} \:\:\:\:\: \\ $$$${g}\:{has}\:{only}\:{one}\:{root}\:{on}\:\left[\frac{−\pi}{\mathrm{2}};\frac{\pi}{\mathrm{2}}\right]\:{let}\:{named}\:{it}\:\:\theta \\ $$$${When}\:{using}\:{the}\:{RT}\:\: \\ $$$${f}\left({a}\right)=\mathrm{2}\pi{iRes}\left(\frac{\mathrm{1}}{{h}},\theta\right)\:=\frac{\mathrm{2}{i}\pi}{{h}'\left(\theta\right)}=\frac{\mathrm{2}{i}\pi}{{i}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){e}^{{itan}\theta} }\: \\ $$$$\theta\:{root}\:{of}\:{h}\:\Rightarrow\:{e}^{{itan}\theta} =−{a}\:\:{and}\:\:{tan}\theta=\pi−{ilna} \\ $$$${So}\:\:{f}\left({a}\right)=\frac{\mathrm{2}\pi}{−{a}\left(\mathrm{1}+\pi^{\mathrm{2}} +\left({lna}\right)^{\mathrm{2}} −\mathrm{2}\pi{ilna}\right)} \\ $$$${we}\:{can}\:{ascertain}\:{that}\:\:{g}\left({a}\right)=\frac{{df}}{{da}}\: \\ $$$${When}\:{stating}\:\:{x}\left({a}\right)=\mathrm{1}+\pi^{\mathrm{2}} +\left({lna}\right)^{\mathrm{2}} \:\:{and}\:\:{y}\left({a}\right)=\mathrm{2}\pi{lna}\: \\ $$$${Re}\left({f}\left({a}\right)\right)=−\frac{\mathrm{2}\pi}{{a}}\:\frac{{x}\left({a}\right)}{{x}^{\mathrm{2}} \left({a}\right)+{y}^{\mathrm{2}} \left({a}\right)}\:\:{and}\:{Im}\left({f}\left({a}\right)\right)=\frac{−\mathrm{2}\pi}{{a}}\:\frac{{y}\left({a}\right)}{{x}^{\mathrm{2}} \left({a}\right)+{y}^{\mathrm{2}} \left({a}\right)}\: \\ $$

Commented by mathmax by abdo last updated on 29/Aug/19

$${thank}\:{you}\:{sir}. \\ $$

Commented by mathmax by abdo last updated on 29/Aug/19

$$\left.\mathrm{1}\right){let}\:{consider}\:{tbe}\:{complex}\:{function}\:{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\left({a}\:+{e}^{{iz}} \right)} \\ $$$$\Rightarrow{W}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−{i}\right)\left({z}+{i}\right)\left({a}+{e}^{{iz}} \right)}\:\:{the}\:{poles}\:{of}\:{W}\:{are}\:{i}\:{and}\:−{i} \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{i}\right) \\ $$$${Res}\left({W},{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\left({z}−{i}\right){W}\left({z}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)\left({a}+{e}^{−\mathrm{1}} \right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{\mathrm{1}}{\mathrm{2}{i}\left({a}+{e}^{−\mathrm{1}} \right)}\:=\frac{\pi}{{a}+{e}^{−\mathrm{1}} }\:=\frac{\pi{e}}{{ae}\:+\mathrm{1}}\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\frac{\pi{e}}{{ae}\:+\mathrm{1}}\:\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:\:{f}^{'} \left({a}\right)\:=−\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({a}+{e}^{{ix}} \right)^{\mathrm{2}} }\:=−{g}\left({a}\right)\:\Rightarrow \\ $$$${g}\left({a}\right)\:=−{f}^{'} \left({a}\right)\:\:\:{but}\:{f}\left({a}\right)\:=\frac{\pi{e}}{{ae}\:+\mathrm{1}}\:\:\Rightarrow{f}^{'} \left({a}\right)\:=−\frac{\pi{e}^{\mathrm{2}} }{\left({ae}\:+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${g}\left({a}\right)\:=\frac{\pi{e}^{\mathrm{2}} }{\left({ae}\:+\mathrm{1}\right)^{\mathrm{2}} }\:. \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 29/Aug/19

$$\left.\mathrm{3}\right)\:{we}\:{have}\:{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({a}\:+{cosx}\:+{isinx}\right)} \\ $$$$=\int_{−\infty} ^{+\infty} \:\:\frac{{a}+{cosx}−{isinx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\left({a}+{cosx}\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} {x}\right)}{dx} \\ $$$$=\int_{−\infty} ^{+\infty} \:\:\frac{{a}+{cosx}}{\left.\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left\{{a}^{\mathrm{2}} \:+\mathrm{2}{acosx}\:+\mathrm{1}\right)\right\}}{dx}\:−{i}\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{sinx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left\{\:{a}^{\mathrm{2}} \:+\mathrm{2}{acosx}\:+\mathrm{1}\right\}}{dx} \\ $$$$\Rightarrow\:{Re}\left({f}\left({a}\right)\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{a}+{cosx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left\{\:{a}^{\mathrm{2}} \:+\mathrm{2}{acosx}\:+\mathrm{1}\right\}}{dx}\:{and} \\ $$$${Im}\left({f}\left({a}\right)\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{sinx}\:{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left\{{a}^{\mathrm{2}} \:+\mathrm{2}{acosx}\:+\mathrm{1}\right\}}=\mathrm{0}\:{because}\:{the}\: \\ $$$${function}\:{x}\:\rightarrow\frac{{sinx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left\{{a}^{\mathrm{2}} \:+\mathrm{2}{a}\:{cosx}\:+\mathrm{1}\right\}}\:{is}\:{odd}. \\ $$

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