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Question Number 70237 by mathmax by abdo last updated on 02/Oct/19

calculate ∫_0 ^∞ ((xsin(αx))/(1+x^4 ))dx with α real

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{xsin}\left(\alpha{x}\right)}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\:{with}\:\alpha\:{real} \\ $$

Commented by mind is power last updated on 02/Oct/19

nice i deleat my post mistack[ ((zsin(z))/(1+z^4 )) when ∣z∣→∞ is not defind z=re^(ia) a∈[0,π[ sin(z)=((e^(ire^(ia) ) −e^(ire^(ia) ) )/(2i)) the factor e^(−ire^(ia) ) =e^(−irvos(a)) .e^(rsin(a)) ∣e^(−ire^(ia) ) ∣=e^(rsin(a)) witch is get big as we want ⇒lim zf(z)⇏0 jordan lemma dont not applie ∫_(−∞) ^(+∞) f(z)dz≠2iπresid(.....) i have done it quickly without seeing what going on Sorry sir

$${nice}\:\:{i}\:{deleat}\:{my}\:{post}\:{mistack}\left[\right. \\ $$$$\frac{{zsin}\left({z}\right)}{\mathrm{1}+{z}^{\mathrm{4}} }\:\:{when}\:\:\mid{z}\mid\rightarrow\infty\:\:\:{is}\:{not}\:{defind} \\ $$$${z}={re}^{{ia}} \:{a}\in\left[\mathrm{0},\pi\left[\right.\right. \\ $$$${sin}\left({z}\right)=\frac{{e}^{{ire}^{{ia}} } −{e}^{{ire}^{{ia}} } }{\mathrm{2}{i}} \\ $$$${the}\:{factor}\:{e}^{−{ire}^{{ia}} } ={e}^{−{irvos}\left({a}\right)} .{e}^{{rsin}\left({a}\right)} \mid{e}^{−{ire}^{{ia}} } \mid={e}^{{rsin}\left({a}\right)} \:{witch}\:{is}\:{get}\:{big}\:{as}\:{we}\:{want} \\ $$$$\Rightarrow{lim}\:{zf}\left({z}\right)\nRightarrow\mathrm{0}\:{jordan}\:{lemma}\:{dont}\:{not}\:{applie} \\ $$$$\int_{−\infty} ^{+\infty} {f}\left({z}\right){dz}\neq\mathrm{2}{i}\pi{resid}\left(.....\right) \\ $$$${i}\:{have}\:{done}\:{it}\:{quickly}\:\:{without}\:{seeing}\:{what}\:{going}\:{on}\:{Sorry}\:{sir}\: \\ $$

Commented by mathmax by abdo last updated on 02/Oct/19

let A =∫_0 ^∞ ((xsin(αx))/(x^4 +1))dx ⇒2A =∫_(−∞) ^(+∞) ((xsin(αx))/(x^4 +1))dx =Im(∫_(−∞) ^(+∞) ((x e^(iαx) )/(x^4 +1))dx) let ϕ(z)=((z e^(iαz) )/(z^4 +1)) poles of ϕ? ϕ(z) =((z e^(iαz) )/((z^2 −i)(z^2 +i))) =((z e^(iαz) )/((z−(√i))(z+(√i))(z−(√(−i)))(z+(√(−i))))) =((z e^(iαz) )/((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) so the poles of ϕ are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) and ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{Res(ϕ,e^((iπ)/4) )+Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,e^((iπ)/4) )=lim_(z→e^((iπ)/4) ) (z−e^((iπ)/4) )ϕ(z) =((e^((iπ)/4) e^(iαe^((iπ)/4) ) )/(2e^((iπ)/4) (2i))) =(e^(iα((1/((√2) ))+(i/(√2)))) /(4i)) =(1/(4i)) e^((iα)/(√2)) e^(−(α/(√2))) Res(ϕ,−e^(−((iπ)/4)) ) =lim_(z→−e^(−((iπ)/4)) ) (z+e^(−((iπ)/4)) )ϕ(z) =((−e^(−((iπ)/4)) e^(iα(−e^(−((iπ)/4)) )) )/((−2i)(−2e^(−((iπ)/4)) ))) =(1/(4i)) e^(−iα((1/(√2))−(i/(√2)))) =−(1/(4i)) e^(−((iα)/(√( 2)))) e^(−(α/(√2))) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{(1/(4i)) e^(−(α/(√2))) e^((iα)/(√2)) −(1/(4i)) e^(−(α/(√2))) e^(−((iα)/(√( 2)))) } =(π/2) e^(−(α/(√2))) { 2i sin((α/(√2)))} =iπ e^(−(α/(√2))) sin((α/(√2))) ⇒ 2A =π e^(−(α/(√2))) sin((α/(√2))) ⇒ A =(π/2) e^(−(α/(√2))) sin((α/(√2))) .

$${let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{xsin}\left(\alpha{x}\right)}{{x}^{\mathrm{4}} \:+\mathrm{1}}{dx}\:\Rightarrow\mathrm{2}{A}\:=\int_{−\infty} ^{+\infty} \:\frac{{xsin}\left(\alpha{x}\right)}{{x}^{\mathrm{4}} \:+\mathrm{1}}{dx} \\ $$$$={Im}\left(\int_{−\infty} ^{+\infty} \:\frac{{x}\:{e}^{{i}\alpha{x}} }{{x}^{\mathrm{4}} \:+\mathrm{1}}{dx}\right)\:\:{let}\:\varphi\left({z}\right)=\frac{{z}\:{e}^{{i}\alpha{z}} }{{z}^{\mathrm{4}} \:+\mathrm{1}}\:\:{poles}\:{of}\:\varphi? \\ $$$$\varphi\left({z}\right)\:=\frac{{z}\:{e}^{{i}\alpha{z}} }{\left({z}^{\mathrm{2}} −{i}\right)\left({z}^{\mathrm{2}} +{i}\right)}\:=\frac{{z}\:{e}^{{i}\alpha{z}} }{\left({z}−\sqrt{{i}}\right)\left({z}+\sqrt{{i}}\right)\left({z}−\sqrt{−{i}}\right)\left({z}+\sqrt{−{i}}\right)} \\ $$$$=\frac{{z}\:{e}^{{i}\alpha{z}} }{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$$\overset{−} {+}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:{and}\:\overset{−} {+}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:{and}\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)+{Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\varphi\left({z}\right) \\ $$$$=\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} \:{e}^{{i}\alpha{e}^{\frac{{i}\pi}{\mathrm{4}}} } }{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \left(\mathrm{2}{i}\right)}\:=\frac{{e}^{{i}\alpha\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}+\frac{{i}}{\sqrt{\mathrm{2}}}\right)} }{\mathrm{4}{i}}\:=\frac{\mathrm{1}}{\mathrm{4}{i}}\:{e}^{\frac{{i}\alpha}{\sqrt{\mathrm{2}}}} \:\:{e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} \\ $$$${Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:={lim}_{{z}\rightarrow−{e}^{−\frac{{i}\pi}{\mathrm{4}}} } \:\:\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\varphi\left({z}\right) \\ $$$$=\frac{−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:{e}^{{i}\alpha\left(−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} }{\left(−\mathrm{2}{i}\right)\left(−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}\:=\frac{\mathrm{1}}{\mathrm{4}{i}}\:{e}^{−{i}\alpha\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}−\frac{{i}}{\sqrt{\mathrm{2}}}\right)} \:=−\frac{\mathrm{1}}{\mathrm{4}{i}}\:{e}^{−\frac{{i}\alpha}{\sqrt{\:\mathrm{2}}}} {e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\frac{\mathrm{1}}{\mathrm{4}{i}}\:{e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} \:\:{e}^{\frac{{i}\alpha}{\sqrt{\mathrm{2}}}} \:\:\:−\frac{\mathrm{1}}{\mathrm{4}{i}}\:{e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} \:{e}^{−\frac{{i}\alpha}{\sqrt{\:\mathrm{2}}}} \right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\:{e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} \left\{\:\:\mathrm{2}{i}\:{sin}\left(\frac{\alpha}{\sqrt{\mathrm{2}}}\right)\right\}\:={i}\pi\:{e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} \:\:\:{sin}\left(\frac{\alpha}{\sqrt{\mathrm{2}}}\right)\:\Rightarrow \\ $$$$\mathrm{2}{A}\:=\pi\:{e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} \:{sin}\left(\frac{\alpha}{\sqrt{\mathrm{2}}}\right)\:\Rightarrow\:{A}\:=\frac{\pi}{\mathrm{2}}\:{e}^{−\frac{\alpha}{\sqrt{\mathrm{2}}}} \:\:\:{sin}\left(\frac{\alpha}{\sqrt{\mathrm{2}}}\right)\:. \\ $$$$ \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 03/Oct/19

nevermind sir.

$${nevermind}\:{sir}. \\ $$

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