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Question Number 53785 by maxmathsup by imad last updated on 25/Jan/19

let f(x)=∫_0 ^∞ ((tsin(tx))/(1+t^4 ))dt with x>0 1) find a explicit form of f(x) 2) find the value of ∫_0 ^∞ ((tsin(2t))/(1+t^4 ))dt.

$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{tsin}\left({tx}\right)}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:\:{with}\:{x}>\mathrm{0} \\ $$ $$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$ $$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{tsin}\left(\mathrm{2}{t}\right)}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}. \\ $$

Commented bymaxmathsup by imad last updated on 26/Jan/19

1) we have 2f(x)=∫_(−∞) ^(+∞) ((t sin(tx))/(t^4 +1))dt=Im(∫_(−∞) ^(+∞) ((t e^(itx) )/(t^4 +1))dt) let consider the complex function ϕ(z) = ((z e^(ixz) )/(z^4 +1)) ⇒ϕ(z) =((z e^(ixz) )/((z^2 −i)(z^2 +i))) =((z e^(ixz) )/((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) the poles of ϕ are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ , e^((iπ)/4) ) +Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ, e^((iπ)/4) ) = ((e^((iπ)/4) e^(ix((1/(√2))+(i/(√2)))) )/(2e^((iπ)/4) (2i))) =(1/(4i)) e^((ix)/(√2)) .e^(−(t/(√2))) =(e^(−(x/(√2))) /(4i)) e^((ix)/(√2)) Res(ϕ,−e^(−((iπ)/4)) ) =((−e^(−((iπ)/4)) e^(ix(−(1/(√2))+(i/(√2)))) )/((−2ie^(−((iπ)/4)) )(−2i))) =−(1/(4i)) e^(−(x/(√2))) e^(−((ix)/(√2))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ e^(−(x/(√2))) { (1/(4i)) e^((ix)/(√2)) −(1/(4i)) e^(−((ix)/2)) } =(π/2) e^(−(x/(√2))) { 2i sin((x/(√2)))} =iπ e^(−(x/(√2))) sin((x/(√2))) ⇒2f(x)=π e^(−(x/(√(2 )))) sin((x/(√2))) ⇒ ★f(x) =(π/2) e^(−(x/(√2))) sin((x/(√2)))★ 2) this integral is a spacial case ∫_0 ^(+∞) ((tsin(2t))/(1+t^4 )) dt =f(2) = (π/2) e^(−(√2)) sin((√2)) .

$$\left.\mathrm{1}\right)\:{we}\:{have}\:\mathrm{2}{f}\left({x}\right)=\int_{−\infty} ^{+\infty} \:\:\:\frac{{t}\:{sin}\left({tx}\right)}{{t}^{\mathrm{4}} \:+\mathrm{1}}{dt}={Im}\left(\int_{−\infty} ^{+\infty} \:\frac{{t}\:{e}^{{itx}} }{{t}^{\mathrm{4}} \:+\mathrm{1}}{dt}\right)\:{let}\:{consider}\:{the}\:{complex}\:{function} \\ $$ $$\varphi\left({z}\right)\:=\:\frac{{z}\:{e}^{{ixz}} }{{z}^{\mathrm{4}} \:+\mathrm{1}}\:\:\Rightarrow\varphi\left({z}\right)\:=\frac{{z}\:{e}^{{ixz}} }{\left({z}^{\mathrm{2}} −{i}\right)\left({z}^{\mathrm{2}} \:+{i}\right)}\:=\frac{{z}\:{e}^{{ixz}} }{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$ $${the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:\:{and}\:\overset{−} {+}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$ $$\int_{−\infty} ^{+\infty} \:\:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi\:,\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:+{Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$ $${Res}\left(\varphi,\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:=\:\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} \:{e}^{{ix}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+\frac{{i}}{\sqrt{\mathrm{2}}}\right)} }{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \left(\mathrm{2}{i}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}{i}}\:{e}^{\frac{{ix}}{\sqrt{\mathrm{2}}}} \:\:.{e}^{−\frac{{t}}{\sqrt{\mathrm{2}}}} =\frac{{e}^{−\frac{{x}}{\sqrt{\mathrm{2}}}} }{\mathrm{4}{i}}\:{e}^{\frac{{ix}}{\sqrt{\mathrm{2}}}} \\ $$ $${Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:=\frac{−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:{e}^{{ix}\left(−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+\frac{{i}}{\sqrt{\mathrm{2}}}\right)} }{\left(−\mathrm{2}{ie}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left(−\mathrm{2}{i}\right)}\:=−\frac{\mathrm{1}}{\mathrm{4}{i}}\:{e}^{−\frac{{x}}{\sqrt{\mathrm{2}}}} {e}^{−\frac{{ix}}{\sqrt{\mathrm{2}}}} \:\:\:\:\Rightarrow \\ $$ $$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{e}^{−\frac{{x}}{\sqrt{\mathrm{2}}}} \left\{\:\frac{\mathrm{1}}{\mathrm{4}{i}}\:{e}^{\frac{{ix}}{\sqrt{\mathrm{2}}}} \:\:−\frac{\mathrm{1}}{\mathrm{4}{i}}\:{e}^{−\frac{{ix}}{\mathrm{2}}} \right\} \\ $$ $$=\frac{\pi}{\mathrm{2}}\:{e}^{−\frac{{x}}{\sqrt{\mathrm{2}}}} \left\{\:\mathrm{2}{i}\:{sin}\left(\frac{{x}}{\sqrt{\mathrm{2}}}\right)\right\}\:={i}\pi\:{e}^{−\frac{{x}}{\sqrt{\mathrm{2}}}} \:{sin}\left(\frac{{x}}{\sqrt{\mathrm{2}}}\right)\:\Rightarrow\mathrm{2}{f}\left({x}\right)=\pi\:{e}^{−\frac{{x}}{\sqrt{\mathrm{2}\:}}} \:\:{sin}\left(\frac{{x}}{\sqrt{\mathrm{2}}}\right)\:\Rightarrow \\ $$ $$\bigstar{f}\left({x}\right)\:=\frac{\pi}{\mathrm{2}}\:{e}^{−\frac{{x}}{\sqrt{\mathrm{2}}}} \:{sin}\left(\frac{{x}}{\sqrt{\mathrm{2}}}\right)\bigstar \\ $$ $$\left.\mathrm{2}\right)\:{this}\:{integral}\:{is}\:{a}\:{spacial}\:{case}\: \\ $$ $$\int_{\mathrm{0}} ^{+\infty} \:\frac{{tsin}\left(\mathrm{2}{t}\right)}{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt}\:={f}\left(\mathrm{2}\right)\:=\:\frac{\pi}{\mathrm{2}}\:{e}^{−\sqrt{\mathrm{2}}} \:{sin}\left(\sqrt{\mathrm{2}}\right)\:. \\ $$ $$ \\ $$

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