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Question Number 34229 by abdo imad last updated on 03/May/18
calculate ∫_(−∞) ^∞ ((cos(tx))/(1+x^4 )) dx with t≥0 2) calculate ∫_0 ^∞ (dx/(1+x^4 )) .
$${calculate}\:\int_{−\infty} ^{\infty} \:\:\frac{{cos}\left({tx}\right)}{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx}\:\:{with}\:{t}\geqslant\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} }\:. \\ $$
Commented by math khazana by abdo last updated on 04/May/18
let put f(t)= ∫_(−∞) ^(+∞) ((cos(tx))/(1+x^4 ))dx f(t)= Re( ∫_(−∞) ^(+∞) (e^(itx) /(1+x^4 ))dx) let consider the complex function ϕ(z) = (e^(itz) /(1+z^4 )) we have ϕ(z)= (e^(itz) /((z^2 −i)(z^2 +i))) = (e^(itz) /((z −(√i))(z+(√i))( z−(√(−i)))(z+(√(−i))))) = (e^(itz) /((z−e^(i(π/4)) )(z +e^(i(π/4)) )( z −e^(−((iπ)/4)) )(z +e^(−i(π/4)) ))) the poles of ϕ are e^(i(π/4)) ,−e^(i(π/4)) ,e^(−i(π/4)) , −e^(−i(π/4)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ( Res(ϕ,e^(i(π/4)) ) +Res(ϕ,−e^(−i(π/4)) ) let p(z) all poles of ϕ are roots of p and p^′ (z) = 4 z^3 ⇒p^′ (z_k )= 4z_k ^3 = ((4z_k ^4 )/z_k ) = ((−4)/z_k ) Res(ϕ,e^(i(π/4)) ) = (e^(it e^(i(π/4)) ) /(−4)) e^(i(π/4)) =−(1/4) e^(i(π/4)) e^(it e^(i(π/4)) ) =−(1/4) e^(i((π/4) +t( (1/( (√2))) +(i/( (√2)))))) = −(1/4) e^(i((π/4) +(t/( (√2))))−(t/( (√2)))) =−(1/4) e^(−(t/( (√2)))) e^(i((π/4)+(t/( (√2))))) Res(ϕ,−e^(−i(π/4)) )= (1/4) e^(−i(π/4)) e^(−ite^(−i(π/4)) ) = (1/4) e^(−i( (π/4) +t ((1/( (√2))) −(i/( (√2)))))) = (1/4) e^(−i((π/4) +(t/( (√2)))) −(t/( (√2)))) = (1/4) e^(−(t/( (√2)))) e^(−i( (π/4) +(t/( (√2))))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(−(t/( (√2)))) /4)( e^(−i((π/4) +(t/( (√2))))) − e^(i((π/4) +(t/( (√2))))) ) =((iπ)/2) e^(−(t/( (√2)))) ( −2i sin((π/4) +(t/2)))= π e^(−(t/( (√2)))) sin((π/4) +(t/2)) f(t)=Re( ∫_(−∞) ^(+∞) ϕ(z)dz) ⇒ f(t)= π e^(−(t/( (√2)))) sin((π/4) +(t/2)) .
$${let}\:{put}\:{f}\left({t}\right)=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left({tx}\right)}{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$$${f}\left({t}\right)=\:{Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{itx}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\right)\:{let}\:{consider}\:{the}\:{complex} \\ $$$${function}\:\varphi\left({z}\right)\:=\:\frac{{e}^{{itz}} }{\mathrm{1}+{z}^{\mathrm{4}} }\:\:\:{we}\:{have} \\ $$$$\varphi\left({z}\right)=\:\frac{{e}^{{itz}} }{\left({z}^{\mathrm{2}} −{i}\right)\left({z}^{\mathrm{2}} +{i}\right)}\:=\:\frac{{e}^{{itz}} }{\left({z}\:−\sqrt{{i}}\right)\left({z}+\sqrt{{i}}\right)\left(\:{z}−\sqrt{−{i}}\right)\left({z}+\sqrt{−{i}}\right)} \\ $$$$=\:\:\frac{{e}^{{itz}} }{\left({z}−{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\left({z}\:+{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\left(\:{z}\:−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}\:+{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)} \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:\:{e}^{{i}\frac{\pi}{\mathrm{4}}} ,−{e}^{{i}\frac{\pi}{\mathrm{4}}} ,{e}^{−{i}\frac{\pi}{\mathrm{4}}} \:,\:−{e}^{−{i}\frac{\pi}{\mathrm{4}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:{Res}\left(\varphi,{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\:+{Res}\left(\varphi,−{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)\right. \\ $$$${let}\:{p}\left({z}\right)\:{all}\:{poles}\:{of}\:\varphi\:{are}\:{roots}\:{of}\:{p}\:{and} \\ $$$${p}^{'} \left({z}\right)\:=\:\mathrm{4}\:{z}^{\mathrm{3}} \:\Rightarrow{p}^{'} \left({z}_{{k}} \right)=\:\mathrm{4}{z}_{{k}} ^{\mathrm{3}} =\:\frac{\mathrm{4}{z}_{{k}} ^{\mathrm{4}} }{{z}_{{k}} }\:=\:\frac{−\mathrm{4}}{{z}_{{k}} } \\ $$$${Res}\left(\varphi,{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\:\:=\:\frac{{e}^{{it}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} } }{−\mathrm{4}}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} =−\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\:\:\:{e}^{{it}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{{i}\left(\frac{\pi}{\mathrm{4}}\:+{t}\left(\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\right)} =\:−\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{{i}\left(\frac{\pi}{\mathrm{4}}\:+\frac{{t}}{\:\sqrt{\mathrm{2}}}\right)−\frac{{t}}{\:\sqrt{\mathrm{2}}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{−\frac{{t}}{\:\sqrt{\mathrm{2}}}} \:\:{e}^{{i}\left(\frac{\pi}{\mathrm{4}}+\frac{{t}}{\:\sqrt{\mathrm{2}}}\right)} \\ $$$${Res}\left(\varphi,−{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)=\:\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \:\:{e}^{−{ite}^{−{i}\frac{\pi}{\mathrm{4}}} } \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{−{i}\left(\:\frac{\pi}{\mathrm{4}}\:+{t}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\:−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\right)} \:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{−{i}\left(\frac{\pi}{\mathrm{4}}\:+\frac{{t}}{\:\sqrt{\mathrm{2}}}\right)\:−\frac{{t}}{\:\sqrt{\mathrm{2}}}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{−\frac{{t}}{\:\sqrt{\mathrm{2}}}} \:\:\:{e}^{−{i}\left(\:\frac{\pi}{\mathrm{4}}\:+\frac{{t}}{\:\sqrt{\mathrm{2}}}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{−\frac{{t}}{\:\sqrt{\mathrm{2}}}} }{\mathrm{4}}\left(\:\:{e}^{−{i}\left(\frac{\pi}{\mathrm{4}}\:+\frac{{t}}{\:\sqrt{\mathrm{2}}}\right)} \:−\:{e}^{{i}\left(\frac{\pi}{\mathrm{4}}\:+\frac{{t}}{\:\sqrt{\mathrm{2}}}\right)} \right) \\ $$$$=\frac{{i}\pi}{\mathrm{2}}\:{e}^{−\frac{{t}}{\:\sqrt{\mathrm{2}}}} \left(\:−\mathrm{2}{i}\:{sin}\left(\frac{\pi}{\mathrm{4}}\:+\frac{{t}}{\mathrm{2}}\right)\right)=\:\pi\:{e}^{−\frac{{t}}{\:\sqrt{\mathrm{2}}}} \:{sin}\left(\frac{\pi}{\mathrm{4}}\:+\frac{{t}}{\mathrm{2}}\right) \\ $$$${f}\left({t}\right)={Re}\left(\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\right)\:\Rightarrow \\ $$$${f}\left({t}\right)=\:\pi\:{e}^{−\frac{{t}}{\:\sqrt{\mathrm{2}}}} \:{sin}\left(\frac{\pi}{\mathrm{4}}\:+\frac{{t}}{\mathrm{2}}\right)\:. \\ $$
Commented by math khazana by abdo last updated on 04/May/18
2) let take t=0 ⇒ ∫_(−∞) ^(+∞) (dx/(1+x^4 )) = f(0)= π sin((π/4)) = (π/( (√2))) = ((π(√2))/2) ⇒ ∫_0 ^(+∞) (dx/(1+x^4 )) = ((π(√2))/4) .
$$\left.\mathrm{2}\right)\:{let}\:{take}\:{t}=\mathrm{0}\:\Rightarrow\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} }\:=\:{f}\left(\mathrm{0}\right)=\:\pi\:{sin}\left(\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\:\frac{\pi}{\:\sqrt{\mathrm{2}}}\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Rightarrow\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} }\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:\:. \\ $$
Commented by math khazana by abdo last updated on 04/May/18
error at the final lines ∫_(−∞) ^(+∞) ϕ(z)dz = ((iπ)/2) e^(−(t/( (√2)))) (−2i sin((π/4) +(t/( (√2))))) = π e^(−(t/( (√2)))) sin((π/4) +(t/( (√2)))) ⇒ f(t) = π e^(−(t/( (√2)))) sin( (π/4) +(t/( (√2)))) .
$${error}\:{at}\:{the}\:{final}\:{lines} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\:\frac{{i}\pi}{\mathrm{2}}\:{e}^{−\frac{{t}}{\:\sqrt{\mathrm{2}}}} \left(−\mathrm{2}{i}\:{sin}\left(\frac{\pi}{\mathrm{4}}\:+\frac{{t}}{\:\sqrt{\mathrm{2}}}\right)\right) \\ $$$$=\:\pi\:{e}^{−\frac{{t}}{\:\sqrt{\mathrm{2}}}} \:\:\:{sin}\left(\frac{\pi}{\mathrm{4}}\:+\frac{{t}}{\:\sqrt{\mathrm{2}}}\right)\:\Rightarrow \\ $$$${f}\left({t}\right)\:\:=\:\pi\:{e}^{−\frac{{t}}{\:\sqrt{\mathrm{2}}}} \:\:\:{sin}\left(\:\frac{\pi}{\mathrm{4}}\:+\frac{{t}}{\:\sqrt{\mathrm{2}}}\right)\:. \\ $$
Answered by Joel578 last updated on 03/May/18
∫_0 ^∞ (t^(a − 1) /(1 + t)) dt = (π/(sin (πa))), 0 < a < 1 I = ∫_0 ^∞ (dx/(1 + x^4 )) (u = x^4 → du = 4x^3 dx) = (1/4) ∫_0 ^∞ (x^(−3) /(1 + u)) du = (1/4) ∫_0 ^∞ (u^(−(3/4)) /(1 + u)) du = (1/4) ∫_0 ^∞ (u^((1/4) − 1) /(1 + u)) du = (1/4)((π/(sin (π/4)))) = ((π(√2))/4)
$$\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}\:−\:\mathrm{1}} }{\mathrm{1}\:+\:{t}}\:{dt}\:=\:\frac{\pi}{\mathrm{sin}\:\left(\pi{a}\right)},\:\:\:\mathrm{0}\:<\:{a}\:<\:\mathrm{1} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\mathrm{1}\:+\:{x}^{\mathrm{4}} }\:\:\left({u}\:=\:{x}^{\mathrm{4}} \:\:\rightarrow\:\:{du}\:=\:\mathrm{4}{x}^{\mathrm{3}} \:{dx}\right) \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{−\mathrm{3}} }{\mathrm{1}\:+\:{u}}\:{du} \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{u}^{−\frac{\mathrm{3}}{\mathrm{4}}} }{\mathrm{1}\:+\:{u}}\:{du}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{u}^{\frac{\mathrm{1}}{\mathrm{4}}\:−\:\mathrm{1}} }{\mathrm{1}\:+\:{u}}\:{du} \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\pi}{\mathrm{sin}\:\left(\pi/\mathrm{4}\right)}\right) \\ $$$$\:\:\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/May/18
∫_0 ^∞ (dx/(1+x^4 )) =∫_0 ^∞ ((1/x^2 )/(x^2 +(1/x^2 ) ))dx
$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} }\:=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:}{dx} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/May/18
(1/2)∫_0 ^∞ (((1+(1/(x^2 )))−(1−(1/x^2 )))/(x^2 +(1/x^2 )))dx (1/2)∫_0 ^∞ (((1+(1/x^2 )))/((x−(1/x))^2 +2)) dx−(1/2)∫_0 ^∞ (((1−(1/x^2 )))/((x+(1/x))^2 −2)) (1/2)∫_0 ^∞ (dt_1 /(t_1 ^2 +((√(2 )) )^2 ))−(1/2)∫_0 ^∞ (dt_2 /(t_2 ^2 −((√2))^2 )) =(1/(2(√2) ))∣tan^(−1) ((t_1 /( (√2))))∣_0 ^∞ −(1/2).(1/(2(√2)))∣ln∣(((√2) −t_2 )/( (√2) +t_2 ))∣
$$\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:}\right)−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}}\:{dx}−\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}_{\mathrm{1}} }{{t}_{\mathrm{1}} ^{\mathrm{2}} +\left(\sqrt{\mathrm{2}\:}\:\right)^{\mathrm{2}} \:}−\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}_{\mathrm{2}} }{{t}_{\mathrm{2}} ^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\:}\mid{tan}^{−\mathrm{1}} \left(\frac{{t}_{\mathrm{1}} }{\:\sqrt{\mathrm{2}}}\right)\underset{\mathrm{0}} {\overset{\infty} {\mid}}−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mid{ln}\mid\frac{\sqrt{\mathrm{2}}\:−{t}_{\mathrm{2}} }{\:\sqrt{\mathrm{2}}\:+{t}_{\mathrm{2}} }\mid \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/May/18
=(1/(2(√2))).(Π/2)−(1/(4(√2))).ln∣((((√2)/t_2 )−1)/(((√2)/t_2 )+1))∣→its value is zero
$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}.\frac{\Pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}.{ln}\mid\frac{\frac{\sqrt{\mathrm{2}}}{{t}_{\mathrm{2}} }−\mathrm{1}}{\frac{\sqrt{\mathrm{2}}}{{t}_{\mathrm{2}} }+\mathrm{1}}\mid\rightarrow{its}\:{value}\:{is}\:{zero} \\ $$

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