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Question Number 37634 by math khazana by abdo last updated on 16/Jun/18
find ∫_0 ^(+∞) e^(−(t^2 +(1/t^2 ))) dt
$${find}\:\int_{\mathrm{0}} ^{+\infty} \:\:{e}^{−\left({t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)} {dt} \\ $$
Commented by prof Abdo imad last updated on 17/Jun/18
let I = ∫_0 ^∞ e^(−(t^2 +(1/t^2 ))) dt 2I = ∫_(−∞) ^∞ e^(−{ ( t−(1/t))^2 +2}) dt = e^(−2) ∫_(−∞) ^(+∞) e^(−(t−(1/t))^2 ) dt changement t−(1/t)=x give t^2 −1=xt ⇒t^2 −xt−1=0 Δ =x^2 +4 ⇒ t_1 =((x +(√(x^2 +4)))/2) and t_2 = ((x−(√(x^2 +4)))/2) let take t =((x +(√(x^2 +4)))/2) ⇒ dt =(1/2)( 1+ (x/( (√(x^2 +4))))) dx⇒ 2I = (e^(−2) /2)∫_(−∞) ^(+∞) e^(−x^2 ) ( 1+(x/( (√(x^2 +4)))))dx =(e^(−2) /2)∫_(−∞) ^(+∞) e^(−x^2 ) dx +(e^(−2) /2) ∫_(−∞) ^(+∞) ((x e^(−x^2 ) )/( (√(x^2 +4))))dx =(((√π) e^(−2) )/2) +0 because the function x→ ((x e^(−x^2 ) )/( (√(x^2 +4)))) is odd ⇒ I = ((e^(−2) (√π))/4) .
$${let}\:{I}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)} {dt} \\ $$$$\mathrm{2}{I}\:=\:\int_{−\infty} ^{\infty} \:\:\:{e}^{−\left\{\:\left(\:{t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} \:+\mathrm{2}\right\}} {dt} \\ $$$$=\:{e}^{−\mathrm{2}} \:\:\int_{−\infty} ^{+\infty} \:\:{e}^{−\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} } {dt}\:\:{changement} \\ $$$${t}−\frac{\mathrm{1}}{{t}}={x}\:{give}\:{t}^{\mathrm{2}} \:−\mathrm{1}={xt}\:\Rightarrow{t}^{\mathrm{2}} \:−{xt}−\mathrm{1}=\mathrm{0} \\ $$$$\Delta\:={x}^{\mathrm{2}} \:+\mathrm{4}\:\Rightarrow\:{t}_{\mathrm{1}} =\frac{{x}\:+\sqrt{{x}^{\mathrm{2}} \:+\mathrm{4}}}{\mathrm{2}}\:\:{and} \\ $$$${t}_{\mathrm{2}} =\:\frac{{x}−\sqrt{{x}^{\mathrm{2}} \:+\mathrm{4}}}{\mathrm{2}}\:\:{let}\:{take}\:{t}\:=\frac{{x}\:+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\:\Rightarrow \\ $$$$\:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\:\mathrm{1}+\:\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{4}}}\right)\:{dx}\Rightarrow \\ $$$$\mathrm{2}{I}\:=\:\frac{{e}^{−\mathrm{2}} }{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\:{e}^{−{x}^{\mathrm{2}} } \left(\:\mathrm{1}+\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{4}}}\right){dx} \\ $$$$=\frac{{e}^{−\mathrm{2}} }{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\:{e}^{−{x}^{\mathrm{2}} } {dx}\:\:\:+\frac{{e}^{−\mathrm{2}} }{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}\:{e}^{−{x}^{\mathrm{2}} } }{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{4}}}{dx} \\ $$$$=\frac{\sqrt{\pi}\:{e}^{−\mathrm{2}} }{\mathrm{2}}\:+\mathrm{0}\:{because}\:\:{the}\:{function} \\ $$$${x}\rightarrow\:\frac{{x}\:{e}^{−{x}^{\mathrm{2}} } }{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{4}}}\:{is}\:{odd}\:\Rightarrow \\ $$$${I}\:\:=\:\frac{{e}^{−\mathrm{2}} \sqrt{\pi}}{\mathrm{4}}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jun/18
=(1/2)∫_0 ^∞ (1−(1/t^2 )+1+(1/(t^2 )))e^(−(t^2 +(1/t^2 ))) dt =(1/2)∫_0 ^∞ (1−(1/t^2 ))e^(−{(t+(1/t))^2 −2}) dt+ (1/2)∫_0 ^∞ (1+(1/t^2 ))e^(−{(t−(1/t))^2 +2}) dt =(1/2)∫_0 ^∞ (1−(1/t^2 ))e^(−{(t+(1/t))^2 }) ×e^2 dt+ (1/2)∫_0 ^∞ (1+(1/t^2 ))e^(−{(t−(1/t))^2 }) ×e^(−2) dt =(e^2 /2)∫_∞ ^∞ e^(−k_1 ^2 ) dk_1 +(e^(−2) /2)∫_(−∞) ^∞ e^(−k_2 ^2 ) dk_(2 ) so first intregsl value=0 2nd intregal=(e^(−2) /2)×2∫_0 ^∞ e^(−k_2 ^2 ) dk_2 =e^(−2) ×(((√Π) )/2) formula∫_0 ^∞ e^(−x^2 ) dx=(((√Π) )/2) i have done a small error marking with red and later corrected...this red marked 2 should not be there..correct answer...(e^(−2) /2)×((√Π)/2)
$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \left(\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:}\right){e}^{−\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \left(\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){e}^{−\left\{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}\right\}} {dt}+ \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){e}^{−\left\{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}\right\}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \left(\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){e}^{−\left\{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} \right\}} ×{e}^{\mathrm{2}} {dt}+ \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){e}^{−\left\{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} \right\}} ×{e}^{−\mathrm{2}} {dt} \\ $$$$=\frac{{e}^{\mathrm{2}} }{\mathrm{2}}\int_{\infty} ^{\infty} {e}^{−{k}_{\mathrm{1}} ^{\mathrm{2}} } \:{dk}_{\mathrm{1}} \:\:+\frac{{e}^{−\mathrm{2}} }{\mathrm{2}}\int_{−\infty} ^{\infty} {e}^{−{k}_{\mathrm{2}} ^{\mathrm{2}} } {dk}_{\mathrm{2}\:} \\ $$$${so}\:{first}\:{intregsl}\:\:{value}=\mathrm{0} \\ $$$$\mathrm{2}{nd}\:{intregal}=\frac{{e}^{−\mathrm{2}} }{\mathrm{2}}×\mathrm{2}\int_{\mathrm{0}} ^{\infty} {e}^{−{k}_{\mathrm{2}} ^{\mathrm{2}} } {dk}_{\mathrm{2}} \\ $$$$={e}^{−\mathrm{2}} ×\frac{\sqrt{\Pi}\:}{\mathrm{2}}\:\:{formula}\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\Pi}\:}{\mathrm{2}} \\ $$$${i}\:{have}\:{done}\:{a}\:{small}\:{error}\:{marking}\:{with}\:{red} \\ $$$${and}\:{later}\:{corrected}…{this}\:{red}\:{marked}\:\mathrm{2}\:{should} \\ $$$${not}\:{be}\:{there}..{correct}\:{answer}…\frac{{e}^{−\mathrm{2}} }{\mathrm{2}}×\frac{\sqrt{\Pi}}{\mathrm{2}} \\ $$

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