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Question Number 41513 by maxmathsup by imad last updated on 08/Aug/18

let A_n = ∫_0 ^∞ e^(−nx^2 ) cos(x^2 ) dx and B_n =∫_0 ^∞ e^(−nx^2 ) sin(x^2 )dx (n∈ N^★ ) 1) calculate A_n and B_n 2) find lim_(n→+∞) (A_n /B_n )

$${let}\:\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{nx}^{\mathrm{2}} } {cos}\left({x}^{\mathrm{2}} \right)\:{dx}\:\:{and}\:\:{B}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{nx}^{\mathrm{2}} } {sin}\left({x}^{\mathrm{2}} \right){dx}\:\:\:\:\left({n}\in\:{N}^{\bigstar} \right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\:{A}_{{n}} \:{and}\:\:{B}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:\:\frac{{A}_{{n}} }{{B}_{{n}} } \\ $$

Answered by alex041103 last updated on 09/Aug/18

Let A_n =2A_n =∫_(−∞) ^∞ e^(−nx^2 ) cos(x^2 ) dx and B_n =2B_n =∫_(−∞) ^∞ e^(−nx^2 ) sin(x^2 ) dx Let Z_n =A_n +iB_n =∫_(−∞) ^∞ e^(−(n−i)x^2 ) dx Then Z_n =(√(∫_(−∞) ^∞ ∫_(−∞) ^∞ e^(−(n−i)(x^2 +y^2 )) dxdy)) We change to polar coordinates Z_n =(√(∫_0 ^(2π) ∫_0 ^∞ re^(−(n−i)r^2 ) drdθ)) Z_n ^2 =−((e^(−(n−i)r^2 ) ∣_(r=0) ^(r=∞) )/(2n−2i))∫_0 ^(2π) dθ= =(π/(n−i))=Z_n ^2 =((π(n+i))/(n^2 +1)) ⇒Z_n =(√(π/(n^2 +1)))(√(n+i)) ⇒A_n =(1/2)(√(π/(n^2 +1)))Re((√(n+i))) B_n =(1/2)(√(π/(n^2 +1)))Im((√(n+i)))

$${Let}\:\mathcal{A}_{{n}} =\mathrm{2}{A}_{{n}} =\int_{−\infty} ^{\infty} {e}^{−{nx}^{\mathrm{2}} } {cos}\left({x}^{\mathrm{2}} \right)\:{dx}\:{and} \\ $$$$\mathcal{B}_{{n}} =\mathrm{2}{B}_{{n}} =\int_{−\infty} ^{\infty} {e}^{−{nx}^{\mathrm{2}} } {sin}\left({x}^{\mathrm{2}} \right)\:{dx} \\ $$$${Let}\:{Z}_{{n}} =\mathcal{A}_{{n}} +{i}\mathcal{B}_{{n}} =\int_{−\infty} ^{\infty} {e}^{−\left({n}−{i}\right){x}^{\mathrm{2}} } {dx} \\ $$$${Then}\:{Z}_{{n}} =\sqrt{\int_{−\infty} ^{\infty} \int_{−\infty} ^{\infty} {e}^{−\left({n}−{i}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} {dxdy}} \\ $$$${We}\:{change}\:{to}\:{polar}\:{coordinates} \\ $$$${Z}_{{n}} =\sqrt{\int_{\mathrm{0}} ^{\mathrm{2}\pi} \int_{\mathrm{0}} ^{\infty} {re}^{−\left({n}−{i}\right){r}^{\mathrm{2}} } {drd}\theta} \\ $$$${Z}_{{n}} ^{\mathrm{2}} =−\frac{{e}^{−\left({n}−{i}\right){r}^{\mathrm{2}} } \mid_{{r}=\mathrm{0}} ^{{r}=\infty} }{\mathrm{2}{n}−\mathrm{2}{i}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} {d}\theta= \\ $$$$=\frac{\pi}{{n}−{i}}={Z}_{{n}} ^{\mathrm{2}} =\frac{\pi\left({n}+{i}\right)}{{n}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow{Z}_{{n}} =\sqrt{\frac{\pi}{{n}^{\mathrm{2}} +\mathrm{1}}}\sqrt{{n}+{i}} \\ $$$$\Rightarrow{A}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{{n}^{\mathrm{2}} +\mathrm{1}}}{Re}\left(\sqrt{{n}+{i}}\right) \\ $$$$\:\:\:\:\:{B}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{{n}^{\mathrm{2}} +\mathrm{1}}}{Im}\left(\sqrt{{n}+{i}}\right) \\ $$

Commented by alex041103 last updated on 09/Aug/18

For 2) (A_n /B_n )=((Re((√(n+i))))/(Im((√(n+i)))))=cot(arg((√(n+i))))= =cot((1/2)arccot(n)) ⇒lim_(n→∞) cot((1/2)arccot(n))= =cot((1/2)arccot(+∞))= =cot((1/2)×0^+ )=cot(0^+ )=+∞ ⇒lim_(n→∞) (A_n /B_n )=+∞

$$\left.{For}\:\mathrm{2}\right) \\ $$$$\frac{{A}_{{n}} }{{B}_{{n}} }=\frac{{Re}\left(\sqrt{{n}+{i}}\right)}{{Im}\left(\sqrt{{n}+{i}}\right)}={cot}\left({arg}\left(\sqrt{{n}+{i}}\right)\right)= \\ $$$$={cot}\left(\frac{\mathrm{1}}{\mathrm{2}}{arccot}\left({n}\right)\right) \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {{lim}cot}\left(\frac{\mathrm{1}}{\mathrm{2}}{arccot}\left({n}\right)\right)= \\ $$$$={cot}\left(\frac{\mathrm{1}}{\mathrm{2}}{arccot}\left(+\infty\right)\right)= \\ $$$$={cot}\left(\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{0}^{+} \right)={cot}\left(\mathrm{0}^{+} \right)=+\infty \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {{lim}}\frac{{A}_{{n}} }{{B}_{{n}} }=+\infty \\ $$

Answered by math khazana by abdo last updated on 09/Aug/18

we have A_n +iB_n = ∫_0 ^∞ e^(−nx^2 +ix^2 ) dx = ∫_0 ^∞ e^(−(n−i)x^2 ) dx ⇒2A_n +2i B_n =∫_(−∞) ^(+∞) e^(−(n−i)x^2 ) dx =_((√(n−i))x=t) ∫_(−∞) ^(+∞) e^(−t^2 ) (dt/(√(n−i))) =((√π)/(√(n−i))) but n−i =(√(n^2 +1)){ (n/(√(n^2 +1))) −(i/(√(n^2 +1)))}=r e^(iθ) ⇒ r =(√(n^2 +1)) and tanθ =−(1/n) ⇒θ=−arctan((1/n))⇒ (√(n−i))=(n^2 +1)^(1/4) e^((iθ)/2) =(n^2 +1)^(1/4) e^(−i((arctan((1/n)))/2)) =^4 (√(n^2 +1)){ cos((1/2) arctan((1/n)) −i sin((1/2)arctan((1/n))) =^4 (√(n^2 +1)){ cos((π/4) −((arctan(n))/2))−isin((π/4)−((arctan(n))/2))⇒ A_n =^4 (√(n^2 +1))cos(−(π/4) +((arctan(n))/2)) and B_n =^4 (√(n^2 +1))sin(−(π/4) +((arctan(n))/2)) .

$${we}\:{have}\:{A}_{{n}} \:+{iB}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{nx}^{\mathrm{2}} \:+{ix}^{\mathrm{2}} } {dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−\left({n}−{i}\right){x}^{\mathrm{2}} } {dx}\:\Rightarrow\mathrm{2}{A}_{{n}} \:+\mathrm{2}{i}\:{B}_{{n}} =\int_{−\infty} ^{+\infty} \:{e}^{−\left({n}−{i}\right){x}^{\mathrm{2}} } {dx} \\ $$$$=_{\sqrt{{n}−{i}}{x}={t}} \:\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{t}^{\mathrm{2}} } \:\frac{{dt}}{\sqrt{{n}−{i}}}\:=\frac{\sqrt{\pi}}{\sqrt{{n}−{i}}} \\ $$$${but}\:{n}−{i}\:=\sqrt{{n}^{\mathrm{2}} +\mathrm{1}}\left\{\:\frac{{n}}{\sqrt{{n}^{\mathrm{2}} \:+\mathrm{1}}}\:−\frac{{i}}{\sqrt{{n}^{\mathrm{2}} \:+\mathrm{1}}}\right\}={r}\:{e}^{{i}\theta} \:\Rightarrow \\ $$$${r}\:=\sqrt{{n}^{\mathrm{2}} \:+\mathrm{1}}\:\:\:\:{and}\:\:{tan}\theta\:=−\frac{\mathrm{1}}{{n}}\:\Rightarrow\theta=−{arctan}\left(\frac{\mathrm{1}}{{n}}\right)\Rightarrow \\ $$$$\sqrt{{n}−{i}}=\left({n}^{\mathrm{2}} \:+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}\theta}{\mathrm{2}}} \:\:\:=\left({n}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−{i}\frac{{arctan}\left(\frac{\mathrm{1}}{{n}}\right)}{\mathrm{2}}} \\ $$$$=^{\mathrm{4}} \sqrt{{n}^{\mathrm{2}} \:+\mathrm{1}}\left\{\:{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left(\frac{\mathrm{1}}{{n}}\right)\:−{i}\:{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{n}}\right)\right)\right.\right. \\ $$$$=^{\mathrm{4}} \sqrt{{n}^{\mathrm{2}} \:+\mathrm{1}}\left\{\:{cos}\left(\frac{\pi}{\mathrm{4}}\:−\frac{{arctan}\left({n}\right)}{\mathrm{2}}\right)−{isin}\left(\frac{\pi}{\mathrm{4}}−\frac{{arctan}\left({n}\right)}{\mathrm{2}}\right)\Rightarrow\right. \\ $$$${A}_{{n}} =^{\mathrm{4}} \sqrt{{n}^{\mathrm{2}} \:+\mathrm{1}}{cos}\left(−\frac{\pi}{\mathrm{4}}\:+\frac{{arctan}\left({n}\right)}{\mathrm{2}}\right)\:{and} \\ $$$${B}_{{n}} =^{\mathrm{4}} \sqrt{{n}^{\mathrm{2}} \:+\mathrm{1}}{sin}\left(−\frac{\pi}{\mathrm{4}}\:+\frac{{arctan}\left({n}\right)}{\mathrm{2}}\right)\:. \\ $$

Commented by math khazana by abdo last updated on 10/Aug/18

error at the final lines A_n = ((√π)/2) ^4 (√(n^2 +1))cos(((arctan(n))/2)−(π/4)) and B_n =((√π)/2)^4 (√(n^2 +1))sin(((arctan(n))/2) −(π/4)) .

$${error}\:{at}\:{the}\:{final}\:{lines} \\ $$$${A}_{{n}} =\:\frac{\sqrt{\pi}}{\mathrm{2}}\:\:^{\mathrm{4}} \sqrt{{n}^{\mathrm{2}} \:+\mathrm{1}}{cos}\left(\frac{{arctan}\left({n}\right)}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)\:{and} \\ $$$${B}_{{n}} =\frac{\sqrt{\pi}}{\mathrm{2}}\:^{\mathrm{4}} \sqrt{{n}^{\mathrm{2}} \:+\mathrm{1}}{sin}\left(\frac{{arctan}\left({n}\right)}{\mathrm{2}}\:−\frac{\pi}{\mathrm{4}}\right)\:. \\ $$

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