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Question Number 37291 by math khazana by abdo last updated on 11/Jun/18
calculate g(θ) = ∫_(−∞) ^(+∞) e^(−x^2 ) sin(sinθ x^2 )dx .
$${calculate}\:{g}\left(\theta\right)\:=\:\int_{−\infty} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } \:{sin}\left({sin}\theta\:{x}^{\mathrm{2}} \right){dx}\:. \\ $$
Commented by math khazana by abdo last updated on 16/Jun/18
let λ =sinθ ⇒g(θ)=∫_(−∞) ^(+∞) e^(−x^2 ) sin(λx^2 )dx =Im { ∫_(−∞) ^(+∞) e^(−x^2 +iλx^2 ) dx} but ∫_(−∞) ^(+∞) e^(−x^2 +iλx^2 ) dx = ∫_(−∞) ^(+∞) e^(−(1−iλ)x^2 ) dx =_((√(1−iλ)) x =t) ∫_(−∞) ^(+∞) e^(−t^2 ) (dt/( (√(1−iλ)))) = (1/( (√(1−iλ)))) (√π) but 1−iλ =(√(1+λ^2 )) { (1/( (√(1+λ^2 )))) +i ((−λ)/( (√(1+λ^2 ))))}=r e^(iϕ) ⇒ r =(√(1+λ^2 )) and cosϕ= (1/( (√(1+λ^2 )))) and sinϕ=((−λ)/( (√(1+λ^2 )))) ⇒ tanϕ =−λ ⇒ ϕ =−arctan(λ) 1−iλ =(√(1+λ^2 )) e^(−iartan(λ)) ⇒ (√(1−iλ_ )) = (1+λ^2 )^(1/4) e^(−(i/2)arctan(λ)) ⇒ ∫_(−∞) ^(+∞) e^(−x^2 +iλx^2 ) dx = (√π)(1+λ^2 )^(−(1/4)) e^((i/2) arctan(λ)) =(√π)(1+λ^2 )^(−(1/4)) { cos(((arctan(λ))/2)) +i sin(((arctan(λ))/2)} g(θ) = (√π)(1+sin^2 θ)^(−(1/4)) sin( ((arctan(sinθ)/2)) .
$${let}\:\lambda\:={sin}\theta\:\Rightarrow{g}\left(\theta\right)=\int_{−\infty} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {sin}\left(\lambda{x}^{\mathrm{2}} \right){dx} \\ $$$$={Im}\:\left\{\:\int_{−\infty} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} \:+{i}\lambda{x}^{\mathrm{2}} } {dx}\right\}\:\:{but} \\ $$$$\int_{−\infty} ^{+\infty} \:\:{e}^{−{x}^{\mathrm{2}} \:+{i}\lambda{x}^{\mathrm{2}} } {dx}\:=\:\int_{−\infty} ^{+\infty} \:\:{e}^{−\left(\mathrm{1}−{i}\lambda\right){x}^{\mathrm{2}} } {dx} \\ $$$$=_{\sqrt{\mathrm{1}−{i}\lambda}\:{x}\:={t}} \:\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{t}^{\mathrm{2}} } \:\:\frac{{dt}}{\:\sqrt{\mathrm{1}−{i}\lambda}} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{i}\lambda}}\:\sqrt{\pi}\:\:\:\:{but} \\ $$$$\mathrm{1}−{i}\lambda\:=\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\left\{\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}\:+{i}\:\frac{−\lambda}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}\right\}={r}\:{e}^{{i}\varphi} \:\:\Rightarrow \\ $$$${r}\:=\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} \:\:}\:\:\:{and}\:\:{cos}\varphi=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }}\:\:{and}\:{sin}\varphi=\frac{−\lambda}{\:\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }} \\ $$$$\Rightarrow\:{tan}\varphi\:=−\lambda\:\Rightarrow\:\varphi\:=−{arctan}\left(\lambda\right) \\ $$$$\mathrm{1}−{i}\lambda\:=\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\:{e}^{−{iartan}\left(\lambda\right)} \:\Rightarrow \\ $$$$\sqrt{\mathrm{1}−{i}\lambda_{\:} }\:\:=\:\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\lambda\right)} \:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:{e}^{−{x}^{\mathrm{2}} \:+{i}\lambda{x}^{\mathrm{2}} } {dx}\:=\:\sqrt{\pi}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \:\:{e}^{\frac{{i}}{\mathrm{2}}\:{arctan}\left(\lambda\right)} \\ $$$$=\sqrt{\pi}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \left\{\:{cos}\left(\frac{{arctan}\left(\lambda\right)}{\mathrm{2}}\right)\:+{i}\:{sin}\left(\frac{{arctan}\left(\lambda\right)}{\mathrm{2}}\right\}\right. \\ $$$${g}\left(\theta\right)\:=\:\sqrt{\pi}\left(\mathrm{1}+{sin}^{\mathrm{2}} \theta\right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \:{sin}\left(\:\frac{{arctan}\left({sin}\theta\right.}{\mathrm{2}}\right)\:. \\ $$

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