Question Number 6686 by FilupSmith last updated on 13/Jul/16
$$\mathrm{Show}\:\mathrm{that}: \\ $$$$\int_{\mathrm{0}} ^{\:\infty} {e}^{−{ix}^{\mathrm{2}} } {dx}=\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{{i}}{\mathrm{2}}\right)\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$
Commented by FilupSmith last updated on 14/Jul/16
![Is this correct??? I=∫_0 ^( ∞) e^(−ix^2 ) dx I^2 =∫_0 ^( ∞) e^(−ix^2 ) dx∫_0 ^( ∞) e^(−iy^2 ) dy I^2 =∫_0 ^( ∞) ∫_0 ^( ∞) e^(−i(x^2 +y^2 )) dxdy r^2 =x^2 +y^2 , 0≤r≤∞ x=rcosθ y=rsinθ 0≤θ≤π/2 Jacobian J(r, θ)= [((∂x/∂r),(∂x/∂θ)),((∂y/∂r),(∂y/∂θ)) ] J(r, θ)=(∂x/∂r) (∂y/∂θ)−(∂x/∂θ) (∂y/∂r) =cosθ×rcosθ+rsinθ×sinθ =r(cos^2 θ+sin^2 θ) =r ∴I^2 =∫_0 ^( (π/2)) ∫_0 ^( ∞) re^(−ir^2 ) drdθ](https://www.tinkutara.com/question/Q6691.png)
$$\mathrm{Is}\:\mathrm{this}\:\mathrm{correct}??? \\ $$$${I}=\int_{\mathrm{0}} ^{\:\infty} {e}^{−{ix}^{\mathrm{2}} } {dx} \\ $$$${I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\:\infty} {e}^{−{ix}^{\mathrm{2}} } {dx}\int_{\mathrm{0}} ^{\:\infty} {e}^{−{iy}^{\mathrm{2}} } {dy} \\ $$$${I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} {e}^{−{i}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} {dxdy} \\ $$$${r}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} ,\:\:\mathrm{0}\leqslant{r}\leqslant\infty \\ $$$${x}={r}\mathrm{cos}\theta \\ $$$${y}={r}\mathrm{sin}\theta \\ $$$$\mathrm{0}\leqslant\theta\leqslant\pi/\mathrm{2} \\ $$$$\mathrm{Jacobian}\:{J}\left({r},\:\theta\right)=\begin{bmatrix}{\frac{\partial{x}}{\partial{r}}}&{\frac{\partial{x}}{\partial\theta}}\\{\frac{\partial{y}}{\partial{r}}}&{\frac{\partial{y}}{\partial\theta}}\end{bmatrix} \\ $$$${J}\left({r},\:\theta\right)=\frac{\partial{x}}{\partial{r}}\:\frac{\partial{y}}{\partial\theta}−\frac{\partial{x}}{\partial\theta}\:\frac{\partial{y}}{\partial{r}} \\ $$$$=\mathrm{cos}\theta×{r}\mathrm{cos}\theta+{r}\mathrm{sin}\theta×\mathrm{sin}\theta \\ $$$$={r}\left(\mathrm{cos}^{\mathrm{2}} \theta+\mathrm{sin}^{\mathrm{2}} \theta\right) \\ $$$$={r} \\ $$$$ \\ $$$$\therefore{I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\:\infty} {re}^{−{ir}^{\mathrm{2}} } {drd}\theta \\ $$
Commented by prakash jain last updated on 14/Jul/16
$$\mathrm{Limit}\:\mathrm{for}\:\theta\:\mathrm{should}\:\mathrm{be}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\pi/\mathrm{2} \\ $$$$\mathrm{Since}\:\mathrm{limits}\:\mathrm{for}\:{x}\:\mathrm{and}\:\:{y}\:\mathrm{cover}\:\mathrm{complete} \\ $$$$\mathrm{quadrant}\:\mathrm{I}. \\ $$
Commented by FilupSmith last updated on 14/Jul/16
$$\mathrm{Ah},\:\mathrm{thank}\:\mathrm{you}. \\ $$$$\mathrm{Is}\:\mathrm{the}\:\mathrm{jacobian}\:\mathrm{correct}? \\ $$