Show-that-0-e-ix-2-dx-1-2-i-2-pi-2-

Question Number 6686 by FilupSmith last updated on 13/Jul/16

Show that :

\int_{0}^{\infty} e^{- {i x}^{2}} d x = (\frac{1}{2} - \frac{i}{2}) \sqrt{\frac{π}{2}}

Commented by FilupSmith last updated on 14/Jul/16

Is this correct ? ? ?

I = \int_{0}^{\infty} e^{- {i x}^{2}} d x

I^{2} = \int_{0}^{\infty} e^{- {i x}^{2}} d x \int_{0}^{\infty} e^{- {i y}^{2}} d y

I^{2} = \int_{0}^{\infty} \int_{0}^{\infty} e^{- i (x^{2} + y^{2})} d x d y

r^{2} = x^{2} + y^{2}, 0 ⩽ r ⩽ \infty

x = r \cos θ

y = r \sin θ

0 ⩽ θ ⩽ π / 2

Jacobian J (r, θ) = [\begin{matrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial θ} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial θ} \end{matrix}]

J (r, θ) = \frac{\partial x}{\partial r} \frac{\partial y}{\partial θ} - \frac{\partial x}{\partial θ} \frac{\partial y}{\partial r}

= \cos θ \times r \cos θ + r \sin θ \times \sin θ

= r (\cos^{2} θ + \sin^{2} θ)

= r

∴ I^{2} = \int_{0}^{\frac{π}{2}} \int_{0}^{\infty} {r e}^{- {i r}^{2}} d r d θ

Commented by prakash jain last updated on 14/Jul/16

Limit for θ should be from 0 to π / 2

Since limits for x and y cover complete

quadrant I .

Commented by FilupSmith last updated on 14/Jul/16

Ah, thank you .

Is the jacobian correct ?