find-e-z-t-2-dt-with-z-r-e-i-C-

Question Number 34297 by math khazana by abdo last updated on 03/May/18

f i n d \int_{- \infty}^{+ \infty} e^{- z t^{2}} d t w i t h z = r e^{i θ} \in C .

Commented by abdo mathsup 649 cc last updated on 04/May/18

l e t p u t I = \int_{- \infty}^{+ \infty} e^{- {z t}^{2}} d t \Rightarrow I = \int_{- \infty}^{+ \infty} e^{- {r e}^{i θ} t^{2}} d t

I = \int_{- \infty}^{+ \infty} e^{- {(\sqrt{r} e^{i \frac{θ}{2}} t)}^{2}} d t c h a n g e m e n t \sqrt{r} e^{i \frac{θ}{2}} t = u

g i v e I = \frac{1}{\sqrt{r}} e^{- i \frac{θ}{2}} \int_{- \infty}^{+ + \infty} e^{- u^{2}} d u

I = \frac{\sqrt{π}}{\sqrt{r}} e^{- i \frac{θ}{2}} = \frac{\sqrt{π}}{\sqrt{r}} (c o s (\frac{θ}{2}) - i s i n (\frac{θ}{2})) (r > 0)

Answered by candre last updated on 03/May/18

I = \underset{- \infty}{\int^{+ \infty}} e^{- {z t}^{2}} d t

I^{2} = {(\underset{- \infty}{\int^{+ \infty}} e^{- {z t}^{2}} d t)}^{2} = \underset{- \infty}{\int^{+ \infty}} e^{- {z x}^{2}} d x \underset{- \infty}{\int^{+ \infty}} e^{- {z y}^{2}} d y

= \underset{- \infty}{\int^{+ \infty}} \underset{- \infty}{\int^{+ \infty}} e^{- z (x^{2} + y^{2})} d x d y

(x, y) = ρ (\cos φ, \sin φ)

\frac{\partial (x, y)}{\partial (ρ, φ)} d ρ d φ = | \begin{matrix} \frac{\partial x}{\partial ρ} & \frac{\partial y}{\partial ρ} \\ \frac{\partial x}{\partial φ} & \frac{\partial y}{\partial φ} \end{matrix} | d ρ d φ = | \begin{matrix} \cos φ & \sin φ \\ - ρ \sin φ & ρ \cos φ \end{matrix} | d ρ d φ

= ρ d ρ d φ

- \infty < x < + \infty, - \infty < y < + \infty \equiv 0 ⩽ ρ < \infty, 0 ⩽ φ < 2 π

I^{2} = \underset{0}{\int^{2 π}} \underset{0}{\int^{+ \infty}} e^{- z ρ^{2}} ρ d ρ d φ

= \underset{0}{\int^{2 π}} d φ \underset{0}{\int^{\infty}} e^{- z ρ^{2}} ρ d ρ

\underset{0}{\int^{2 π}} d φ = 2 π

\underset{0}{\int^{\infty}} e^{- z ρ^{2}} ρ d ρ = \frac{1}{2 z} \underset{- \infty}{\int^{0}} e^{u} d u = \frac{1}{2 z}

u = - z ρ^{2}

d u = - 2 z ρ d ρ

I^{2} = \frac{π}{z}

I = \sqrt{\frac{π}{z}} (ℜ (z) > 0)

Commented by math khazana by abdo last updated on 04/May/18

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