calculate-0-1-e-zx-2-x-2-dx-with-z-from-C-and-Re-z-lt-0-

Question Number 90292 by mathmax by abdo last updated on 22/Apr/20

c a l c u l a t e \int_{0}^{\infty} \frac{1 - e^{{z x}^{2}}}{x^{2}} d x w i t h z f r o m C a n d R e (z) < 0

Commented by mathmax by abdo last updated on 23/Apr/20

l e t f (z) = \int_{0}^{\infty} \frac{1 - e^{{z x}^{2}}}{x^{2}} d x \Rightarrow f^{^{'}} (z) = - \int_{0}^{\infty} e^{{z x}^{2}} d x

= - \int_{0}^{\infty} e^{- {(\sqrt{- z} x)}^{2}} d x =_{\sqrt{- z} x = u} - \int_{0}^{\infty} e^{- u^{2}} \frac{d u}{\sqrt{- z}}

= - \frac{1}{\sqrt{- z}} \int_{0}^{\infty} e^{- u^{2}} d u = - \frac{π}{2 \sqrt{- z}} \Rightarrow f (z) = π \sqrt{- z} + c

f (0) = 0 = c \Rightarrow f (z) = π \sqrt{- z} i f z = - α + i β (α > 0)

- z = α - i β = \sqrt{α^{2} + β^{2}} e^{- i a r c t a n (\frac{β}{α})} \Rightarrow \sqrt{- z} = {(α^{2} + β^{2})}^{\frac{1}{4}} e^{- \frac{i}{2} a r c t a n (\frac{β}{α})}

\Rightarrow f (z) =^{4} \sqrt{α^{2} + β^{2}} e^{- \frac{i}{2} a r c t a n (\frac{β}{α})}