find-0-cos-zx-2-dx-with-z-C-

Question Number 61528 by maxmathsup by imad last updated on 04/Jun/19

f i n d \int_{0}^{\infty} c o s ({z x}^{2}) d x w i t h z \in C .

Commented by maxmathsup by imad last updated on 04/Jun/19

l e t I = \int_{0}^{\infty} c o s ({z x}^{2}) d x \Rightarrow 2 I = \int_{- \infty}^{+ \infty} c o s ({z x}^{2}) l e t z = a + i b \Rightarrow

2 I = \int_{- \infty}^{+ \infty} c o s ((a + i b) x^{2}) b u t c o s (z) = c h (i z) \Rightarrow

2 I = \int_{- \infty}^{+ \infty} c h (i (a + i b) x^{2}) d x = \int_{- \infty}^{+ \infty} c h ({i a x}^{2} - {b x}^{2}) d x

= \int_{- \infty}^{+ \infty} \frac{e^{i ({i a x}^{2} - {b x}^{2})} + e^{- i ({i a x}^{2} - {b x}^{2})}}{2} d x

= \int_{- \infty}^{+ \infty} \frac{e^{- {a x}^{2} - {i b x}^{2}} + e^{{a x}^{2} + {i b x}^{2}}}{2} d x = \frac{1}{2} \int_{- \infty}^{+ \infty} e^{- (a + i b) x^{2}} d x + \frac{1}{2} \int_{- \infty}^{+ \infty} e^{- (- a - i b) x^{2}} d x

\int_{- \infty}^{+ \infty} e^{- (a + i b) x^{2}} d x =_{\sqrt{a + i b} x = t} \int_{- \infty}^{+ \infty} e^{- t^{2}} \frac{d t}{\sqrt{a + i b}} = \frac{π}{\sqrt{a + i b}} = \frac{π}{\sqrt{z}}

\int_{- \infty}^{+ \infty} e^{- (- a - i b) x^{2}} d x =_{\sqrt{- a - i b} x = t} \int_{- \infty}^{+ \infty} e^{- t^{2}} \frac{d t}{\sqrt{- a - i b}} = \frac{π}{\sqrt{- a - i b}}

= \frac{π}{\sqrt{- 1} \sqrt{a + i b}} = \frac{π}{i \sqrt{z}} \Rightarrow 2 I = \frac{π}{2 \sqrt{z}} + \frac{π}{2 i \sqrt{z}} = \frac{π}{2 \sqrt{z}} - \frac{i π}{2 \sqrt{z}} \Rightarrow I = \frac{π}{4} \frac{1 - i}{\sqrt{z}}

i f w e t a k e z = {r e}^{i θ} w e g e t I = \frac{π}{4} \frac{1 - i}{\sqrt{r} e^{i \frac{θ}{2}}} = \frac{π}{4 \sqrt{r}} \sqrt{2} e^{- \frac{i π}{4}} e^{- \frac{i θ}{2}} = \frac{π \sqrt{2}}{4 \sqrt{r}} e^{i (- \frac{π}{4} - \frac{θ}{2})}

I = \frac{π \sqrt{2}}{4 \sqrt{r}} e^{- i (\frac{π}{4} + \frac{θ}{2})} .

Commented by Smail last updated on 05/Jun/19

Y o u m e a n \int_{- \infty}^{\infty} e^{- t^{2}} d t = \sqrt{π} n o t π

Commented by maxmathsup by imad last updated on 08/Jun/19

y e s s i r \dots .

Commented by maxmathsup by imad last updated on 08/Jun/19

d u e t o \int_{- \infty}^{+ \infty} e^{- t^{2}} d t = \sqrt{π} t h e f i n a l a n s w e r i s I = \frac{\sqrt{2 π}}{4 \sqrt{r}} e^{- i (\frac{π}{4} + \frac{θ}{2})}