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Question Number 45705 by Sanjarbek last updated on 15/Oct/18

Commented by maxmathsup by imad last updated on 16/Oct/18
$∫ sin(x^2 )dx =(((√π)((√2)+i(√2))erf{ ((√2)+i(√2))(x/2)}+(√π)((√2)−i(√2))erf{((√(2−))i(√2))(x/2)})/8) this formulae is given by integral calculator so give me time to prof this...$
$\int s i n (x^{2}) d x = \frac{\sqrt{π} (\sqrt{2} + i \sqrt{2}) e r f {(\sqrt{2} + i \sqrt{2}) \frac{x}{2}} + \sqrt{π} (\sqrt{2} - i \sqrt{2}) e r f {(\sqrt{2 -} i \sqrt{2}) \frac{x}{2}}}{8}$ $t h i s f o r m u l a e i s g i v e n b y i n t e g r a l c a l c u l a t o r s o g i v e m e t i m e t o p r o f t h i s . . .$
Commented by maxmathsup by imad last updated on 16/Oct/18

$e r f (x) = \frac{2}{\sqrt{π}} \int_{0}^{x} e^{- t^{2}} d t a n d t h i s f u n c t i o n i s u s e d i n p r o b a l i t y a n d$ $s t a t i s t i c s . . . .$
Commented by maxmathsup by imad last updated on 16/Oct/18

$l e t f (x) = \int_{0}^{x} s i n (t^{2}) d t a n d g (x) = \int_{0}^{x} c o s (t^{2}) d t \Rightarrow$ $g (x) - i f (x) = \int_{0}^{x} e^{- {i t}^{2}} d t c h a n g e m e n t \sqrt{i} t = u g i v e$ $\int_{0}^{x} e^{- {i t}^{2}} d t = \int_{0}^{x \sqrt{i}} e^{- u^{2}} \frac{d u}{\sqrt{i}} = \frac{1}{e^{i \frac{π}{4}}} \int_{0}^{x \frac{1 + i}{\sqrt{2}}} e^{- u^{2}} d u$ $= e^{- \frac{i π}{4}} \frac{π}{2} e r f (\frac{x}{\sqrt{2}} (1 + i)) = \frac{π}{2} (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) e r f (\frac{x (1 + i)}{\sqrt{2}})$ $= \frac{π}{2 \sqrt{2}} (1 - i) e r f (\frac{x (1 + i)}{\sqrt{2}}) . . . . b e c o n t i n u e d . . . .$
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