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Question Number 41678 by math khazana by abdo last updated on 11/Aug/18

p r o v e t h a t \int_{0}^{\infty} c o s (x^{2}) d x = \int_{0}^{\infty} s i n (x^{2}) d x b y u s i n g

o n l y s e r i e s .

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Aug/18

p + i q = \int_{0}^{\infty} e^{{i x}^{2}} d x

- t = {i x}^{2} i^{2} t = {i x}^{2}

i t = x^{2}

2 x d x = i d t

d x = \frac{i d t}{2 x} = \frac{i d t}{2 \sqrt{i t}} = \frac{\sqrt{i} t^{\frac{- 1}{2}}}{2} d t

\int_{0}^{\infty} e^{- t} . \frac{\sqrt{i} t^{\frac{- 1}{2}}}{2} d t

\frac{\sqrt{i}}{2} \int_{0}^{\infty} e^{- t} . t^{\frac{1}{2} - 1} d t

= \frac{\sqrt{i}}{2} ⌈ (\frac{1}{2}) = \frac{\sqrt{Π}}{2} \times \frac{\sqrt{i}}{}

n o w c a l c u l a t i n g t h e v a l u e o f \sqrt{i}

\sqrt{i}

= \frac{1}{\sqrt{2}} \times \sqrt{1 - 1 + 2 i}

= \frac{1}{\sqrt{2}} \times \sqrt{1^{2} + i^{2} + 2 \times 1 \times i}

= \frac{1}{\sqrt{2}} \times \sqrt{{(1 + i)}^{2}}

= \frac{1}{\sqrt{2}} \times (1 + i)

= \frac{1}{\sqrt{2}} + i \times \frac{1}{\sqrt{2}}

s o p + i q = \int_{0}^{\infty} e^{{i x}^{2}} d x = \frac{\sqrt{Π}}{2} \times \sqrt{i} = \frac{\sqrt{Π}}{2} (\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}})

p = \frac{\sqrt{Π}}{2 \sqrt{2}} q = \frac{\sqrt{Π}}{2 \sqrt{2}}