∫_0 ^( ∞) e^(−ix^2 ) dx=?? plz..


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Question Number 45841 by Rauny last updated on 17/Oct/18

$\int_{0}^{\infty} e^{- {i x}^{2}} d x = ? ?$ $plz . .$
Commented by MJS last updated on 17/Oct/18

$\frac{\sqrt{(2 π)}}{4} - \frac{\sqrt{(2 π)}}{4} i$
Commented by maxmathsup by imad last updated on 17/Oct/18

$\int_{0}^{\infty} e^{- {i x}^{2}} d x = \int_{0}^{\infty} e^{- {(x \sqrt{i})}^{2}} d x c h a n g e m e n t x \sqrt{i} = t g i v e$ $\int_{0}^{\infty} e^{- {i x}^{2}} d x = \frac{1}{\sqrt{i}} \int_{0}^{\infty} e^{- t^{2}} d t = \frac{1}{\sqrt{i}} \frac{\sqrt{π}}{2} = \frac{\sqrt{π}}{2} \frac{1}{e^{i \frac{π}{4}}} = \frac{\sqrt{π}}{2} e^{- \frac{i π}{4}}$ $= \frac{\sqrt{π}}{2} (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) = \frac{\sqrt{π}}{2 \sqrt{2}} - i \frac{\sqrt{π}}{2 \sqrt{2}} = \frac{\sqrt{2 π}}{4} - i \frac{\sqrt{2 π}}{4}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 17/Oct/18
	$t=ix^2 dt=2ixdx dx=(dt/(2ix))=(dt/(2i×(√t)))×(√i) ∫_0 ^∞ e^(−t) ×t^(−(1/2)) ×(1/(2(√i))) (1/(2(√i)))∫_0 ^∞ e^(−t) ×t^((1/2)−1) dt ←gamma function (1/(2(√i)))×⌈((1/2)) (1/(2(√i)))×(√π) {since ⌈((1/2))=(√π) } =(1/2)×(((√i) )/i)×(√π) now calculation of (√i) =(1/((√)2))×(√(1−1+2i)) =(1/(√2))×(√(1^2 +i^2 +2×i×1)) =(1/(√2))×(√((1+i)^2 )) =(1/(√2))×(1+i) so the ans is =(1/2)×(1/((√2) ))×((1+i)/i)×(√π) =(1/(2(√2) ))×(1+(1/i))×(√π) =(1/(2(√2)))×(1−i)×(√π) {∫_0 ^∞ e^(−y) ×y^(n−1) dy=⌈(n)}$
	$t = {i x}^{2} d t = 2 i x d x$ $d x = \frac{d t}{2 i x} = \frac{d t}{2 i \times \sqrt{t}} \times \sqrt{i}$ $\int_{0}^{\infty} e^{- t} \times t^{- \frac{1}{2}} \times \frac{1}{2 \sqrt{i}}$ $\frac{1}{2 \sqrt{i}} \int_{0}^{\infty} e^{- t} \times t^{\frac{1}{2} - 1} d t \leftarrow g a m m a f u n c t i o n$ $\frac{1}{2 \sqrt{i}} \times ⌈ (\frac{1}{2})$ $\frac{1}{2 \sqrt{i}} \times \sqrt{π} {s i n c e ⌈ (\frac{1}{2}) = \sqrt{π}}$ $= \frac{1}{2} \times \frac{\sqrt{i}}{i} \times \sqrt{π}$ $n o w c a l c u l a t i o n o f \sqrt{i}$ $= \frac{1}{\sqrt{} 2} \times \sqrt{1 - 1 + 2 i}$ $= \frac{1}{\sqrt{2}} \times \sqrt{1^{2} + i^{2} + 2 \times i \times 1}$ $= \frac{1}{\sqrt{2}} \times \sqrt{{(1 + i)}^{2}}$ $= \frac{1}{\sqrt{2}} \times (1 + i)$ $s o t h e a n s i s$ $= \frac{1}{2} \times \frac{1}{\sqrt{2}} \times \frac{1 + i}{i} \times \sqrt{π}$ $= \frac{1}{2 \sqrt{2}} \times (1 + \frac{1}{i}) \times \sqrt{π}$ $= \frac{1}{2 \sqrt{2}} \times (1 - i) \times \sqrt{π}$ ${\int_{0}^{\infty} e^{- y} \times y^{n - 1} d y = ⌈ (n)}$
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