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Question Number 167647 by mathlove last updated on 22/Mar/22

	Answered by RoswelCod2003 last updated on 22/Mar/22

	$Using ‘ ‘ {Feynman}^{'} s Rule of {Integration}^{″}$ $I (k) = \int_{0}^{\infty} e^{- x^{2}} \cos (k x) d x ∴ where k = 2$ $\frac{d}{d k} I (k) = \frac{d}{d k} \int_{0}^{\infty} e^{- x^{2}} \cos (k x) d x$ $\frac{d I}{d k} = \int_{0}^{\infty} \frac{\partial}{\partial k} (e^{- x^{2}} \cos (k x)) d x$ $\frac{d I}{d k} = \int_{0}^{\infty} {x e}^{- x^{2}} \sin (k x) d x ∴ \| \begin{matrix} u = \sin (k x); d v = - {x e}^{- x^{2}} \\ d u = - k \cos (k x); v = \frac{1}{2} e^{- x^{2}} \end{matrix} \|$ $\lim_{x \to \infty} (\frac{1}{2} e^{- x^{2}} \sin (k x)) - \lim_{x \to 0} (\frac{1}{2} e^{- x^{2}} \sin (k x)) - \frac{k}{2} I (k)$ $- \frac{k}{2} I \to \frac{d I}{d k} = - \frac{k}{2} I$ $\frac{d I}{I} = - \frac{k}{2} d k \to \int \frac{d I}{I} = \int - \frac{k}{2} d k$ $\ln ∣ I ∣ = - \frac{k^{2}}{4} + C_{1} \to e^{\ln ∣ I ∣} = e^{- \frac{k^{2}}{4} + C_{1}}$ $I (k) = C_{2} e^{- \frac{k^{2}}{4}}$ $\int_{0}^{\infty} e^{- x^{2}} \cos (k x) d x = C_{2} e^{- \frac{k^{2}}{4}} ∴ l e t k = 0$ $\int_{0}^{\infty} e^{- x^{2}} d x \to \frac{\sqrt{π}}{2} = C_{2} \to \frac{\sqrt{π}}{2} e^{- \frac{k^{2}}{4}} ∴ k = 2$ $\int_{0}^{\infty} e^{- x^{2}} \cos (k x) d x = \frac{\sqrt{π}}{2} e^{- 1} o r \frac{\sqrt{π}}{2 e}$ $F i n a l A n s w e r :$ $\int_{0}^{\infty} e^{- x^{2}} \cos (2 x) d x = \frac{\sqrt{π}}{2 e}$ $R o s w e l B e r n a r d o :)$
	Commented by mkam last updated on 22/Mar/22

	$s i r c a n y o u g i v e n m e t h e F e y m a n s R u l e f o r m u l l a$
	Commented by mkam last updated on 22/Mar/22

	$s i r c a n y o u g i v e n m e t h e F e y m a n s R u l e f o r m u l l a$
	Commented by RoswelCod2003 last updated on 22/Mar/22

	$A h m m m . . . T h e {F e y n m a n}^{'} s R u l e o r {F e y n m a n}^{'} s T r i c k$ $i s p e r f o r m e d b y d i f f e r e n t i a t i n g u n d w r t h e i n t e g r a l s i g n .$ $E x a m p l e :$ $\int_{- 1}^{1} (2 k x + k^{2}) d x$ $s e t t h i s i n t e g r a l a s f u n c t i o n o f a n y v a r i a b l e, {l e t}^{'} s ‘ ‘ k^{″}$ $F (k) = \int_{- 1}^{1} (2 k x + k^{2}) d x$ $g e t t h e d e r i v a t i v e b o t h s i d e s w i t h r e s p e c t t o k$ $\frac{d}{d k} F (k) = \frac{d}{d k} \int_{- 1}^{1} (2 k x + k^{2}) d x ∴ t h e d e r i v a t i v e o f F (k) i s s i m p l y F^{'} (k) o r \frac{d F}{d k}$ $\frac{d F}{d k} = \frac{d}{d k} \int_{- 1}^{1} (2 k x + k^{2}) d x ∴ b e f o r e y o u g e t t h e d e r i v a t i v e o f (2 k x + k^{2})$ $p u t f i r s t t h e d e r i v a t i v e n o t i o n i n s i d e t h e i n t e g r a l s i g n b u t t r a n s f o r m i t$ $i n t o partial derivative notation :$ $\frac{d F}{d k} = \int_{- 1}^{1} \frac{\partial}{\partial k} (2 k x + k^{2}) d x ∴ t h e n t h e d e r i v a t i v e o f t h e f u n c t i o n u n d e r t h e$ $i n t e g r a l s i g n .$ $w e w i l l h a v e : \frac{d F}{d k} = \int_{- 1}^{1} (2 x + 2 k) d x ∴ n o w i n t e g r a t e w i t h$ $r e s p e c t t o X :$ $\frac{d F}{d k} = \frac{2 x^{1 + 1}}{1 + 1} \to \frac{2 x^{2}}{2} = x^{2} + 2 k$ $2 k w i l l r e m a i n t o 2 k b e c a u s e w e a r e i n t e g r a t i n g$ $w i t h r e s p e c t t o x n o t k$ $w e w i l l h a v e : {[x^{2} + 2 k]}_{- 1}^{1}$ $a p p l y n o w t h e f u n d a m e n t a l t h e o r e m o f c a l c u l u s$ $\int_{a}^{b} f (x) d x = f (b) - f (a)$ $[f (b) = (x^{2} + 2 k)] - [f (a) = (x^{2} + 2 k)]$ $[f (1) = (1 + 2 k)] - [f (- 1) = (1 + 2 k)]$ $1 + 2 k - 1 + 2 k = 4 k$ $Final Answer :$ $\int_{- 1}^{1} (2 k x + k^{2}) d x = 4 k$ $h o p e y o u l e a r n e d s o m e t h i n g, y o u c a n a c t u a l l y$ $l e a r n t h i s f r o m y o u t u b e :)$ $Roswel Bernardo at your service < 3$
	Answered by Mathspace last updated on 22/Mar/22

	$\int_{0}^{\infty} e^{- x^{2}} c o s (2 x) d x$ $= \frac{1}{2} \int_{- \infty}^{+ \infty} e^{- x^{2}} c o s (2 x) d x$ $= \frac{1}{2} R e (\int_{- \infty}^{+ \infty} e^{- x^{2} + 2 i x} d x) a n d$ $\int_{- \infty}^{+ \infty} e^{- x^{2} + 2 i x} d x = \int_{- \infty}^{+ \infty} e^{- (x^{2} - 2 i x + i^{2} - i^{2})} d x$ $= e^{- 1} \int_{- \infty}^{+ \infty} e^{- {(x - i)}^{2}} d x (x - i = t)$ $= e^{- 1} \int_{- \infty}^{+ \infty} e^{- t^{2}} d t$ $= e^{- 1} . \sqrt{π} \Rightarrow$ $\int_{0}^{\infty} e^{- x^{2}} c o s (2 x) d x = \frac{e^{- 1} \sqrt{π}}{2}$ $= \frac{\sqrt{π}}{2 e}$
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