does the rule of odd and even functions can be applied with improper integration? I=∫_(−∞) ^∞ xe^(−x^2 ) dx while f(x)= xe^(−x^2 ) is odd then I =0


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Question Number 208842 by NasaSara last updated on 24/Jun/24

$d o e s t h e r u l e o f o d d a n d e v e n f u n c t i o n s$ $c a n b e a p p l i e d w i t h i m p r o p e r i n t e g r a t i o n ?$ $I = \int_{- \infty}^{\infty} {x e}^{- x^{2}} d x$ $w h i l e f (x) = {x e}^{- x^{2}} i s o d d$ $t h e n I = 0$
Commented by Frix last updated on 24/Jun/24

$Yes .$
	Answered by mathzup last updated on 24/Jun/24

	$I = {l i m}_{a \to + \infty} \int_{- a}^{a} {x e}^{- x^{2}} d x = 0$ $p u i s q u e l a f o n c t i o n x \to {x e}^{- x^{2}} e s t i m p a i r e$
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