How do you evaluate the integral #int e^(-absx)# from #-oo# to #oo#?

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Answer 1 · 2016-07-07T14:21:26

2

graph{e^(- | x|) [-10, 10, -5, 5]}

use the symmetry so that it becomes

#color{red}{2 times} int_0^oo dx qquad e^{-x}#

ie we are integrating in the region #x>=0# using the fact that #|x| = x#

#= 2 [- e^{-x}]_0^oo#

#= 2 [ e^{-x}]_oo^0#

2

to test for the symmetry use the even funcition test ie does #f(-x) = f(x)#

here

#f(-x) = e^{- abs (-x)} = e^{- abs x}= f(x)#