How do you test the improper integral #int absx(x^2+1)^-3 dx# from #(-oo, oo)# and evaluate if possible?

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Answer 1 · 2018-06-05T12:43:46

#=1/2#

#int_(-oo)^(oo) \ absx (x^2+1)^-3 \ dx#

The integrand is even:

... and for an even function:

#implies 2 int_(0)^(oo) \ absx (x^2+1)^-3 \ dx#

#= 2 int_(0)^(oo) \ color(red)(x) (x^2+1)^-3 \ dx#

# = 2 int_(0)^(oo) \ d(-1/4(x^2+1)^-2) #

# = -1/2 [ 1/(x^2+1)^2 ]_(0)^(x to oo) = 1/2#

Answer 2 · 2018-06-05T13:33:37

#int_(-oo)^oo absx(x^2+1)^-3"d"x=1/2#

We note that #|-x|((-x)^2+1)^-3=|x|(x^2+1)^-3# for all #x# so

#int_(-oo)^oo absx(x^2+1)^-3"d"x=int_0^oo2x(x^2+1)^-3"d"x#

Now let #u=x^2+1# and #"d"u=2"d"x#; #u(0)=1# and #u(oo)=oo#

Then

#int_0^oo2x(x^2+1)^-3"d"x=int_1^oou^-3"d"u=lim_(a->oo)[-1/(2u^2)]_1^a=lim_(a->oo)-1/(2a^2)+1/2=1/2#