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

Question

Ezra Adams · Accepted Answer

#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#

Ezekiel Alexander · Answer

#=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#

Isaiah Allen · Answer

undefined To test the convergence of the improper integral $ \int_{-\infty}^{\infty} |x| (x^2+1)^{-3} dx $ and evaluate if possible, we can use the limit comparison test.

1. First, we choose a function to compare with. Since the integrand involves $ |x| $, we choose a function that behaves similarly near infinity. One such function is $ \frac{1}{x^2} $.

2. We calculate the limit of the ratio of the two functions as $ x $ approaches infinity:

$$
\lim_{{x \to \infty}} \frac{\frac{1}{x^2}}{|x|(x^2+1)^{-3}}
$$

3. Simplifying the expression, we get:

$$
\lim_{{x \to \infty}} \frac{1}{x^3(x^2+1)^{-3}}
$$

4. Using L'Hôpital's Rule if necessary, we evaluate the limit. After evaluating, if the limit is a positive finite number, then both integrals either converge or diverge.

5. If the limit is finite and positive, then the original integral converges. Otherwise, it diverges.

Note: If the limit is infinite or undefined, then we may need to choose a different function to compare with.

The evaluation of the integral itself involves integration techniques which can be carried out once convergence is established.