How do you evaluate the integral #int1/(1+x^2)^2 dx# from #-oo# to #oo#?

1 Answer
Oct 21, 2016

#int_(-oo)^oo1/(1+x^2)^2dx=pi/2#

Explanation:

First, for simplicity's sake, examining without the bounds:

#I=int1/(1+x^2)^2dx#

Apply the trig substitution #x=tantheta#. This implies that #dx=sec^2thetad theta#. Thus:

#I=intsec^2theta/(1+tan^2theta)^2d theta#

Since #1+tan^2theta=sec^2theta#:

#I=intsec^2theta/sec^4thetad theta=intcos^2thetad theta#

This can be solved through the cosine double angle formula: since #cos2theta=2cos^2theta-1#, we see that #cos^2theta=1/2(cos2theta+1)#.

#I=1/2int(cos2theta+1)d theta=1/2intcos2thetad theta+1/2intd theta#

The first integral can be solved either through another substitution or by inspection.

#I=1/4sin2theta+1/2theta+C#

Using the identity #sin2theta=2sinthetacostheta#:

#I=1/2sinthetacostheta+1/2theta+C#

Here, note that #sinthetacostheta=tantheta/sec^2theta=tantheta/(1+tan^2theta)#. Thus:

#I=1/(2+2tan^2theta)+1/2theta+C#

Using #x=tantheta# we see that:

#I=1/(2+2x^2)+1/2arctanx+C#

Reapplying the bounds:

#J=int_(-oo)^oo1/(1+x^2)^2dx=[1/(2+2x^2)+1/2arctanx]_(-oo)^oo#

Taking the limits at both infinities:

#J=(lim_(xrarroo)1/(2+2x^2)+1/2lim_(xrarroo)arctanx)-(lim_(xrarr-oo)1/(2+2x^2)+1/2lim_(xrarr-oo)arctanx)#

#J=(0+1/2(pi/2))-(0+1/2(-pi/2))#

#J=pi/4+pi/4=pi/2#