Question

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

Answer 1

`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`