Question

How do you evaluate the integral `int 1/(x^2+16)dx` from `-oo` to `oo`?

Answer 1

The answer is `=pi/4`

Explanation:

We need

`1+tan^2theta=sec^2theta`

`(tantheta)'=sec^2theta`

We apply the substitution

`x=4tantheta`

`dx=4sec^2theta d theta`

Therefore,

We compute the indefinite integral first

`intdx/(x^2+16)`

`=int(4sec^2theta d theta)/(16tan^2theta+16)`

`=int4/16d theta`

`=1/4theta`

`=1/4arctan(x/4) +C`

Now,

we compute the boundaries

`int_(-oo)^(+oo)dx/(x^2+16)=[1/4arctan (x/4)]_-oo^(+oo)`

`=1/4(lim_(x->+oo)arctan(x/4)-lim_(x->-oo)arctan(x/4))`

`lim_(x->+oo)(1/4arctan(x/4))=1/4lim_(x->+oo)(arctan(x/4))`

`=1/4*pi/2=pi/8`

`lim_(x->-oo)(1/4arctan(x/4))=1/4lim_(x->-oo)(arctan(x/4))`

`=-1/4*pi/2=-pi/8`

Therefore,

`int_(-oo)^(+oo)dx/(x^2+16)=pi/8-(-pi/8)=pi/4`

Answer 2

`pi/4`

Explanation:

`1/(x^2+16)=i/8(1/(x+i4)-1/(x-i4))` so

`int (dx)/(x^2+16) = i/8 int(1/(x+i4)-1/(x-i4))dx`

`=i/8(log(4i+x)-log(4i-x))=log((4i+x)/(4i-x))^(i/8)`

but `4ipm x = sqrt(16+x^2)e^(pm i phi)` where `phi= arctan(4/x)` so

`log((4i+x)/(4i-x))^(i/8)=log((e^(2iphi))^(i/8))=log(e^(-phi/4))=-phi/4`

now

`int_0^oo (dx)/(x^2+16)=lim_(x->oo)(-phi/4)-lim_(x->0)(-phi/4) = 0-1/4(-pi/2)=pi/8`

so

`int_oo^oo (dx)/(x^2+16)=pi/4`