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

2 Answers
Mar 18, 2017

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#

Mar 18, 2017

#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#