How do you determine if the improper integral converges or diverges #int [e^(1/x)] / [x^3]# from 0 to 1?
1 Answer
May 12, 2017
The integral diverges.
Explanation:
First focusing on the integral without bounds:
#inte^(1/x)/x^3dx#
Let
#=-inte^(1/x)(1/x)(-1/x^2)dx=-intte^tdt#
Performing integration by parts, letting:
#{(u=t,=>,du=dt),(dv=e^tdt,=>,v=e^t):}#
#=-(te^t-inte^tdt)=-e^t(t+1)=e^(1/x)(1-1/x)#
So:
#int_0^1e^(1/x)/x^3dx=e^(1/x)(1-1/x)| _ 0^1#
#=e(1-1) - lim _ (xrarr0^+)e^(1/x)(1-1/x)#
As