How do you determine if the improper integral converges or diverges #int [e^(1/x)] / [x^3]# from 0 to 1?

Question

12213 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-12T19:02:50

The integral diverges.

First focusing on the integral without bounds:

#inte^(1/x)/x^3dx#

Let #t=1/x# this implies that #dt=-1/x^2dx#, so the integral can be rewritten as:

#=-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 #xrarr0#, #1/xrarroo# and #e^(1/x)rarroo#. Together, the limit will also be #oo#, so the integral diverges.