How do you integrate #int e^(3/x)/x^2dx# from #[0,1]#?

1 Answer
Jan 14, 2017

Answer:

The integral does not converge.

Explanation:

Notice that #e^(3/x)/x^2# is undefined at #x=0#, so we are working with an improper integral. Instead of having a bound of #0#, we will take the limit of the integral as some variable approaches #0#.

Before doing so, first find the general antiderivative of the function.

#inte^(3/x)/x^2dx#

Let #u=3/x# so #du=-3/x^2dx#.

#=-1/3inte^(3/x)(-3/x^2dx)=-1/3inte^udu=-1/3e^u=-1/3e^(3/x)#

So:

#int_0^1e^(3/x)/x^2dx=lim_(brarr0^+)int_b^1e^(3/x)/x^2dx=lim_(brarr0^+)[-1/3e^(3/x)]_b^1#

Evaluating:

#=-1/3e^3+lim_(brarr0^+)1/3e^(3/b)#

As #brarr0# from the right, we see that #3/brarroo#. Thus #lim_(brarr0^+)1/3e^(3/b)=oo# as well and the integral will not converge.