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

1 Answer
mason m
Jan 14, 2017

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.

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