# How To Evaluate This Integral?

• Feb 22nd 2010, 05:28 PM
AlphaRock
How To Evaluate This Integral?
Integral of (e^(1/x))/(x^2) dx from 0 to 1. (1 is above curly S, 0 is below curly S)

My work:
= [-e^(1/x)] from 0 to 1.
= [-e^(1/1)] - [-e^(1/0)] (This is where I get stuck, since 1/0 is DNE.
= -e - ???
• Feb 22nd 2010, 05:52 PM
Prove It
Quote:

Originally Posted by AlphaRock
Integral of (e^(1/x))/(x^2) dx from 0 to 1. (1 is above curly S, 0 is below curly S)

My work:
= [-e^(1/x)] from 0 to 1.
= [-e^(1/1)] - [-e^(1/0)] (This is where I get stuck, since 1/0 is DNE.
= -e - ???

This is solved using a $u$ substitution.

Note that

$\int{\frac{e^{\frac{1}{x}}}{x^2}\,dx} = \int{e^{x^{-1}}\,x^{-2}\,dx}$

$= -\int{e^{x^{-1}}(-x^{-2})\,dx}$.

Now let $u = x^{-1}$ so that $\frac{du}{dx} = -x^{-2}$.

So the integral becomes

$-\int{e^u\,\frac{du}{dx}\,dx}$

$= -\int{e^u\,du}$

$= -e^u + C$

$= -e^{x^{-1}} + C$

$= -e^{\frac{1}{x}} + C$.
• Feb 22nd 2010, 06:32 PM
AlphaRock
Quote:

Originally Posted by Prove It
This is solved using a $u$ substitution.

Note that

$\int{\frac{e^{\frac{1}{x}}}{x^2}\,dx} = \int{e^{x^{-1}}\,x^{-2}\,dx}$

$= -\int{e^{x^{-1}}(-x^{-2})\,dx}$.

Now let $u = x^{-1}$ so that $\frac{du}{dx} = -x^{-2}$.

So the integral becomes

$-\int{e^u\,\frac{du}{dx}\,dx}$

$= -\int{e^u\,du}$

$= -e^u + C$

$= -e^{x^{-1}} + C$

$= -e^{\frac{1}{x}} + C$.

Thanks. We got the same answer. However, how do we evaluate it from 0 to 1? Inputting it directly will give us [-e^(1/1)] - [-e^(1/0)]. The first part ([-e^(1/1)]) will give us e, however, the 2nd part [-e^(1/0)] is where I am stuck. How do we evaluate it from 0 to 1?
• Feb 22nd 2010, 06:39 PM
Prove It
Quote:

Originally Posted by AlphaRock
Thanks. We got the same answer. However, how do we evaluate it from 0 to 1? Inputting it directly will give us [-e^(1/1)] - [-e^(1/0)]. The first part ([-e^(1/1)]) will give us e, however, the 2nd part [-e^(1/0)] is where I am stuck. How do we evaluate it from 0 to 1?

This is known as an "improper integral", since the function is not defined at one of the end points.

So we need to see what happens CLOSE to that end point.

$\int_0^1{\frac{e^{\frac{1}{x}}}{x^2}\,dx} = \lim_{\epsilon \to 0}\left[-e^{\frac{1}{x}}\right]_\epsilon^1$

$= - e^{\frac{1}{1}} - \lim_{\epsilon \to 0}\left(-e^{\frac{1}{\epsilon}}\right)$.

If this limit converges, then the integral will converge and can be evaluated. If the limit does not exist or tends to $\pm \infty$, the integral will diverge.