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 $\displaystyle u$ substitution.
Note that
$\displaystyle \int{\frac{e^{\frac{1}{x}}}{x^2}\,dx} = \int{e^{x^{-1}}\,x^{-2}\,dx}$
$\displaystyle = -\int{e^{x^{-1}}(-x^{-2})\,dx}$.
Now let $\displaystyle u = x^{-1}$ so that $\displaystyle \frac{du}{dx} = -x^{-2}$.
So the integral becomes
$\displaystyle -\int{e^u\,\frac{du}{dx}\,dx}$
$\displaystyle = -\int{e^u\,du}$
$\displaystyle = -e^u + C$
$\displaystyle = -e^{x^{-1}} + C$
$\displaystyle = -e^{\frac{1}{x}} + C$.
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.
$\displaystyle \int_0^1{\frac{e^{\frac{1}{x}}}{x^2}\,dx} = \lim_{\epsilon \to 0}\left[-e^{\frac{1}{x}}\right]_\epsilon^1$
$\displaystyle = - e^{\frac{1}{1}} - \lim_{\epsilon \to 0}\left(-e^{\frac{1}{\epsilon}}\right)$.
If this limit converges, then the integral will converge and can be evaluated. If the limit does not exist or tends to $\displaystyle \pm \infty$, the integral will diverge.