How To Evaluate This Integral?

**AlphaRock** · February 22nd 2010, 05:28 PM

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 - ???

**Prove It** · February 22nd 2010, 05:52 PM

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

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

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

$= -e^u + C$

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

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

**AlphaRock** · February 22nd 2010, 06:32 PM

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?

**Prove It** · February 22nd 2010, 06:39 PM

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.

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