How do you determine if the improper integral converges or diverges $int [e^(1/x)] / [x^3]$ from 0 to 1?

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Answer 1 · 2017-05-12T19:02:50

The integral diverges.

First focusing on the integral without bounds:

$inte^(1/x)/x^3dx$

Let $t=1/x$ this implies that $dt=-1/x^2dx$ , so the integral can be rewritten as:

$=-inte^(1/x)(1/x)(-1/x^2)dx=-intte^tdt$

Performing integration by parts, letting:

${(u=t,=>,du=dt),(dv=e^tdt,=>,v=e^t):}$

$=-(te^t-inte^tdt)=-e^t(t+1)=e^(1/x)(1-1/x)$

So:

$int_0^1e^(1/x)/x^3dx=e^(1/x)(1-1/x)| _ 0^1$

$=e(1-1) - lim _ (xrarr0^+)e^(1/x)(1-1/x)$

As $xrarr0$ , $1/xrarroo$ and $e^(1/x)rarroo$ . Together, the limit will also be $oo$ , so the integral diverges.

How do you determine if the improper integral converges or diverges int [e^(1/x)] / [x^3]int [e^(1/x)] / [x^3] from 0 to 1?