evaluate the improper integral?

Question

the integral from o to infinity of [ e^ (-1/x) / (x^2) ] dx.

Answer 1

First find the integral and then evaluate the limits.

integral [ e^ (-1/x) / (x^2) ] dx
= integral [ e^ (-1/x) * dx/(x^2) ]

Let 1/x = t
Differentiate both the sides.
gives -1/x^2 * dx = dt
which is dx/(x^2) = -dt
Substituting the above values in the integral form.

Gives, integral [ e^(-t) * -dt ]
= - integral [ e^(-t) dt]

I think you know that, integral [e^(ax) dx ] = e^(ax)/a

This implies, integral [ e^ (-1/x) / (x^2) ] dx = - integral [ e^(-t) dt]
= - [ -e^(-t) } = e^-t = 1/e^t.

Now applying the limits.
The limits of x are from 0 to infinity.
The limits of t wii be from infinity to 0. According to the equation (1)

Therefore evaluating the limits,
{1/e^(infinity) } - {1/e^0}

We know that e^(infinity) = infinity and also 1/infinity = 0.

this gives, {0} - {1/1} = 0 - 1 = -1.

Therefore The integral from o to infinity of [ e^ (-1/x) / (x^2) ] dx is -1.

Hope you can understand this explaination.

Answer 2

let u=-1/x. Then du=(1/x^2)dx.

Thus ∫[e^(-1/x)/x^2]dx= ∫(e^u)du= e^u=e^(-1/x)

evaluating from 0 to ∞ is the same as evaluating lim (m->0+) {lim (n->∞) from m to n of e^(-1/x)}= lim (m->0+) {lim (n->∞) [e^(-1/n)-e^(-1/m)]} = lim (m->0+) {1-e^(-1/m)}= 1-0=1.