Question

Integral

Answer 1

of?

Answer 2

\[\int\limits_{0}^{1}\frac{ e ^{1/x} }{ x^3 }\]

Answer 3

Well first thing you want to do is observe the 1/x and 1/x^3 and see if it has a asymptotic nature at x = 0.

Answer 4

Can you show me steps?

Answer 5

\[u=1/x=x^-1\]
\[du=-x^-2dx=-dx/x^2\]

Answer 6

\[\int\limits1/xe ^{1/x}\frac{ dx }{ x^2 }\]

Answer 7

This integral involves a u-sub, parts, and maybe a l'hospital rule I believe.

Answer 8

\[-\int\limits ue^udu\]

Now do parts

Answer 9

is that
\[\int\limits \frac{ e^{1/2} dx }{ x * x^2}\]

Answer 10

oops 1/x laughing out loud

Answer 11

when you're doing u-sub, shouldn't the limit change as well?

Answer 12

You can change the limit but you don't have to. basically you can find the indefinite integral first with the u-sub and then sub back in for u and use the normal limits. or u can do a u-sub and change the limits if u wanna compute the definite integral with respect to u.

Answer 13

Like @iambatman said, at this point u should do parts. But there is one neat thing to note here, \(u\) can be differentiate to 0 and \(e^u\) can be integrated infinitely many times which means that one can use Tabular Integration (a shorthand method which is essentially integration by parts). This will allow you to find the indefinite integral faster.

Answer 14

So lets use s and t now for parts

\[s=u \]
\[ds=du\]
\[t=e^u\]
\[dt=e^udu\]

\[=-ue^u+\int\limits e^udu\]

\[=-ue^u+e^u+C\]

\[=-\frac{ 1 }{ x }e ^{1/x}+e ^{1/x}+C\]

\[\int\limits_{0}^{1}\frac{ e ^{1/x} }{ x^3 }dx = \lim_{t \rightarrow 0^{+}}\int\limits_{0}^{1} \frac{ e ^{1/x} }{ x^3 }dx\]

\[\lim_{t \rightarrow 0^{+}}(\frac{ -e ^{1/x} }{ x }+e ^{1/x}) _{t}^{1}\]

\[\lim_{t \rightarrow 0^{+}}\frac{ -e^1 }{ 1 }+e^1+\frac{ e ^{1/t} }{ t }-e ^{1/t}\]

\[\lim_{t \rightarrow 0^{+}}e ^{1/t}(\frac{ 1 }{ t }-1) \]

\[=\infty \]


Therefore it's divergent

Answer 15

So I guess you didn't need to do L'H :p

Answer 16

Yup that's what I got too. It diverges.

Answer 17

:)

Answer 18

I got it does not converge ;)