1. ## Improper Integral

I need to integrate
$
\int_1^4 \frac{1}{x \sqrt[3]{lnx}}dx
$

The integral is impoper because x cannot be 1. Since 1 is at the left hand, I rewrite it to...

$
\lim t \rightarrow 1^+ \int_t^4 \frac{1}{x \sqrt[3]{lnx}}dx
$

Then I integrate and get...
Let u=lnx, du=1/x dx

$
\int_1^4 \frac{1}{x \sqrt[3]{lnx}}dx = \frac{3}{2} (lnx)^{2/3} +C
$

Then
$
\lim t \rightarrow 1^+ \int_t^4 \frac{1}{x \sqrt[3]{lnx}}dx
$

=
$
\lim t \rightarrow 1^+ [\frac{3}{2}*[(ln4)^{2/3} - (lnt)^{2/3}]]
$

Now I do not know how to find out what the answer is. Lnt would be Ln1+, so it gets closer and closer to ln1. Ln1 = 0. Does the term (lnt)^2/3 gets zero then?
And the limit goes to 3/2((ln4)^{2/3} - 0 ?

Thanks for any help.

2. Originally Posted by DBA
I need to integrate
$
\int_1^4 \frac{1}{x \sqrt[3]{lnx}}dx
$

The integral is impoper because x cannot be 1. Since 1 is at the left hand, I rewrite it to...

$
\lim t \rightarrow 1^+ \int_t^4 \frac{1}{x \sqrt[3]{lnx}}dx
$

Then I integrate and get...
Let u=lnx, du=1/x dx

$
\int_1^4 \frac{1}{x \sqrt[3]{lnx}}dx = \frac{3}{2} (lnx)^{2/3} +C
$

Then
$
\lim t \rightarrow 1^+ \int_t^4 \frac{1}{x \sqrt[3]{lnx}}dx
$

=
$
\lim t \rightarrow 1^+ [\frac{3}{2}*[(ln4)^{2/3} - (lnt)^{2/3}]]
$

Now I do not know how to find out what the answer is. Lnt would be Ln1+, so it gets closer and closer to ln1. Ln1 = 0. Does the term (lnt)^2/3 gets zero then?
And the limit goes to 3/2((ln4)^{2/3} - 0 ?

Thanks for any help.
Its pretty easy. $(lnt)^{\frac{2}{3}} \rightarrow 0$ as $t \rightarrow 1$.
$0^{\frac{2}{3}} = 0$.

Originally Posted by DBA
And the limit goes to 3/2((ln4)^{2/3} - 0 ?
Yes. So the improper integral converges to $\frac{3}{2} ln^{\frac{2}{3}}(4)$.

Originally Posted by DBA
$
\int_1^4 \frac{1}{x \sqrt[3]{lnx}}dx = \frac{3}{2} (lnx)^{2/3} +C
$
This is wrong!
1- as you said its improper integral. You can not evaluate it directly.
2- Its and definite integral. And you solve it as an indefinite integral!
You should write it as a indefinite integral. Without the upper/lower limits.

3. Originally Posted by DBA
I need to integrate
$
\int_1^4 \frac{1}{x \sqrt[3]{lnx}}dx
$

The integral is impoper because x cannot be 1. Since 1 is at the left hand, I rewrite it to...

$
\lim t \rightarrow 1^+ \int_t^4 \frac{1}{x \sqrt[3]{lnx}}dx
$

Then I integrate and get...
Let u=lnx, du=1/x dx

$
\int_1^4 \frac{1}{x \sqrt[3]{lnx}}dx = \frac{3}{2} (lnx)^{2/3} +C
$

Then
$
\lim t \rightarrow 1^+ \int_t^4 \frac{1}{x \sqrt[3]{lnx}}dx
$

=
$
\lim t \rightarrow 1^+ [\frac{3}{2}*[(ln4)^{2/3} - (lnt)^{2/3}]]
$

Now I do not know how to find out what the answer is. Lnt would be Ln1+, so it gets closer and closer to ln1. Ln1 = 0. Does the term (lnt)^2/3 gets zero then?
And the limit goes to 3/2((ln4)^{2/3} - 0 ?

Thanks for any help.

First, Note that the integral via substitution should look like this:

$\lim_{t \to 0^+} \int_t^{ln(4)} \frac{1}{\sqrt[3]{u}} ~du$

You must change the limits, so you get:

$u(1) = ln(1) = 0$
$u(4) = ln(4)$

So, your final solution looks like this:

$\lim_{t \to 0^+} \left[\frac{3u^{\frac{2}{3}}}{2}\right]^{\ln{(4)}}_t$

$= \frac{3}{2}\lim_{t \to 0^+} (ln(4)^{\frac{2}{3}} - t^{\frac{2}{3}})$