Find the integral of ln(x)/sqrt(x) as x is between 0 and 1

[suspenc3]

Hi I am trying to figure out the following..Ive tried substitutions but can't find any that work:

$$\int_0^1 \frac{lnx}{ \sqrt x}dx$$

HA @ 0 so
$$\int_0^1 \frac{lnx}{ \sqrt x} dx$$
=
$$\lim_{t\rightarrow 0^+} \int_t^1 \frac{lnx}{\sqrt x}dx$$

 

[StatusX]

Try the substitution u=sqrt(x).

 

[suspenc3]

so...2du=1/root(x)dx.. it becomes just ln(u^2)?

 

[suspenc3]

if this is right I end up with

$$\lim_{t\rightarrow 0^+} 2 - \frac{s\sqrt t}{t}$$

so would this be divergent?

 

[StatusX]

I don't know what s is there, but yes, this is divergent. Another way to find this is to note that this function is less than ln(x), whose integral diverges to negative infinity on this interval.

 

[suspenc3]

ok, the s should be a two...so basically if the integral ever has $$\frac{a}{0}$$...then it is called divergent?..or whenever an answer cannot be found..it is divergent?

 

[StatusX]

This integral is divergent because the limit you mention in your first post is infinity. "if the integral ever has a/0" isn't precise enough an expression for me to say anything meaningful about. In general, for improper integrals like this one, you have to take a limit as you approach vertical asymptopes or as [itex]x \rightarrow \pm \infty[/tex]. If the result after taking all these limits is infinity, the integral is said to diverge. Whether you can find the answer or not is your problem, not the integral's.[/itex]