Thread: improper integrals!

the integral from 0 to 1 of lnx/(x^1/2)


thanks!

You got a typo in the upper integration limit, fix it and then we'll talk.

Originally Posted by luckeyduck

the integral from 0 to 1 of lnx/(x^1/2)


thanks!

try integration by parts...

$\displaystyle u=ln(x)$ and $\displaystyle dv=\frac{1}{\sqrt{x}}dx$

$\displaystyle du=\frac{1}{x}dx$ and $\displaystyle v=2\sqrt{x}$

so we get

$\displaystyle \int\frac{ln(x)}{\sqrt{x}}=2\sqrt{x}ln(x)-\int 2 \sqrt{x}\frac{1}{x}dx$

$\displaystyle \int\frac{ln(x)}{\sqrt{x}}=2\sqrt{x}ln(x)-2\int x^{-1/2}dx$


You should be able to get it from here.

I'm pretty sure that luckeyduck is not ready to digest the following solution, but I'd like to show this attempt, since double integration tricks make (sometimes) these things easier:

$\displaystyle \begin{aligned}
\int_0^1 {\frac{{\ln x}}
{{\sqrt x }}\,dx} &= - \int_0^1 {\int_x^1 {\frac{{dy\,dx}}
{{y\sqrt x }}} } \hfill \\
&= - \int_0^1 {\underbrace {\int_0^y {\frac{{dx}}
{{\sqrt x }}} }_{2\sqrt y }\,\frac{{dy}}
{y}} \hfill \\
&= - 2\int_0^1 {\frac{1}
{{\sqrt y }}\,dy} \hfill \\
&= - 4.
\end{aligned}$