improper integrals!

**luckeyduck** · March 6th 2008, 05:21 PM

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

thanks!

**Krizalid** · March 6th 2008, 05:22 PM

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

**TheEmptySet** · March 6th 2008, 05:45 PM

Originally Posted by luckeyduck

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

thanks!

try integration by parts...

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

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

so we get

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

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

You should be able to get it from here.

**Krizalid** · March 6th 2008, 05:59 PM

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:

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

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