# Compute an integral

• Sep 20th 2009, 12:43 PM
Krizalid
Compute an integral
Find $\displaystyle \int_0^1\frac{\ln x}{\sqrt{x-x^2}}\,dx.$
• Sep 20th 2009, 01:21 PM
NonCommAlg
Quote:

Originally Posted by Krizalid
Find $\displaystyle \int_0^1\frac{\ln x}{\sqrt{x-x^2}}\,dx.$

substitute $\displaystyle x = \sin^2t$ to get $\displaystyle \int_0^1\frac{\ln x}{\sqrt{x-x^2}}\,dx=4 \int_0^{\frac{\pi}{2}} \ln(\sin t) \ dt = -2\pi \ln 2.$
• Sep 20th 2009, 01:56 PM
Krizalid
yes, but can you find another way to solve it?
• Sep 21st 2009, 12:14 AM
halbard
Another way? What does (s)he want? Hmm ...

Let $\displaystyle I=\int_0^1\frac{\ln x}{\sqrt{x-x^2}}\mathrm dx$ and let $\displaystyle J=\int_0^1\frac{\ln x}{\sqrt{1-x^2}}\mathrm dx$.

The substitution $\displaystyle x=u^2$ shows that $\displaystyle I=4J$.

The substituion $\displaystyle y=1-x$ shows that $\displaystyle I=\int_0^1\frac{\ln (1-x)}{\sqrt{x-x^2}}\mathrm dx$ and so $\displaystyle 2I=\int_0^1\frac{\ln x+\ln(1-x)}{\sqrt{x-x^2}}\mathrm dx=\int_0^1\frac{\ln(x-x^2)}{\sqrt{x-x^2}}\mathrm dx$.

Thus $\displaystyle I=\frac12\int_0^1\frac{\ln(x-x^2)}{\sqrt{x-x^2}}\mathrm dx =\int_0^{1/2}\frac{\ln(x-x^2)}{\sqrt{x-x^2}}\mathrm dx$ by symmetry.

Now put $\displaystyle u=2\sqrt{x-x^2}$ so that $\displaystyle 1-u^2=(1-2x)^2$, $\displaystyle 1-2x=\sqrt{1-u^2}$ for $\displaystyle 0\leq x\leq{\textstyle\frac12}$ and $\displaystyle 0\leq u\leq 1$, and $\displaystyle \mathrm dx=\frac{u\,\mathrm du}{2\sqrt{1-u^2}}$.

Thus $\displaystyle I=\int_0^1\frac{\ln (u^2/4)}{u/2}\frac{u\,\mathrm du}{2\sqrt{1-u^2}}=\int_0^1\frac{2\ln u-\ln 4}{\sqrt{1-u^2}}\mathrm du=2J-\ln 4\bigl[\arcsin u\bigr]_0^1={\textstyle\frac12}I-\pi\ln 2$.

Thus $\displaystyle I=-2\pi\ln 2$.

Wait a moment, this is merely a disguised version of the $\displaystyle \int_0^{\pi/2}\ln(\sin x)\,\mathrm dx$ method. It's no better than NonCommAlg's rejected solution.

(Thinks: Perhaps I should have used an infinite series, or some kind of limit, or introduced a parameter, or ... (Mind boggles at this point))

An answer that is not really an answer? This is trickier than it looks.

Hmm ...
• Sep 21st 2009, 06:00 AM
Krizalid
Quote:

Originally Posted by halbard

Wait a moment, this is merely a disguised version of the $\displaystyle \int_0^{\pi/2}\ln(\sin x)\,\mathrm dx$ method. It's no better than NonCommAlg's rejected solution.

I never rejected NCA's solution, it's just that the integral is quite trivial by using the fact you stated.

It's a well-known fact and easy to prove, so that's why he left details omitted.

I have another solution, I'll post it later, but I like yours.
• Sep 22nd 2009, 03:32 AM
simplependulum

$\displaystyle \int_0^1 \frac{\ln(x)}{\sqrt{x-x^2}}~dx$

$\displaystyle = \lim_{a\to0} \frac{1}{a} [ \int_0^1 \frac{x^a-1}{\sqrt{x-x^2}}~dx ]$

$\displaystyle = \lim_{a\to0} \frac{1}{a} \sqrt{\pi} [ \frac{ \Gamma(a + \frac{1}{2})}{\Gamma(a + 1) } - \sqrt{\pi} ]$

From the Lengre's Identity

$\displaystyle = \lim_{a\to0} \frac{1}{a} \sqrt{\pi} [ \frac{\sqrt{\pi} \Gamma(2a) 2^{-2a+1}}{ \Gamma(a) \Gamma(a + 1) } - \sqrt{\pi} ]$

$\displaystyle = \pi \lim_{a\to0} \frac{1}{a}[ 2^{-2a+1} / 2 - 1]$

$\displaystyle = \pi \lim_{a\to0} \frac{1}{a}[ 2^{-2a} - 1]$

$\displaystyle = \pi \ln(2) (-2) = - 2 \pi \ln(2)$

I think it is the best method by substituting $\displaystyle x = \sin^2{\theta}$ ..