Compute an integral

**Krizalid** · September 20th 2009, 12:43 PM

Find $\int_0^1\frac{\ln x}{\sqrt{x-x^2}}\,dx.$

**NonCommAlg** · September 20th 2009, 01:21 PM

Originally Posted by Krizalid

Find $\int_0^1\frac{\ln x}{\sqrt{x-x^2}}\,dx.$

substitute $x = \sin^2t$ to get $\int_0^1\frac{\ln x}{\sqrt{x-x^2}}\,dx=4 \int_0^{\frac{\pi}{2}} \ln(\sin t) \ dt = -2\pi \ln 2.$

**Krizalid** · September 20th 2009, 01:56 PM

yes, but can you find another way to solve it?

**halbard** · September 21st 2009, 12:14 AM

Another way? What does (s)he want? Hmm ...

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

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

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

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

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

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

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

Wait a moment, this is merely a disguised version of the $\int_0^{\pi/2}\ln(\sin x)\,\mathrm dx$ 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 ...

**Krizalid** · September 21st 2009, 06:00 AM

Originally Posted by halbard

Wait a moment, this is merely a disguised version of the $\int_0^{\pi/2}\ln(\sin x)\,\mathrm dx$ 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.

**simplependulum** · September 22nd 2009, 03:32 AM

What about this ?

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

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

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

From the Lengre's Identity

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

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

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

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

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

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