[SOLVED] using beta or gamma function

**akane** · Jan 5th 2010, 12:08 PM

I have a problem solving this:

$\int_{-1}^{1} \sqrt {\frac {1+x}{1-x}} dx$

The result is $\pi$ .
Could someone please help me

**Drexel28** · Jan 5th 2010, 12:38 PM

Originally Posted by akane

I have a problem solving this:

$\int_{-1}^{1} \sqrt {\frac {1+x}{1-x}} dx$

The result is $\pi$ .
Could someone please help me

Let $I=\int_{-1}^{1}\sqrt{\frac{1+x}{1-x}}dx$ . Furthermore, let $\sqrt{\frac{1+x}{1-x}}=y\implies x=\frac{y^2-1}{y^2+1}\implies dx=\frac{4y}{(y^2+1)^2}$ . So that $I=\int_0^{\infty}\frac{4y^2}{(y^2+1)^2}\text{ }dy$ . This integral is easy and does come out to $\pi$ . No need for gamma function here. I guess, you could use the reflection formula for that last bit, but that just seems superfluous.

**Drexel28** · Jan 5th 2010, 01:00 PM

Alternatively, if you are one of "those types" that insists on using the gamma and beta function in everything...

Define $I$ as before and make the substitution $1+x=2z$ . We see that: $x=2z-1,dx=2\text{ }dz,1-x=2\left(1-z\right)$ and our limits of integration are $x\to-1\implies z\to0,x\to1\implies z\to 1$ . Therefore, we may finally conclude that $I=2\int_0^1 \sqrt{2z}\left(2\left(1-z\right)\right)^{\frac{-1}{2}}\text{ }dz=2\text{B}\left(\tfrac{3}{2},\tfrac{-1}{2}\right)=2\cdot\frac{\pi}{2}=\pi$ ...but that just seems stupid.

**Krizalid** · Jan 5th 2010, 01:05 PM

$\int_{-1}^{1}{\sqrt{\frac{1+x}{1-x}}\,dx}=\int_{-1}^{1}{\frac{\left| 1+x \right|}{\sqrt{1-x^{2}}}\,dx}=\int_{-1}^{1}{\frac{dx}{\sqrt{1-x^{2}}}}+\int_{-1}^{1}{\frac{x}{\sqrt{1-x^{2}}}\,dx}.$

The integral equals $\pi,$ since the second piece is a bounded with two finite discontinuities points, hence integrable and since it's odd over that symmetric interval, that integral is zero.

**akane** · Jan 5th 2010, 10:48 PM

Originally Posted by Drexel28

Alternatively, if you are one of "those types" that insists on using the gamma and beta function in everything...

Define $I$ as before and make the substitution $1+x=2z$ . We see that: $x=2z-1,dx=2\text{ }dz,1-x=2\left(1-z\right)$ and our limits of integration are $x\to-1\implies z\to0,x\to1\implies z\to 1$ . Therefore, we may finally conclude that $I=2\int_0^1 \sqrt{2z}\left(2\left(1-z\right)\right)^{\frac{-1}{2}}\text{ }dz=2\text{B}\left(\tfrac{3}{2},\tfrac{-1}{2}\right)=2\cdot\frac{\pi}{2}=\pi$ ...but that just seems stupid.

We have just learnt beta and gamma function so that's the reason for using them, thank you very much

**simplependulum** · Jan 5th 2010, 11:02 PM

$\int_{-1}^1 \sqrt{ \frac{1 + x }{ 1- x } } ~dx$

$= \int_0^1 \sqrt{ \frac{1 + x }{ 1- x } } ~dx + \int_{-1}^0 \sqrt{ \frac{1 + x }{ 1- x } } ~dx$

Substitute $x = -t$ in the second integral .

$= \int_0^1 \sqrt{ \frac{1 + x }{ 1- x } } ~dx + \int_0^1 \sqrt{ \frac{1 - x }{ 1+ x } } ~dx$

$= \int_0^1 \left ( \sqrt{ \frac{1 + x }{ 1- x } } + \sqrt{ \frac{1 - x }{ 1 + x } } \right ) ~dx$

$= \int_0^1 \frac{ 1 + x + 1 - x }{\sqrt{1 - x^2 } }~dx$

$= \int_0^1 \frac{2}{\sqrt{ 1 - x^2}}~dx$

$= 2 \left [ \sin^{-1}(x) \right ]_0^1$

$= 2 \frac{\pi}{2} = \pi$

Thread: [SOLVED] using beta or gamma function

LinkBack

Thread Tools

Display

[SOLVED] using beta or gamma function

Similar Math Help Forum Discussions

Gamma Beta Function

[SOLVED] beta function

please help me the problem using beta and gamma

Another integral using the gamma and beta function

If b\cos{\beta} = c\cos{\gamma} prove that

Search tags for this page

solved examples on beta and gamma functions

beta gamma functions solved problems

beta and gamma integrals examples

beta gamma function examples

beta function problems with solutions

beta and gamma functions problems

gamma function problems

problems of gamma & beta function

problems of beta and gama function

solved questions on beta and gamma functions

solved problems on beta and gamma function

beta and gamma functions solved problems

problems on beta gamma functions

solved problems on beta functions

solved gamma function

Search Tags