# [SOLVED] using beta or gamma function

• Jan 5th 2010, 12:08 PM
akane
[SOLVED] using beta or gamma function
I have a problem solving this:

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

The result is $\displaystyle \pi$.
• Jan 5th 2010, 12:38 PM
Drexel28
Quote:

Originally Posted by akane
I have a problem solving this:

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

The result is $\displaystyle \pi$.

Let $\displaystyle I=\int_{-1}^{1}\sqrt{\frac{1+x}{1-x}}dx$. Furthermore, let $\displaystyle \sqrt{\frac{1+x}{1-x}}=y\implies x=\frac{y^2-1}{y^2+1}\implies dx=\frac{4y}{(y^2+1)^2}$. So that $\displaystyle I=\int_0^{\infty}\frac{4y^2}{(y^2+1)^2}\text{ }dy$. This integral is easy and does come out to $\displaystyle \pi$. No need for gamma function here. I guess, you could use the reflection formula for that last bit, but that just seems superfluous.
• Jan 5th 2010, 01:00 PM
Drexel28
Alternatively, if you are one of "those types" that insists on using the gamma and beta function in everything...

Define $\displaystyle I$ as before and make the substitution $\displaystyle 1+x=2z$. We see that: $\displaystyle x=2z-1,dx=2\text{ }dz,1-x=2\left(1-z\right)$ and our limits of integration are $\displaystyle x\to-1\implies z\to0,x\to1\implies z\to 1$. Therefore, we may finally conclude that $\displaystyle I=2\int_0^1 \sqrt{2z}\left(2\left(1-z\right)\right)^{\frac{-1}{2}}\text{ }dz=2\text{B}\left(\tfrac{3}{2},\tfrac{-1}{2}\right)=2\cdot\frac{\pi}{2}=\pi$...but that just seems stupid.
• Jan 5th 2010, 01:05 PM
Krizalid
$\displaystyle \int_{-1}^{1}{\sqrt{\frac{1+x}{1-x}}\,dx}=\int_{-1}^{1}{\frac{\left| 1+x \right|}{\sqrt{1-x^{2}}}\,dx}=\int_{-1}^{1}{\frac{dx}{\sqrt{1-x^{2}}}}+\int_{-1}^{1}{\frac{x}{\sqrt{1-x^{2}}}\,dx}.$

The integral equals $\displaystyle \pi,$ 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.
• Jan 5th 2010, 10:48 PM
akane
Quote:

Originally Posted by Drexel28
Alternatively, if you are one of "those types" that insists on using the gamma and beta function in everything...

Define $\displaystyle I$ as before and make the substitution $\displaystyle 1+x=2z$. We see that: $\displaystyle x=2z-1,dx=2\text{ }dz,1-x=2\left(1-z\right)$ and our limits of integration are $\displaystyle x\to-1\implies z\to0,x\to1\implies z\to 1$. Therefore, we may finally conclude that $\displaystyle I=2\int_0^1 \sqrt{2z}\left(2\left(1-z\right)\right)^{\frac{-1}{2}}\text{ }dz=2\text{B}\left(\tfrac{3}{2},\tfrac{-1}{2}\right)=2\cdot\frac{\pi}{2}=\pi$...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 (Wink)
• Jan 5th 2010, 11:02 PM
simplependulum
$\displaystyle \int_{-1}^1 \sqrt{ \frac{1 + x }{ 1- x } } ~dx$

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

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

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

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

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

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

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

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