[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 .

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

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

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

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

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

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

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