I have a problem solving this:

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

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

Originally Posted by akane

I have a problem solving this:

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

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

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.

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.

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

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.

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

$\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 $