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An exercise in an analysis textbook says to show that $ \displaystyle \int_{-1}^{1} \frac{(1+x)^{2m-1}(1-x)^{2n-1}}{(1+x^{2})^{m+n}} \ dx = 2^{m+n-2} B(m,n) \ (m,n >0)$ and then deduce that for $\alpha$ not an integer multiple of $\pi$ that $\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \Big( \frac{\cos x + \sin x}{\cos x - \sin x} \Big)^{\cos \alpha} \ dx = \frac{\pi}{2 \sin \left( \pi \cos^{2} \frac{\alpha}{2}\right)}$.

I managed to figure out the second part but not the first part.

$ \displaystyle \int_{-1}^{1} \frac{(1+x)^{2m-1}(1-x)^{2n-1}}{(1+x^{2})^{m+n}} \ dx $

$ \displaystyle = \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{(1 + \tan u)^{2m-1}(1-\tan u)^{2n-1}}{\sec^{2(m+n)}u} \ \sec^{2} u \ du $

$ \displaystyle= \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} (\cos u + \sin u)^{2m-1}(\cos u - \sin u)^{2n-1} \ du $

If we let $\displaystyle m = \frac{1 + \cos a}{2}$ and $\displaystyle n= \frac{1- \cos a}{2}$, we get the above integral.

So $\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \Big( \frac{\cos x + \sin x}{\cos x - \sin x} \Big)^{\cos \alpha} \ dx = 2^{-1} \frac{\Gamma \left(\frac{1 + \cos a}{2} \right) \Gamma \left(\frac{1 - \cos a}{2} \right)}{\Gamma (1)}$

$ \displaystyle= 2^{-1} \Gamma \left(\frac{1 + \cos a}{2} \right) \Gamma \left( 1- \frac{ 1 + \cos a }{2} \right) $

$ \displaystyle= 2^{-1} \frac{\pi}{\sin \left( \pi \frac{1+\cos a}{2} \right)}$ (reflection formula)

$\displaystyle = \frac{\pi}{2 \sin \left( \pi \cos^{2} \frac{a}{2}\right)}$

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1 Answer 1

up vote 4 down vote accepted

Let's prove the first part, you need change the variables.

  • $y:=\frac{(1+x)^2}{2(1+x^2)}$, then $1-y=\frac{(1-x)^2}{2(1+x^2)}$, and $$\mathrm{d}y=\frac{1-x^2}{(1+x^2)^2}=2\cdot y^{\frac{1}{2}}\cdot (1-y)^{\frac{1}{2}}\cdot \frac{1}{1+x^2}\mathrm{d}x$$ Theorfore, \begin{eqnarray} &&\int_{-1}^{1}\frac{(1+x)^{2m-1}(1-x)^{2n-1}}{(1+x^2)^{m+n}}\mathrm{d}x \\ &=&\int_0^1 (2y)^{m-\frac{1}{2}}\cdot (2(1-y))^{n-\frac{1}{2}}\cdot\frac{1}{1+x^2} \big(2\cdot y^{\frac{1}{2}}\cdot (1-y)^{\frac{1}{2}}\cdot\frac{1}{1+x^2}\big)^{-1}\mathrm{d}y\\ &=&2^{m+n-2}\cdot \int_{0}^{1}y^{m-1}\cdot(1-y)^{n-1}\mathrm{d}y\\ &=&2^{m+n-2}\cdot B(m,n) \end{eqnarray}
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Doesn't $2 y^{\frac{1}{2}} (1-y)^{\frac{1}{2}} = \frac{1-x^{2}}{1+x^{2}}$? –  Random Variable Mar 2 '13 at 20:59
$dy = 2 \frac{y^{\frac{1}{2}}(1-y)^{\frac{1}{2}}}{1+x^{2}} \ dx $ and the rest follows –  Random Variable Mar 2 '13 at 21:07
@RandomVariable Yes, thanks. I've edited it just now. –  Coiacy Mar 2 '13 at 21:52

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