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There are many books with a proof of Legendre's duplication formula:

$$2^{1-2z}\sqrt{\pi}\Gamma(2z)=\Gamma(z)\Gamma\left(z+\frac{1}{2}\right).$$

which start with the Beta function in the form:

$$B(z_1,z_2)=\int_{0}^{1}u^{z_1-1}(1-u)^{z_2-1}\text{d}u$$

Is there, however, a way to prove it from the trigonometric definition of the Beta function:

$$B(z_1,z_2)=2\int_0^{\frac {\pi}{2}}\sin ^{2z_1-1}\theta\cos^{2z_2-1}\theta\text{d}\theta$$

without going back to the non-trig version of the function?

My attempt so far:

Let $z_1=z_2=z$

$$B(z,z)=2\int_0^{\frac {\pi}{2}}\sin ^{2z-1}\theta\cos^{2z-1}\theta\text{d}\theta$$

$$=2\int_0^{\frac {\pi}{2}}\sin ^{2z_1-1}\theta\cos^{2z_2-1}\theta\text{d}\theta$$

$$=2\int_0^{\frac {\pi}{2}}\left(\frac {\sin 2\theta}{2}\right)^{2z-1}\text{d}\theta$$

$$=2^{-2z}\int_0^{\frac {\pi}{2}}\left(\sin 2\theta\right)^{2z-1}\text{d}\theta$$

and I'm not sure where to go from here.

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  • $\begingroup$ U can show that ur last integral equals another beta function with different parameters. from there it is just massaging $\endgroup$ – tired Mar 9 '16 at 15:01
  • $\begingroup$ @tired which parameters work here? $\endgroup$ – John Mar 9 '16 at 15:23
  • $\begingroup$ $B(z,z)$ and $B(z,1/2)$ $\endgroup$ – tired Mar 9 '16 at 15:24
  • $\begingroup$ Does that not leave the problem of the $\sin 2\theta$ instead of $\sin \theta$? $\endgroup$ – John Mar 9 '16 at 15:26
  • $\begingroup$ sub $\theta \rightarrow \phi/2$ and use symmetrz $\endgroup$ – tired Mar 9 '16 at 15:27

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