Question: Show that if $c_n = \int_{-1}^1 (1-t^2)^\frac{n-1}{2}dt$ then $c_n = \frac{n-1}{n}c_{n-2}$. I have tried two things both of which I didn't get super far with. First, I tried dividing n into two cases odd and even and applying the binomial theorem and second I tried integration by parts with $u = (1-t^2)^\frac{n-1}{2}$ and $dv = dt$. Maybe one of these approaches are right and I just screwed up somewhere but some help would be appreciated.

  • $\begingroup$ $1= t^2+(1-t^2)$ and integration by parts. $\endgroup$ – Jack D'Aurizio Mar 1 '17 at 1:26
  • $\begingroup$ You can rearrange this to show that $c_n +\frac{1}{n} c_{n-2} = 1$ is equivalent to your identity. It could potentially be easier to work with this. $\endgroup$ – Mark Mar 1 '17 at 1:29
  • 2
    $\begingroup$ This can be inspiring: math.stackexchange.com/questions/2162309/… $\endgroup$ – Olivier Oloa Mar 1 '17 at 1:31

Observe that $$ c_n = \int_{-1}^{0}(1-t^2)^{\frac{n-1}{2}}dt + \int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt = \\ =2\int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt, $$

where you make the substitution $t\mapsto -t$ in the first integral (or just observe that the function being integrated is even). Then, we do partial integration:

$$ \int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt = t(1-t^2)^{\frac{n-1}{2}}\big|_{0}^{1} + (n-1)\int_{0}^{1}t^2(1-t^2)^{\frac{n-3}{2}}dt = \\ =(n-1) \int_{0}^{1}t^2(1-t^2)^{\frac{n-3}{2}}dt. $$

Then, we conclude that $$ (n-1)\int_{0}^{1}(1-t^2)^{\frac{n-3}{2}}dt - \int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt = (n-1)\int_{0}^{1}(1-t^{2})^{\frac{n-1}{2}}dt, $$

which means $$ \int_{0}^{1}(1-t^2)^{\frac{n-1}{2}}dt = \frac{n-1}{n}\int_{0}^{1}(1-t^2)^{\frac{n-3}{2}}dt. $$

Then, it follows that $$ \frac{c_n}{2}=\frac{n-1}{n}\frac{c_{n-2}}{2} \implies c_n = \frac{n-1}{n}c_{n-2}. $$


Let $t=\cos x$ with $x\in [0,\pi].$

For $n\geq 1$ we have $$c_n=\int_{\pi}^0 (\sin^2 x)^{(n-1)/2}\;d\cos x=\int_{\pi}^0(\sin x)^{n-1}(-\sin x)\;dx=$$ $$=\int_0^{\pi}\sin^n x\;dx.$$ Now for $n\geq 2$ we have, integrating by parts, $$c_n=\int_0^{\pi}\sin^{n-1}x\;d (-\cos x)=$$ $$=(\sin^{n-1}x)(-\cos x)|_{x=0}^{x=\pi}\;-\int_0^{\pi}(-\cos x)(n-1)(\sin^{n-2}x) (\cos x)\;dx.$$ And since $0=(\sin^{n-1}0)(\cos 0)=(\sin^{n-1}\pi)(\cos \pi)$ when $n\geq 2$, we have $$c_n= \int_0^{\pi}(n-1)(\cos^2 x)(\sin^{n-2}x)\; dx=\int_0^{\pi}(n-1)(1-\sin^2x)(\sin^{n-2}x)\;dx=$$ $$=\int_0^{\pi} (n-1)(\sin^{n-2}x-\sin^nx)\;dx=$$ $$=(n-1)c_{n-2}-(n-1)c_n.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.