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Let $\omega_d$ denote the surface area of the $d$ dimensional unit sphere in $\mathbf{R}^{d+1}$. I want to show that

$$ \frac{\omega_d}{2} = \int_0^{\pi/2} \omega_{d-1} \sin(t)^{d-1}\; dt. $$

My calculations seem to have an error, but I am not sure where. Using Fubini's theorem we can write

$$ \frac{\omega_d}{2} = \int_0^1 \omega_{d-1} (1 - r^2)^{(d-1)/2}\; dr. $$

Now we apply the substitution $r = \cos(t)$, with $dr = \sin(t)\; dt$ so we conclude

$$ \int_0^1 (1 - r^2)^{(d-1)/2}\; dr = \int_0^{\pi/2} (1 - \cos(t)^2)^{(d-1)/2} \sin(t)\; dt = \int_0^{\pi/2} \sin(t)^d\; dt$$

Why do I have a $d$ in the power of $\sin$ instead of $d-1$?

Jacob Denson
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1 Answers1

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Knowing that $$ \omega_{d-1} = \frac{2 \pi^{\frac{d}{2}}}{\Gamma\left(\frac{d}{2}\right)} $$ we can prove by strong induction that $$ \frac{\omega_{d-1}}{2} = \omega_{d-2}\int_{0}^{\frac{\pi}{2}}\sin^{d-2}(t) \\ \implies \frac{ \pi^{\frac{d}{2}}}{\Gamma\left(\frac{d}{2}\right)} = \frac{2 \pi^{\frac{d-1}{2}}}{\Gamma\left(\frac{d-1}{2}\right)}\int_{0}^{\frac{\pi}{2}}\sin^{d-2}(t) \ dt $$

The base cases can be easily verified knowing the formulas for the area of a sphere and circle (and also for a point and line). For the induction step ($d-1 \implies d$), we recall the reduction formula for $\sin^n(t)$, which tells is that $$ \int_0^{\frac{\pi}{2}} \sin^{d}(t) \ dt = \underbrace{\frac{-\cos (t) \sin^{d-1}(t)}{d}\Bigg\vert_{0}^{\frac{\pi}{2}}}_{\color{blue}{0}} + \frac{d-1}{d} \int_{0}^{\frac{\pi}{2}} \sin^{d-2} (t) \ dt \tag{1} $$ Which means \begin{align*} \omega_{d}\int_{0}^{\frac{\pi}{2}}\sin^{d}(t) & = \frac{2 \pi^{\frac{d+1}{2}}}{\Gamma\left(\frac{d+1}{2}\right)}\int_{0}^{\frac{\pi}{2}}\sin^{d}(t)\\ & \overset{\color{blue}{(1)}}{=}\frac{2 \pi^{\frac{d-1}{2}+1}}{\Gamma\left(\frac{d-1}{2} +1\right)}\frac{d-1}{d} \int_{0}^{\frac{\pi}{2}} \sin^{d-2} (t) \ dt\\ & = \frac{\pi (d-1)}{d \left(\frac{d-1}{2}\right)}\frac{2 \pi^{\frac{d-1}{2}}}{\Gamma\left(\frac{d-1}{2}\right)}\int_{0}^{\frac{\pi}{2}}\sin^{d-2}(t) \ dt\\ & \overset{\color{blue}{Induction}}{=} \frac{\pi}{ \left(\frac{d}{2}\right)}\frac{ \pi^{\frac{d}{2}}}{\Gamma\left(\frac{d}{2}\right)}\\ & = \frac{1}{2} \frac{2 \pi^{\frac{d+2}{2}}}{\Gamma\left(\frac{d+2}{2}\right)}\\ & = \frac{\omega_{d+1}}{2} \end{align*} Notice here we need to check $2$ adjacent base cases since we require to go back to $d-2$ in the inductive step.

I'm unfamiliar with the theorem you quote in your attempt, but it seems to be a similar approach used in this solution for calculating the volume of the $n$-sphere, so maybe that helps. Good day!

Robert Lee
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