# Integrating a jacobian to find the volume.

I want to solve the following:

Prove that

$$\displaystyle \int_R \sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2} d\theta d\phi_1\cdots d\phi_{n-2} = \frac{2\pi^{n/2}}{\Gamma(n/2)}$$

where $R=[0,2\pi] \times [0,\pi]^{n-2}$. Hint: Compute $\int_{\mathbb R^n}e^{-|x|^2}dx$ in spherical coordinates.

So I am having problems to calculate the latter integral in spherical coordinates because I dont know how to integrate (in finite steps) $sin^{n}(x)$ and I dont know how this results to be a division of integrals.Can you help me to solve this please?, Thanks a lot in advance :)

You don't want to do the integral directly: hence the hint. One way to integrate is to split $$e^{-|x|^2} = e^{-(x_1^2+x_2^2+\dotsb+x_n^2)} = e^{-x_1^2} e^{-x_2^2} \dotsm e^{-x_n^2} ,$$ which is a product of $n$ one-dimensional integrals, all of which we know evaluate to $$\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}.$$ On the other hand, if you substitute in spherical coordinates, you have $e^{-|x|^2}=e^{-r^2}$, and the volume element is $$r^{n-1}\sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2} \, dr \, d\theta \, d\phi_1\cdots d\phi_{n-2},$$ which means that the integral splits this time as: $$\left( \int_0^{\infty} r^{n-1} e^{-r^2} \, dr \right) \left( \int_R \sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2} \, d\theta \, d\phi_1\cdots d\phi_{n-2} \right).$$ The second integral is the one you want, and you can do the first using the Gamma function, which will give you the answer you want when you divide: that is, you have two different evaluations of the same quantity, so they must be equal, $$\pi^{n/2} = \int_{\mathbb{R}^n} e^{-|x|^2} \, dx = \left( \int_0^{\infty} r^{n-1} e^{-r^2} \, dr \right) \left( \int_R \sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2} \, d\theta \, d\phi_1\cdots d\phi_{n-2} \right),$$ so $$\int_R \sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2} \, d\theta \, d\phi_1\cdots d\phi_{n-2} = \frac{\pi^{n/2}}{\int_0^{\infty} r^{n-1} e^{-r^2} \, dr}.$$

• Thanks a lot for your answer, but How can I divide? Commented Apr 16, 2015 at 14:26
• The integral's just a number, and not zero? Commented Apr 16, 2015 at 14:32
• this is $$\left( \int_0^{\infty} r^{n-1} e^{-r^2} \, dr \right) \left( \int_R \sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2} \, d\theta \, d\phi_1\cdots d\phi_{n-2} \right)=...$$to divide :) Commented Apr 16, 2015 at 14:33
• Well I dont understand your above comment Commented Apr 16, 2015 at 14:34
• Set $u=r^2$, so $\frac{du}{u} = 2\frac{dr}{r}$. Commented Apr 16, 2015 at 14:47

Hint all the variables $\phi_i$ can be separted in the integral, so you can transform the integral into: $$\displaystyle \int_R \sin^{n-2}\phi_1 \sin^{n-3}\phi_2\cdots\sin \phi_{n-2} d\theta d\phi_1\cdots d\phi_{n-2}=\int_{0}^{2\pi}d\theta \prod_{i=1}^{n-2}\int_{0}^{\pi}\sin^i\phi_i d\phi_i$$

• thats right I have get there but how do I can have: $\frac{integral}{integral}$ Commented Apr 16, 2015 at 14:17
• what do you mean by $\frac{\text{integral}}{\text{integral}}$ Commented Apr 16, 2015 at 14:23
• Yes because $\Gamma$ is an integral and the result $\pi^{n/2}$ is another one :) Commented Apr 16, 2015 at 14:24
• I'm sorry to say this but Chappers'answer is better, I tried to evaluate the integral without using the hint, if you want to use this method you can just compute the integral of powers of $\sin$ which is known in the literature, and the product in the final will give you the result Commented Apr 16, 2015 at 14:27
• How is that, can you post it please? Commented Apr 16, 2015 at 14:30