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It is easier to take the derivative, and consider the volume of the n-1-sphere (i.e., the "surface area" of the boundary of the ball).

Start with the integral $\int\_{\mathbb{R}^n} e^{-x\_1^2 - ... - x\_n^2} dx\_1 \dots dx\_n$. Fubini's theorem lets you decompose this into a product of 1-dimensional integrals, and you get $\pi^{n/2}$. Since the integrand is spherically symmetric, you can change to the integral $\int\_0^\infty vol(S^{n-1}(r)) e^{-r^2} dr$, where $S^{n-1}(r)$ is the unit n-1-sphere of radius r. The volume of this sphere is $r^{n-1}$ times the volume of the unit sphere, so solving for that, you get $\frac{\pi^{n/2}}{\int_0^\infty r^{n-1} e^{-r^2} dr}$. A change of coordinates (u = r²) in the denominator yields the integral defining $\Gamma(n/2)$.

It is easier to take the derivative, and consider the volume of the $(n-1)$-sphere (i.e., the "surface area" of the boundary of the ball).

Start with the integral $\int_{\mathbb{R}^n} e^{-x_1^2 - \cdots - x_n^2} dx_1 \cdots dx_n$. Fubini's theorem lets you decompose this into a product of $1$-dimensional integrals, and you get $\pi^{n/2}$. Since the integrand is spherically symmetric, you can change to the integral $\int_0^\infty \mathrm{vol}(S^{n-1}(r)) \, e^{-r^2} dr$, where $S^{n-1}(r)$ is the unit $(n-1)$-sphere of radius $r$. The volume of this sphere is $r^{n-1}$ times the volume of the unit sphere, so solving for that, you get $\frac{\pi^{n/2}}{\int_0^\infty r^{n-1} e^{-r^2} dr}$. A change of coordinates ($u = r^2$) in the denominator yields the integral defining $\Gamma(n/2)$.

It is easier to take the derivative, and consider the volume of the n-1-sphere.

Start with the integral of e^{-x_1^2 - ... - x_n^2} over Rⁿ. Fubini's theorem lets you decompose this into a product of 1-dimensional integrals, and you get pi^n/2. Since the integrand is spherically symmetric, you can change to an integral of the volume of the n-1-sphere of radius r times e^{-r^2} dr. Solving for the volume of the unit sphere, you get a power of pi divided by the integral of r^n-1 e^{-r^2} dr from 0 to infinity. A change of coordinates (u = r²) yields the integral defining Gamma(n/2) .

It is easier to take the derivative, and consider the volume of the n-1-sphere (i.e., the "surface area" of the boundary of the ball).

Start with the integral $\int\_{\mathbb{R}^n} e^{-x\_1^2 - ... - x\_n^2} dx\_1 \dots dx\_n$. Fubini's theorem lets you decompose this into a product of 1-dimensional integrals, and you get $\pi^{n/2}$. Since the integrand is spherically symmetric, you can change to the integral $\int\_0^\infty vol(S^{n-1}(r)) e^{-r^2} dr$, where $S^{n-1}(r)$ is the unit n-1-sphere of radius r. The volume of this sphere is $r^{n-1}$ times the volume of the unit sphere, so solving for that, you get $\frac{\pi^{n/2}}{\int_0^\infty r^{n-1} e^{-r^2} dr}$. A change of coordinates (u = r²) in the denominator yields the integral defining $\Gamma(n/2)$.

It is easier to take the derivative, and consider the volume of the n-1-sphere.

Start with the integral of e^{x_1^2 + ... _ x_n^2} over Rⁿ. Fubini yields an answer of pi^n/2. Since the integrand is spherically symmetric, you can change to an integral of the volume of the n-1-sphere of radius r times e^{-r^2} dr. Solving for the volume of the unit sphere, you get a power of pi divided by integral of r^n-1 e^{-r^2} dr from 0 to infinity. A change of coordinates (u = r²) yields the integral defining Gamma(n/2) .

It is easier to take the derivative, and consider the volume of the n-1-sphere.

Start with the integral of e^{-x_1^2 - ... - x_n^2} over Rⁿ. Fubini's theorem lets you decompose this into a product of 1-dimensional integrals, and you get pi^n/2. Since the integrand is spherically symmetric, you can change to an integral of the volume of the n-1-sphere of radius r times e^{-r^2} dr. Solving for the volume of the unit sphere, you get a power of pi divided by the integral of r^n-1 e^{-r^2} dr from 0 to infinity. A change of coordinates (u = r²) yields the integral defining Gamma(n/2) .