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Aymen Hafeez

The factorial operator can be defined as:

n!=n⋅(n−1)⋅(n−2)⋅(n−3)⋯3⋅2⋅1n!=n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdots3\cdot2\cdot1

Or in a more compact form:

n!=∏k=1nkn!=\prod_{k=1}^nk

However, this is only true for integer values of nn. The factorial for non-integer values can be defined using the gamma function. The integral form of the gamma function is

Γ(z)=∫0∞xz−1e−xdx\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}dx

From here we’ll inductively derive an expression for finding the factorial of any real or complex zz, with ℜ(z)>0\Re(z)>0. We do this by first finding Γ(1)\Gamma(1):

Γ(1)=∫0∞e−xdx=[−e−x]0∞=(lim⁡x→∞−e−x)+e−0=1 \begin{aligned} \Gamma(1)&=\int_0^{\infty}e^{-x}dx \\ &=\left[-e^{-x}\right]_0^{\infty} \\ &=\left(\lim_{x\to\infty}-e^{-x}\right)+e^{-0} \\ &=1 \end{aligned}

Next, we find Γ(z+1)\Gamma(z+1):

Γ(z+1)=[−xze−x]0∞+z∫0∞xz−1e−xdx=[−xze−x]0∞+zΓ(z)=(lim⁡x→ ∞−xze−x)+zΓ(z)=zΓ(z) \begin{aligned} \Gamma(z+1)&=\left[-x^ze^{-x}\right]_0^\infty+z\int_0^{\infty}x^{z-1}e^{-x}dx \\ &=\left[-x^ze^{-x}\right]_0^\infty+z\Gamma(z) \\ &=\left(\lim_{x\to\ \infty}-x^ze^{-x}\right)+z\Gamma(z) \\ &=z\Gamma(z) \end{aligned}

Combining this with Γ(1)=1\Gamma(1)=1 we can see that

Γ(z)=(z−1)Γ(z−1)=(z−1)(z−2)Γ(z−2)=(z−1)(z−2)⋯(z−(z−2))(z−(z−1))Γ(z−(z−1))=(z−1)(z−2)⋯2⋅1 \begin{aligned} \Gamma(z)&=(z-1)\Gamma(z-1) \\ &=(z-1)(z-2)\Gamma(z-2) \\ &=(z-1)(z-2)\cdots(z-(z-2))(z-(z-1))\Gamma(z-(z-1)) \\ &=(z-1)(z-2)\cdots2\cdot1 \end{aligned}

And so, we get that

Γ(z)=(z−1)!\Gamma(z)=(z-1)!

For (12)!(\frac{1}{2})! let z=32z=\frac{3}{2}:

Γ(32)=(12)!\Gamma\left(\frac{3}{2}\right)=\left(\frac{1}{2}\right)!

Γ(32)\Gamma(\frac{3}{2}) can be given in terms of Γ(12)\Gamma(\frac{1}{2}) by the property Γ(z+1)=zΓ(z)\Gamma(z+1)=z\Gamma(z):

Γ(32)=(12)Γ(12)=12∫0∞x−12e−xdx \begin{aligned} \Gamma\left(\frac{3}{2}\right)&=\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right) \\ &=\frac{1}{2}\int_0^{\infty}x^{-\frac{1}{2}}e^{-x}dx \end{aligned}

If we substitute x=u2x=u^2, with dx=2ududx=2udu, we get

Γ(32)=∫0∞e−u2du\Gamma\left(\frac{3}{2}\right)=\int_0^{\infty}e^{-u^2}du

The right hand side is known as the Gaussian integral. To evaluate this integral we must first square both sides. This may seem strange, however, in order to solve it the integral must be transformed from cartesian to polar coordinates.

(Γ(32))2=(∫0∞e−u2du)2=(∫0∞e−u2du)(∫0∞e−v2dv)=∫0∞∫0∞e−(u2+v2)dudv \begin{aligned} \left(\Gamma\left(\frac{3}{2}\right)\right)^2&=\left(\int_0^{\infty}e^{-u^2}du\right)^2 \\ &=\left(\int_0^{\infty}e^{-u^2}du\right)\left(\int_0^{\infty}e^{-v^2}dv\right) \\ &=\int_0^{\infty}\int_0^{\infty}e^{-(u^2+v^2)}dudv \end{aligned}

The transformation to polar coordinates comes from making the substitution u=rcos⁡θu=r\cos\theta and v=rsin⁡θv=r\sin\theta. And so, in terms of the new variables, (r,θ)(r, \theta), the integral is taken over 0⩽r⩽∞0\leqslant r\leqslant\infty and 0⩽θ⩽π20\leqslant\theta\leqslant\frac{\pi}{2}, and dudvdudv becomes rdrdθrdrd\theta. This linear transformation can be represented by the Jacobian matrix:

JF(r,θ)=[∂u∂r∂u∂θ∂v∂r∂v∂θ]=[cos⁡θ−rsin⁡θsin⁡θrcos⁡θ]\begin{aligned} \textbf{J}_{\textbf{F}}(r,\theta)= \begin{bmatrix} \frac{\partial u}{\partial r}&\frac{\partial u}{\partial\theta}\\ \frac{\partial v}{\partial r}&\frac{\partial v}{\partial\theta}\\ \end{bmatrix} = \begin{bmatrix} \cos\theta&-r\sin\theta\\ \sin\theta&r\cos\theta\\ \end{bmatrix} \\ \end{aligned}

Taking the determinant of JF\textbf{J}_{\textbf{F}} gives

∣JF(r,θ)∣=rcos⁡2θ−(−rsin⁡2θ)=r |\textbf{J}_{\textbf{F}}(r,\theta)|=r\cos^2\theta-(-r\sin^2\theta)=r

And so, the integral becomes

(Γ(32))2=∫0∞∫0π2e−r2rdrdθ\left(\Gamma\left(\frac{3}{2}\right)\right)^2=\int_0^{\infty}\int_0^{\frac{\pi}{2}}e^{-r^2}rdrd\theta

The way we’ve transformed the coordinates may seem a bit confusing and abstract. I have put some links at the bottom of the page which better explain how this transformation is made and what’s actually going on when the Jacobian matrix is used.

Splitting the double integral makes it easier to see how it can be evaluated:

(Γ(32))2=(∫0π2dθ)(∫0∞e−r2rdr)=([θ]0π2)(∫0∞e−r2rdr)=(π2)(∫0∞e−r2rdr) \begin{aligned} \left(\Gamma\left(\frac{3}{2}\right)\right)^2&=\left(\int_0^{\frac{\pi}{2}}d\theta\right)\left(\int_0^{\infty}e^{-r^2}rdr\right) \\ &=\left(\left[\theta\right]_0^{\frac{\pi}{2}}\right)\left(\int_0^{\infty}e^{-r^2}rdr\right) \\ &=\left(\frac{\pi}{2}\right)\left(\int_0^{\infty}e^{-r^2}rdr\right) \\ \end{aligned}

To evaluate the second integral, we make the substitution t=−r2t=-r^2, with dt=−2rdrdt=-2rdr:

(Γ(32))2=π2(−∫0−∞etrdt2r)=−π4(∫0−∞etdt)=−π4[(lim⁡t→−∞et)−e0]=−π4(0−1)=π4 \begin{aligned} \left(\Gamma\left(\frac{3}{2}\right)\right)^2&=\frac{\pi}{2}\left(-\int_0^{-\infty}e^tr\frac{dt}{2r}\right) \\ &=-\frac{\pi}{4}\left(\int_0^{-\infty}e^tdt\right) \\ &=-\frac{\pi}{4}\left[\left(\lim_{t\to-\infty}e^t\right)-e^0\right] \\ &=-\frac{\pi}{4}(0-1) \\ &=\frac{\pi}{4} \end{aligned}

Taking the square root of both sides gives

Γ(32)=π2\Gamma\left(\frac{3}{2}\right)=\frac{\sqrt{\pi}}{2}

We showed earlier that Γ(32)=(12)!\Gamma(\frac{3}{2})=(\frac{1}{2})!. And so, using this we get:

(12)!=π2\left(\frac{1}{2}\right)!=\frac{\sqrt{\pi}}{2}

Though the method of getting to the above result was pretty convoluted, I think with just how mind-blowing the result is, it’s definitely worth working through.

We showed earlier that Γ(32)=(12)Γ(12)\Gamma\left(\frac{3}{2}\right)=\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right). And so, from the result we just proved, it follows that

Γ(12)=π\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}

If we recall the property Γ(z)=(z−1)!\Gamma(z)=(z-1)!, then we see that

(−12)!=π\left(-\frac{1}{2}\right)!=\sqrt{\pi}

Useful links: