Question 4.5: The d-Dimensional Gaussian Integral Derive (4.65) for intege......
The d-Dimensional Gaussian Integral
Derive (4.65) for integer d.
\int\mathrm{d}^{d}k_{E}f(k_{E\mu})=\int\mathrm{d}^{d}k_{E}f(k_{E\mu}+p_{E\mu})\ . (4.65)
Omitting the subscript ” E ” in this exercise, we define for integer d and Euclidian k_{\mu}, \mu=0,1,2, \ldots, d-1,
\begin{aligned} k_{0} & =k \cos \vartheta_{1}, \\ k_{1} & =k \sin \vartheta_{1} \cos \vartheta_{2}, \\ k_{2} & =k \sin \vartheta_{1} \sin \vartheta_{2} \cos \vartheta_{3}, \\ \vdots & \\ k_{d-2} & =k \sin \vartheta_{1} \sin \vartheta_{2} \ldots \sin \vartheta_{d-2} \cos \vartheta_{d-1}, \\ k_{d-1} & =k \sin \vartheta_{1} \sin \vartheta_{2} \ldots \sin \vartheta_{d-2} \sin \vartheta_{d-1} & (1) \end{aligned}
and calculate the Jacobian row by row
\begin{aligned} & J=\frac{\partial\left(k_{0} k_{1} k_{2} \ldots k_{d-1}\right)}{\partial\left(k, \vartheta_{1}, \vartheta_{2} \ldots \vartheta_{d-1}\right)} \\ & =\left|\begin{array}{cccc} \cos \vartheta_{1} & -k \sin \vartheta_{1} & 0 & \ldots \\ \sin \vartheta_{1} \cos \vartheta_{2} & k \cos \vartheta_{1} \cos \vartheta_{2} & -k \sin \vartheta_{1} \sin \vartheta_{2} & \ldots \\ \sin \vartheta_{1} \sin \vartheta_{2} \cos \vartheta_{3} & k \cos \vartheta_{1} \sin \vartheta_{2} \cos \vartheta_{3} & k \sin \vartheta_{1} \cos \vartheta_{2} \cos \vartheta_{3} & \ldots \\ \vdots & \vdots & \vdots & \ddots \end{array}\right| \\ & =k^{d-1}\left\{\cos \vartheta_{1} \cos \vartheta_{1}\left(\sin \vartheta_{1}\right)^{d-2}+\sin \vartheta_{1} \sin \vartheta_{1}\left(\sin \vartheta_{1}\right)^{d-2}\right\} \\ & \times\left|\begin{array}{ccc} \cos \vartheta_{2} & -\sin \vartheta_{2} & \ldots \\ \sin \vartheta_{2} \cos \vartheta_{3} & \cos \vartheta_{2} \cos \vartheta_{3} & \cdots \\ \vdots & \vdots & \ddots \end{array}\right|=\ldots \\ & =k^{d-1}\left(\sin \vartheta_{1}\right)^{d-2}\left(\sin \vartheta_{2}\right)^{d-3} \ldots\left(\sin \vartheta_{d-3}\right)^{2} \\ & \times\left|\begin{array}{ccc} \cos \vartheta_{d-2} & -\sin \vartheta_{d-2} & 0 \\ \sin \vartheta_{d-2} \cos \vartheta_{d-1} & \cos \vartheta_{d-2} \cos \vartheta_{d-1} & -\sin \vartheta_{d-2} \sin \vartheta_{d-1} \\ \sin \vartheta_{d-2} \sin \vartheta_{d-1} & \cos \vartheta_{d-2} \sin \vartheta_{d-1} & \sin \vartheta_{d-2} \cos \vartheta_{d-1} \end{array}\right| \\ & =k^{d-1}\left(\sin \vartheta_{1}\right)^{d-2}\left(\sin \vartheta_{2}\right)^{d-3} \ldots\left(\sin \vartheta_{d-2}\right) \\ & \times\left|\begin{array}{cc} \cos \vartheta_{d-1} & -\sin \vartheta_{d-1} \\ \sin \vartheta_{d-1} & \cos \vartheta_{d-1} \end{array}\right| \\ & =k^{d-1}\left(\sin \vartheta_{1}\right)^{d-2}\left(\sin \vartheta_{2}\right)^{d-3} \ldots\left(\sin \vartheta_{d-2}\right). & (2) \end{aligned}
The Gaussian integral thus becomes
\begin{aligned} & \int\limits_{-\infty}^{\infty} \mathrm{d}^{d} k \mathrm{e}^{-k^{2}} & (3) \\ & \quad=\int\limits_{0}^{\infty} \mathrm{d} k k^{d-1} \mathrm{e}^{-k^{2}} \int\limits_{0}^{\pi}\left(\sin \vartheta_{1}\right)^{d-2} \mathrm{~d} \vartheta_{1} \ldots \int\limits_{0}^{\pi} \sin \vartheta_{d-2} \mathrm{~d} \vartheta_{d-2} \int\limits_{0}^{\pi} \mathrm{d} \vartheta_{d-1} . \end{aligned}
We use partial integration to get
\begin{aligned} & \int\limits_{0}^{\pi}(\sin \vartheta)^{n-1} \sin \vartheta \mathrm{d} \vartheta \\ & \quad=\left[(\sin \vartheta)^{n-1}(-\cos \vartheta)\right]_{0}^{\pi}-(n-1) \int\limits_{0}^{\pi}(\sin \vartheta)^{n-2} \cos \vartheta(-\cos \vartheta) \mathrm{d} \vartheta \\ & \quad=(n-1) \int\limits_{0}^{\pi}\left[(\sin \vartheta)^{n-2}-(\sin \vartheta)^{n}\right] \mathrm{d} \vartheta & (4) \end{aligned}
or
\int\limits_{0}^{\pi}(\sin \vartheta)^{n} \mathrm{~d} \vartheta=\frac{n-1}{n} \int\limits_{0}^{\pi}(\sin \vartheta)^{n-2} \mathrm{~d} \vartheta .
For even n \geq 2 we thus get
\int\limits_{0}^{\pi}(\sin \vartheta)^{n} \mathrm{~d} \vartheta=\frac{(n-1) ! !}{n ! !} \int\limits_{0}^{\pi} \mathrm{d} \vartheta=\frac{(n-1) ! !}{n ! !} \pi , (5)
and for n=0 we have
\int\limits_{0}^{\pi} \mathrm{d} \vartheta=\pi. (6)
For odd n we find
\begin{aligned} \int\limits_{0}^{\pi}(\sin \vartheta)^{n} \mathrm{~d} \vartheta & =\frac{(n-1) ! !}{n ! !} \int\limits_{0}^{\pi} \sin \vartheta \mathrm{d} \vartheta \\ & =\frac{(n-1) ! !}{n ! !} \times 2. & (7) \end{aligned}
Inserting all these equations into (3) we get
\begin{aligned} \int\limits_{-\infty}^{\infty} \mathrm{d}^{d} k \mathrm{e}^{-k^{2}} & =\int\limits_{0}^{\infty} \mathrm{d} k k^{d-1} \mathrm{e}^{-k^{2}} I_{d} \frac{1}{(d-2) ! !}, \\ d \text { even } \Rightarrow I_{d} & =(2 \pi)^{\frac{d}{2}}, \\ d \text { odd } \Rightarrow I_{d} & =2 \times(2 \pi)^{\frac{d-1}{2}} . & (8) \end{aligned}
On the other hand, partial integration gives
\begin{aligned} \int\limits_{0}^{\infty} \mathrm{d} k k^{d-1} \mathrm{e}^{-k^{2}} & =\frac{1}{2} \int\limits_{0}^{\infty} \mathrm{d} k k^{d-2} 2 k \mathrm{e}^{-k^{2}} \\ & =\frac{d-2}{2} \int\limits_{0}^{\infty} \mathrm{d} k k^{d-3} \mathrm{e}^{-k^{2}} \\ & =(d-2) ! ! K_{d}, & (9) \end{aligned}
\begin{aligned} d \text { even } \Rightarrow K_{d} & =\left(\frac{1}{2}\right)^{\frac{d-2}{2}} \int\limits_{0}^{\infty} \mathrm{d} k k \mathrm{e}^{-k^{2}}=\left(\frac{1}{2}\right)^{\frac{d}{2}}, \\ d \text { odd } \Rightarrow K_{d} & =\left(\frac{1}{2}\right)^{\frac{d-1}{2}} \int\limits_{0}^{\infty} \mathrm{d} k \mathrm{e}^{-k^{2}}=\left(\frac{1}{2}\right)^{\frac{d-1}{2}} \frac{\sqrt{\pi}}{2} . & (10) \end{aligned}
All together this gives
\int\limits_{-\infty}^{\infty} \mathrm{d}^{d} k \mathrm{e}^{-k^{2}}=\frac{1}{(d-2) ! !}(d-2) ! ! I_{d} K_{d}=\pi^{\frac{d}{2}} (11)
which completes our proof.