Integral of angular functions over d-dim solid angle

[Einj]

Hello everyone! I have a question about angular integration in arbitrary d dimensions. The interest comes from the need to use dimensional regularization. Suppose I start with a 2-dimensional integral and then I have to move to $d=2-\epsilon$ dimension to regularize my integral. Now, suppose $\theta$ is the polar angle in the original 2 dimensions and now I want to compute the following integrals:

$$
\int d\Omega_d\cos\theta ,
$$
$$
\int d\Omega_d\cos^2\theta.
$$

What is the result? Can I still say that the first integral is zero? What about the second?

Thanks a lot!

 

[jvicens]

Hello! Thank you for your question about angular integration in arbitrary d dimensions. Dimensional regularization is a powerful technique used in theoretical physics to handle divergent integrals in higher dimensions.

To answer your question, let's first define the integral in question as:

$$
I = \int d\Omega_d\cos^k\theta ,
$$

where $k$ is any integer. In this case, $k=1$ for the first integral and $k=2$ for the second.

In dimensional regularization, we introduce a small parameter $\epsilon$ to represent the deviation from the original dimension. In your case, we have $d=2-\epsilon$.

To compute the integral in $d$ dimensions, we can use the following formula:

$$
I_d = \int d\Omega_d\cos^k\theta = \frac{2\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}\int_0^{\pi}\sin^{d-2}\theta\cos^k\theta d\theta ,
$$

where $\Gamma$ is the Gamma function. Plugging in $d=2-\epsilon$ and expanding the integrand in powers of $\epsilon$, we get:

$$
I_{2-\epsilon} = \frac{2\pi^{\frac{2-\epsilon}{2}}}{\Gamma(\frac{2-\epsilon}{2})}\int_0^{\pi}\sin^{\epsilon}\theta\left(1 + \frac{\epsilon}{2}\cos^2\theta + \mathcal{O}(\epsilon^2)\right)d\theta .
$$

Using the fact that $\Gamma(\frac{1}{2}) = \sqrt{\pi}$ and expanding the Gamma function to first order in $\epsilon$, we can simplify the above expression to:

$$
I_{2-\epsilon} = \frac{\pi}{\sqrt{\pi}}\left(1 - \frac{\epsilon}{2}\ln\frac{\pi}{2} + \mathcal{O}(\epsilon^2)\right)\left(\frac{\pi}{2} + \mathcal{O}(\epsilon)\right) .
$$

Finally, taking the limit $\epsilon \rightarrow 0$, we get the result:

$$
I_2 = \frac{\pi}{2} .
$$

Therefore, for the first integral, we have