In this post, we’ll prove a very interesting identity: \int_0^\pi \left(\sin(\theta) \right)^{\alpha-1} e^{i \beta \theta} \; d\theta = \frac{\pi e^{\frac{i \pi}{2} \beta}}{\alpha 2^{\alpha-1} B \left(\frac{\alpha+\beta+1}{2}, \frac{\alpha-\beta+1}{2} \right)} where \beta + 1> \alpha > 0 and B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} is the Beta function.
The idea is to integrate the principal branch of f(z) = (1-z^2)^{\alpha-1} (-iz)^{\beta - \alpha} around the following contour:
where
- C_\epsilon is an arc parameterized by e^{it}, where \varphi_\epsilon\leq t \leq \pi - \varphi_\epsilon and \varphi_\epsilon = \text{arctan}\left( \frac{\epsilon \sqrt{4-\epsilon^2}}{2-\epsilon^2}\right).
- I_{1,\epsilon}, \; I_{2,\epsilon}, \; I_{3,\epsilon} are circular indents of radius \epsilon around the branch points -1, 1 and 0, respectively.
- S_{1,\epsilon} is a line joining -1+\epsilon and -\epsilon.
- S_{2,\epsilon} is a line joining \epsilon and 1-\epsilon.
The condition \beta + 1 > \alpha > 0 ensures that: \lim_{\epsilon\to 0^+}\int_{I_{1,\epsilon}}f(z)\; dz = \lim_{\epsilon\to 0^+}\int_{I_{2,\epsilon}}f(z)\; dz = \lim_{\epsilon\to 0^+}\int_{I_{3,\epsilon}}f(z)\; dz = 0 We’ll only prove that the integral around I_{3,\epsilon} tends to 0 as \epsilon \to 0^+. The proof is quite similar for I_{1,\epsilon} and I_{2,\epsilon}. We have \begin{aligned} \left| \int_{I_{3,\epsilon}} f(z) \; dz\right| &= \left|-i\epsilon \int_0^\pi e^{it} f(\epsilon e^{it})\; dt\right| \\ &\leq \epsilon\int_0^\pi |f(\epsilon e^{it})| \; dt \\ &\leq \epsilon^{\beta-\alpha+1} \int_0^\pi |1-\epsilon^2 e^{2it}|^{\alpha-1} \; dt \\ &\leq \pi \epsilon^{\beta-\alpha+1}(1+\epsilon^2)^{\alpha-1} \end{aligned} Hence, \left| \int_{I_{3,\epsilon}} f(z) \; dz\right| \to 0 as \epsilon\to 0^+. Next, by Cauchy’s theorem we have: \begin{aligned} \int_{-1}^1 f(z) \; dz &= -i\int_0^\pi e^{it}f(e^{it})\; dt \end{aligned} The integral on the left hand side is: \begin{aligned}\int_{-1}^1 f(z) \; dz &= 2\cos\left( (\beta-\alpha)\frac{\pi}{2}\right)\int_0^1 (1-z^2)^{\alpha-1} z^{\beta-\alpha}\; dz \\ &= \cos\left( (\beta-\alpha)\frac{\pi}{2}\right) \int_0^1 (1-w)^{\alpha-1} w^{\frac{\beta-\alpha-1}{2}}\; dw \\ &= \cos\left( (\beta-\alpha)\frac{\pi}{2}\right) B\left(\alpha, \frac{\beta-\alpha+1}{2} \right) \\ &= \cos\left( (\beta-\alpha)\frac{\pi}{2}\right) \frac{\Gamma(\alpha) \Gamma\left(\frac{\beta-\alpha+1}{2} \right)}{\Gamma\left(\frac{\beta+\alpha+1}{2} \right)} \\ &= \frac{\pi \Gamma(\alpha)}{\Gamma\left(\frac{\beta+\alpha+1}{2} \right) \Gamma\left(\frac{-\beta+\alpha+1}{2} \right)} \\ &= \frac{\pi}{\alpha B \left(\frac{\alpha+\beta+1}{2}, \frac{\alpha-\beta+1}{2} \right)}\end{aligned} The integral on the right hand side is:
\begin{aligned}-i\int_0^\pi e^{it}f(e^{it})\; dt &= \int_0^\pi (1-e^{2it})^{\alpha-1}(-i e^{it})^{\beta-\alpha+1} \; dt \\ &= \int_0^\pi (2\sin(t))^{\alpha-1}e^{i\beta t -\frac{i\pi}{2}\beta}\; dt \\ &= e^{-\frac{i\pi}{2}\beta} 2^{\alpha-1} \int_0^\pi (\sin (t))^{\alpha-1}e^{i\beta t}\; dt \end{aligned}Therefore, we have:
\begin{aligned}e^{-\frac{i\pi}{2}\beta} 2^{\alpha-1} \int_0^\pi (\sin (t))^{\alpha-1}e^{i\beta t}\; dt &= \frac{\pi}{\alpha B \left(\frac{\alpha+\beta+1}{2}, \frac{\alpha-\beta+1}{2} \right)} \\ \implies \int_0^\pi \left(\sin(t) \right)^{\alpha-1} e^{i \beta t} \; dt &= \frac{\pi e^{\frac{i \pi}{2} \beta}}{\alpha 2^{\alpha-1} B \left(\frac{\alpha+\beta+1}{2}, \frac{\alpha-\beta+1}{2} \right)}\end{aligned} We can use a similar technique to show that: \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\cos(t)\right)^{\alpha-1}e^{i\beta t}\; dt = \frac{\pi}{\alpha 2^{\alpha-1} B \left(\frac{\alpha+\beta+1}{2}, \frac{\alpha-\beta+1}{2} \right)}, \quad \beta+1 > \alpha > 0
One can further argue using the identity theorem that these results still hold even if we relax the requirement: \beta + 1> \alpha.
Many interesting logarithmic integrals can be derived using these identities by differentiating both sides with respect to the parameters \alpha and \beta. One such example is:
\int_0^{\frac{\pi}{2}} \theta^2 \log^2\left(\cos \theta\right)d\theta = \frac{11\pi^5}{1440}+\frac{\pi^3}{24}\log^2(2)+\frac{\pi}{2} \log(2)\zeta(3)