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:
$$ \int_0^{\frac{\pi}{2}} \theta^2 \log^2\left(\cos \theta\right)d\theta = \frac{11\pi^5}{1440}+\frac{\pi^3}{24}\log^2(2)+\frac{\pi}{2} \log(2)\zeta(3) $$