Integral of the day XXV

Integral related to the reciprocal beta function

Today we show the proof of this integral posted by @integralsbot \[ \int_{0}^{\pi} \frac{x\left(\sin\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right)}{\sqrt{\sin x}} dx = 2\pi\ln(2) \] The proof will rely on the integral representation of the reciprocal beta function.

Proof:

Recall the following integral representation of the reciprocal beta function (Proof in the Appendix): \[ \int_{0}^{\pi} \sin^{v-1}x \sin(ax) dx = \frac{\pi \sin\left(\frac{\pi a}{2} \right)}{2^{v-1}vB\left(\frac{1+a+v}{2},\frac{1-a+v}{2}\right)} \quad v,a\in \mathbb{R} , \; \; v>0\] \[ \int_{0}^{\pi} \sin^{v-1}x \cos(ax) dx = \frac{\pi \cos\left(\frac{\pi a}{2} \right)}{2^{v-1}vB\left(\frac{1+a+v}{2},\frac{1-a+v}{2}\right)} \quad v,a\in \mathbb{R} , \; \; v>0\] If we put $\displaystyle v = \frac{1}{2}$ \[ \int_{0}^{\pi} \frac{ \sin(ax)}{\sqrt{\sin x} } dx = \frac{\pi \sin\left(\frac{\pi a}{2} \right)}{2^{\frac{1}{2}-1}vB\left(\frac{1+a}{2}+\frac{1}{4},\frac{1-a}{2}+\frac{1}{4}\right)} \] \[ \int_{0}^{\pi} \frac{ \cos(ax)}{\sqrt{\sin x} } dx = \frac{\pi \cos\left(\frac{\pi a}{2} \right)}{2^{\frac{1}{2}-1}vB\left(\frac{1+a}{2}+\frac{1}{4},\frac{1-a}{2}+\frac{1}{4}\right)} \] Differentiating with respect to $a$: \[ \frac{d}{da} \int_{0}^{\pi} \frac{ \sin(ax)}{\sqrt{\sin x} } dx = \int_{0}^{\pi} \frac{ x\cos(ax)}{\sqrt{\sin x} } dx = \frac{\sqrt{2}\pi^2\cos\left(\frac{\pi a}{2}\right) + \sqrt{2}\pi \sin\left(\frac{\pi a}{2}\right)\psi\left(\frac{3}{4}-\frac{a}{2}\right)-\sqrt{2}\pi \sin\left(\frac{\pi a}{2}\right)\psi\left(\frac{a}{2}+\frac{3}{4}\right)}{B\left(\frac{a}{2}+\frac{3}{4}, \frac{3}{4}-\frac{a}{2}\right)} \] \[ \frac{d}{da} \int_{0}^{\pi} \frac{ \cos(ax)}{\sqrt{\sin x} } dx = -\int_{0}^{\pi} \frac{ x\sin(ax)}{\sqrt{\sin x} } dx = \frac{\sqrt{2}\pi^2\sin\left(\frac{\pi a}{2}\right) - \sqrt{2}\pi \cos\left(\frac{\pi a}{2}\right)\psi\left(\frac{3}{4}-\frac{a}{2}\right)+\sqrt{2}\pi \cos\left(\frac{\pi a}{2}\right)\psi\left(\frac{a}{2}+\frac{3}{4}\right)}{B\left(\frac{a}{2}+\frac{3}{4}, \frac{3}{4}-\frac{a}{2}\right)}\] Taking the limit as $\displaystyle a \to \frac{1}{2}$ \[ \int_{0}^{\pi} \frac{x\cos\left(\frac{x}{2}\right)}{\sqrt{\sin x}} dx = \frac{\pi\gamma+\pi^2+\pi \psi\left(\frac{1}{2}\right)}{2} \] \[ -\int_{0}^{\pi} \frac{ x\sin\left(\frac{x}{2}\right)}{\sqrt{\sin x} } dx = \frac{\pi\gamma-\pi^2+\pi \psi\left(\frac{1}{2}\right)}{2} \] From the formula \[ \psi\left(\frac{1}{2}-J\right) = \psi\left(\frac{1}{2}+J\right) = -\gamma -\ln(4) +\sum_{j=1}^{J} \frac{2}{2j-1} \quad J=0,1,2,...\] if we put $J=0$ we have \[ \psi\left(\frac{1}{2}\right) + \gamma = -\ln(4) \] Therefore \[ \int_{0}^{\pi} \frac{x\cos\left(\frac{x}{2}\right)}{\sqrt{\sin x}} dx = \frac{\pi^2-\pi\ln(4)}{2} \] \[ -\int_{0}^{\pi} \frac{ x\sin\left(\frac{x}{2}\right)}{\sqrt{\sin x} } dx = \frac{-\pi^2-\pi\ln(4)}{2} \] Hence \[ \int_{0}^{\pi} \frac{x\left(\sin\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right)}{\sqrt{\sin x}} dx = \frac{\pi^2}{2} + \frac{\pi\ln(4)}{2} -\frac{\pi^2}{2} +\frac{\pi\ln(4)}{2} = 2\pi\ln(2) \] Therefore we can conclude \[ \boxed{\int_{0}^{\pi} \frac{x\left(\sin\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right)}{\sqrt{\sin x}} dx = 2\pi\ln(2) }\]

Appendix:

Here is the proof of the following integrals with $a,b \in \mathbb{R}$: \[ \int_{0}^{\pi} \sin^{v-1}x \sin(ax) dx = \frac{\pi \sin\left(\frac{\pi a}{2} \right)}{2^{v-1}vB\left(\frac{1+a+v}{2},\frac{1-a+v}{2}\right)} \quad v>0\] \[ \int_{0}^{\pi} \sin^{v-1}x \cos(ax) dx = \frac{\pi \cos\left(\frac{\pi a}{2} \right)}{2^{v-1}vB\left(\frac{1+a+v}{2},\frac{1-a+v}{2}\right)} \quad v>0 \]

Proof:

Let $a,b \in \mathbb{R}$ \begin{align*} I= \int_{0}^{\pi} \sin^{v-1}x \; e^{iax}dx =& \int_{0}^{\pi} e^{iax}\left(\frac{e^{ix}+e^{-ix}}{2i} \right)^{v-1}dx \\ =& \frac{1}{(2i)^{v-1}}\int_{0}^{\pi} e^{iax} \sum_{j=0}^{\infty} \binom{v-1}{j}(-1)^j e^{ix(v-1-2j)} dx\\ =&\frac{1}{(2i)^{v-1}} \sum_{j=0}^{\infty} \binom{v-1}{j}(-1)^j\int_{0}^{\pi} e^{ix(a+v-1-2j)} dx\\ =&\frac{1}{(2i)^{v-1}} \sum_{j=0}^{\infty} \binom{v-1}{j}\frac{(-1)^j}{i(a+v-1-2j)}\int_{0}^{i\pi(a+v-1-2j)} e^{w}dw\\ =&\frac{1}{(2i)^{v-1}} \sum_{j=0}^{\infty} \binom{v-1}{j}\frac{(-1)^j\left(e^{i\pi (a+v-1-2j)}-1\right)}{i(a+v-1-2j)}\\ =&\frac{1}{(2i)^{v-1}} \sum_{j=0}^{\infty} \binom{v-1}{j}\frac{(-1)^j\left(e^{i\pi (a+v-1)}-1\right)}{i(a+v-1-2j)}\\ =&\frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v-1}i^v} \sum_{j=0}^{\infty} \binom{v-1}{j}\frac{(-1)^j}{(a+v-1-2j)}\\ =&\frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v-1}i^v} \sum_{j=0}^{\infty} \frac{(v-j)_{j}}{j!}\frac{(-1)^j}{(a+v-1-2j)}\\ =&\frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v-1}i^v} \sum_{j=0}^{\infty} \frac{(1-v)_{j}}{(a+v-1-2j)j!} \end{align*} Note that \[ 2j-a-v+1 = \frac{-(a+v-1)\left(\frac{3-a-v}{2}\right)_{j}}{\left(\frac{-a-v+1}{2}\right)_{j}}\] Hence \begin{align*} I =&\frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v-1}i^v} \sum_{j=0}^{\infty} \frac{(1-v)_{j}}{(a+v-1-2j)j!}\\ =& \frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v-1}i^v}\frac{1}{(a+v-1)} \sum_{j=0}^{\infty} \frac{(1-v)_{j}\left(\frac{-a-v+1}{2}\right)_{j}}{\left(\frac{3-a-v}{2}\right)_{j}}\\ =& \frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v-1}i^v}\frac{1}{(a+v-1)} F\left(1-v,\frac{-a-v+1}{2};\frac{3-a-v}{2};1 \right) \end{align*} From the Gauss's Hypergeometric Theorem: \[ F(a,b;c;1) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)} \quad c>a+b \] Therefore, supposing $v>0$: \begin{align*} I =& \frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v-1}i^v}\frac{1}{(a+v-1)} \frac{\Gamma\left(\frac{3-a-v}{2}\right)\Gamma(v)}{\Gamma\left(\frac{1-a+v}{2}\right)}\\ =&\frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v}i^v}\frac{1}{-\frac{(1-a-v)}{2}} \frac{\left(\frac{1-a-v}{2}\right)\Gamma\left(-\frac{1-a-v}{2}\right)\Gamma(v)}{\Gamma\left(\frac{1-a+v}{2}\right)}\\ =&-\frac{\left(e^{i\pi (a+v-1)}-1\right)}{2^{v}i^v} \frac{\Gamma\left(\frac{1-a-v}{2}\right)\Gamma(v)}{\Gamma\left(\frac{1-a+v}{2}\right)}\\ \end{align*} Note \[ \Im (e^{i\pi (a+v-1)}-1)i^{-v} = -2\sin\left(\frac{\pi a}{2} \right)\cos\left(\frac{\pi(a+v)}{2}\right)\] \[ \Re (e^{i\pi (a+v-1)}-1)i^{-v} = -2\cos\left(\frac{\pi a}{2} \right)\sin\left(\frac{\pi(a+v)}{2}\right)\] and recall the Euler's reflection formula for $\cos(\pi t)$: \[\Gamma\left(\frac{1}{2}+t\right)\Gamma\left(\frac{1}{2}-t\right) = \frac{\pi}{\cos(\pi t)}\] Therefore \begin{align*} \int_{0}^{\pi} \sin^{v-1}x \sin(ax) dx = \Im \left(\int_{0}^{\pi} \sin^{v-1}x \; e^{iax}dx \right) = & \frac{ \sin\left(\frac{\pi a}{2} \right)\cos\left(\frac{\pi(a+v)}{2}\right)}{2^{v-1}} \frac{\Gamma\left(\frac{1-a-v}{2}\right)\Gamma(v)}{\Gamma\left(\frac{1-a+v}{2}\right)}\\ =&\frac{ \sin\left(\frac{\pi a}{2} \right)}{2^{v-1}} \frac{\pi\Gamma\left(\frac{1-a-v}{2}\right)\Gamma(v)}{\Gamma\left(\frac{1+a+v}{2}\right)\Gamma\left(\frac{1-a+v}{2}\right)\Gamma\left(\frac{1-a+v}{2}\right)} \quad \textrm{(Euler's reflection formula)}\\ =&\frac{ \sin\left(\frac{\pi a}{2} \right)}{2^{v-1}} \frac{\pi \Gamma(v)}{\Gamma\left(\frac{1+a+v}{2}\right)\Gamma\left(\frac{1-a+v}{2}\right)}\\ =&\frac{\pi \sin\left(\frac{\pi a}{2} \right)}{2^{v-1}vB\left(\frac{1+a+v}{2},\frac{1-a+v}{2}\right)} \end{align*} In a very similar way we can show : \[ \int_{0}^{\pi} \sin^{v-1}x \cos(ax) dx = \Re \left(\int_{0}^{\pi} \sin^{v-1}x \; e^{iax}dx \right) = \frac{\pi \cos\left(\frac{\pi a}{2} \right)}{2^{v-1}vB\left(\frac{1+a+v}{2},\frac{1-a+v}{2}\right)} \]