An integral related to Gamma value.

Question

We have: $$\int_{0}^{\pi/2}\frac{\sin{x}\log{(\tan{(x/2))}+x}}{\sqrt{\sin{x}}(\sin{x}+1)}dx=\pi-\frac{\sqrt{2\pi}\Gamma{(1/4)}^{2}}{16}-\frac{\sqrt{2}\pi^{5/2}}{2\Gamma{(1/4)}^{2}}\tag{1}.$$ As other integrals, Mathametica and Wolfram Alpha are unable to deal with it (see Wolfram Alpha response). The proof I have is not elementary and it is based on the following identity: $$\Im\left(K(\sqrt{\frac{2k}{1+k}})\right)=\Im \left(K(\sqrt{\frac{1-k}{1+k}})-K(\sqrt{\frac{1+k}{1-k}})\cdot\sqrt{\frac{1+k}{1-k}}\right),\tag{2}$$ where: $$K(k)=\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^{2}{x}}},\tag{3}$$ is the complete elliptic integral of the first kind. The proof follows from $(2)$ integrating both sides carefully.

Question: Can we prove $(1)$ using only Feynman's trick and Beta function? Thanks for your cooperation.

Answer 1

---

Let $\mathcal{I}$ denote the value of the definite integral in question:

$$\mathcal{I}:=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{\theta+\sin{\left(\theta\right)}\ln{\left(\tan{\left(\frac{\theta}{2}\right)}\right)}}{\left[1+\sin{\left(\theta\right)}\right]\sqrt{\sin{\left(\theta\right)}}}\approx0.141216.$$

---

$$\begin{align} \mathcal{I} &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{\theta+\sin{\left(\theta\right)}\ln{\left(\tan{\left(\frac{\theta}{2}\right)}\right)}}{\left[1+\sin{\left(\theta\right)}\right]\sqrt{\sin{\left(\theta\right)}}}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{2}{1+t^{2}}\cdot\frac{2\arctan{\left(t\right)}+\left(\frac{2t}{1+t^{2}}\right)\ln{\left(t\right)}}{\left(1+\frac{2t}{1+t^{2}}\right)\sqrt{\frac{2t}{1+t^{2}}}};~~~\small{\left[\theta=2\arctan{\left(t\right)}\right]}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{2\left[2\arctan{\left(t\right)}+\left(\frac{2t}{1+t^{2}}\right)\ln{\left(t\right)}\right]}{\left(1+t\right)^{2}\sqrt{\frac{2t}{1+t^{2}}}}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{2\arctan{\left(\frac{1-u}{1+u}\right)}+\left(\frac{1-u^{2}}{1+u^{2}}\right)\ln{\left(\frac{1-u}{1+u}\right)}}{\sqrt{\frac{1-u^{2}}{1+u^{2}}}};~~~\small{\left[t=\frac{1-u}{1+u}\right]}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{2\left[\arctan{\left(1\right)}-\arctan{\left(u\right)}\right]-2\left(\frac{1-u^{2}}{1+u^{2}}\right)\operatorname{artanh}{\left(u\right)}}{\sqrt{\frac{1-u^{2}}{1+u^{2}}}}\\ &=2\int_{0}^{1}\mathrm{d}u\,\frac{\arctan{\left(1\right)}}{\sqrt{\frac{1-u^{2}}{1+u^{2}}}}-2\int_{0}^{1}\mathrm{d}u\,\frac{\arctan{\left(u\right)}}{\sqrt{\frac{1-u^{2}}{1+u^{2}}}}-2\int_{0}^{1}\mathrm{d}u\,\sqrt{\frac{1-u^{2}}{1+u^{2}}}\operatorname{artanh}{\left(u\right)}\\ &=\frac{\pi}{2}\int_{0}^{1}\mathrm{d}u\,\frac{1+u^{2}}{\sqrt{1-u^{4}}}-2\int_{0}^{1}\mathrm{d}u\,\sqrt{\frac{1+u^{2}}{1-u^{2}}}\arctan{\left(u\right)}\\ &~~~~~-2\int_{0}^{1}\mathrm{d}u\,\sqrt{\frac{1-u^{2}}{1+u^{2}}}\operatorname{artanh}{\left(u\right)}\\ &=\frac{\pi}{2}\int_{0}^{1}\mathrm{d}x\,\frac{1+\sqrt{x}}{4x^{3/4}\sqrt{1-x}};~~~\small{\left[u=\sqrt[4]{x}\right]}\\ &~~~~~-2\int_{0}^{1}\mathrm{d}u\,\left[\sqrt{\frac{1+u^{2}}{1-u^{2}}}\arctan{\left(u\right)}+\sqrt{\frac{1-u^{2}}{1+u^{2}}}\operatorname{artanh}{\left(u\right)}\right]\\ &=\frac{\pi}{8}\left[\int_{0}^{1}\mathrm{d}x\,\frac{1}{x^{3/4}\sqrt{1-x}}+\int_{0}^{1}\mathrm{d}x\,\frac{1}{\sqrt[4]{x}\sqrt{1-x}}\right]\\ &~~~~~-2\int_{0}^{1}\mathrm{d}u\,\left[\sqrt{\frac{1+u^{2}}{1-u^{2}}}\int_{0}^{1}\mathrm{d}x\,\frac{u}{1+x^{2}u^{2}}+\sqrt{\frac{1-u^{2}}{1+u^{2}}}\int_{0}^{1}\mathrm{d}x\,\frac{u}{1-x^{2}u^{2}}\right]\\ &=\frac{\pi}{8}\left[\int_{0}^{1}\mathrm{d}x\,x^{-3/4}\left(1-x\right)^{-1/2}+\int_{0}^{1}\mathrm{d}x\,x^{-1/4}\left(1-x\right)^{-1/2}\right]\\ &~~~~~-\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}x\,(2u)\left[\frac{1}{1+x^{2}u^{2}}\sqrt{\frac{1+u^{2}}{1-u^{2}}}+\frac{1}{1-x^{2}u^{2}}\sqrt{\frac{1-u^{2}}{1+u^{2}}}\right]\\ &=\frac{\pi}{8}\left[\operatorname{B}{\left(\frac14,\frac12\right)}+\operatorname{B}{\left(\frac34,\frac12\right)}\right]\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}u\,(2u)\left[\frac{1}{1+x^{2}u^{2}}\sqrt{\frac{1+u^{2}}{1-u^{2}}}+\frac{1}{1-x^{2}u^{2}}\sqrt{\frac{1-u^{2}}{1+u^{2}}}\right]\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}v\,\left[\frac{1}{1+x^{2}v}\sqrt{\frac{1+v}{1-v}}+\frac{1}{1-x^{2}v}\sqrt{\frac{1-v}{1+v}}\right];~~~\small{\left[u^{2}=v\right]}\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}-\int_{0}^{1}\mathrm{d}x\int_{1}^{0}\mathrm{d}w\,\frac{(-2)}{\left(1+w\right)^{2}}\\ &~~~~~\times\left[\frac{1}{1+x^{2}\left(\frac{1-w}{1+w}\right)}\sqrt{\frac{1}{w}}+\frac{1}{1-x^{2}\left(\frac{1-w}{1+w}\right)}\sqrt{w}\right];~~~\small{\left[v=\frac{1-w}{1+w}\right]}\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}w\,\frac{2}{\left(1+w\right)\sqrt{w}}\\ &~~~~~\times\left[\frac{1}{\left(1+w\right)+x^{2}\left(1-w\right)}+\frac{w}{\left(1+w\right)-x^{2}\left(1-w\right)}\right],\\ \end{align}$$

and then,

$$\begin{align} \mathcal{I} &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}w\,\frac{2}{\left(1+w\right)\sqrt{w}}\\ &~~~~~\times\left[\frac{1}{\left(1+x^{2}\right)+\left(1-x^{2}\right)w}+\frac{w}{\left(1-x^{2}\right)+\left(1+x^{2}\right)w}\right]\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}w\,\frac{1}{\left(1+x^{2}\right)}\left[\frac{1}{1+\left(\frac{1-x^{2}}{1+x^{2}}\right)w}+\frac{w}{\left(\frac{1-x^{2}}{1+x^{2}}\right)+w}\right]\frac{2}{\left(1+w\right)\sqrt{w}}\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}w\,\frac{1}{x^{2}}\left[\frac{2}{1+w}-\frac{\left(\frac{1-x^{2}}{1+x^{2}}\right)}{1+\left(\frac{1-x^{2}}{1+x^{2}}\right)w}-\frac{\left(\frac{1-x^{2}}{1+x^{2}}\right)}{\left(\frac{1-x^{2}}{1+x^{2}}\right)+w}\right]\frac{1}{\sqrt{w}}\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{2}{x^{2}}\left[\frac{2}{1+y^{2}}-\frac{\left(\frac{1-x^{2}}{1+x^{2}}\right)}{1+\left(\frac{1-x^{2}}{1+x^{2}}\right)y^{2}}-\frac{\left(\frac{1-x^{2}}{1+x^{2}}\right)}{\left(\frac{1-x^{2}}{1+x^{2}}\right)+y^{2}}\right];~~~\small{\left[\sqrt{w}=y\right]}\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\,\frac{2}{x^{2}}\left[\int_{0}^{1}\mathrm{d}y\,\frac{2}{1+y^{2}}-\int_{0}^{1}\mathrm{d}y\,\frac{\left(\frac{1-x^{2}}{1+x^{2}}\right)}{1+\left(\frac{1-x^{2}}{1+x^{2}}\right)y^{2}}-\int_{0}^{1}\mathrm{d}y\,\frac{\left(\frac{1-x^{2}}{1+x^{2}}\right)}{\left(\frac{1-x^{2}}{1+x^{2}}\right)+y^{2}}\right]\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\,\frac{2}{x^{2}}\left[\frac{\pi}{2}-\sqrt{\frac{1-x^{2}}{1+x^{2}}}\arctan{\left(\sqrt{\frac{1-x^{2}}{1+x^{2}}}\right)}-\sqrt{\frac{1-x^{2}}{1+x^{2}}}\arctan{\left(\frac{1}{\sqrt{\frac{1-x^{2}}{1+x^{2}}}}\right)}\right]\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{2}{x^{2}}\left[\frac{\pi}{2}-\frac{\pi}{2}\sqrt{\frac{1-x^{2}}{1+x^{2}}}\right]\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}-\pi\int_{0}^{1}\mathrm{d}x\,\frac{1}{x^{2}}\left[1-\sqrt{\frac{1-x^{2}}{1+x^{2}}}\right].\\ \end{align}$$

Evaluating the remaining integral by parts, we find

$$\begin{align} \int_{0}^{1}\mathrm{d}x\,\frac{1}{x^{2}}\left[1-\sqrt{\frac{1-x^{2}}{1+x^{2}}}\right] &=\int_{0}^{1}\mathrm{d}x\,\left[1-\sqrt{\frac{1-x^{2}}{1+x^{2}}}\right]\frac{d}{dx}\left[-\frac{1}{x}\right]\\ &=\left[-\frac{1-\sqrt{\frac{1-x^{2}}{1+x^{2}}}}{x}\right]_{x=0}^{x=1}-\int_{0}^{1}\mathrm{d}x\,\left[-\frac{1}{x}\right]\frac{d}{dx}\left[1-\sqrt{\frac{1-x^{2}}{1+x^{2}}}\right];~~~\small{I.B.P.}\\ &=-1+\int_{0}^{1}\mathrm{d}x\,\frac{1}{x}\cdot\frac{2x}{\left(1+x^{2}\right)^{2}\sqrt{\frac{1-x^{2}}{1+x^{2}}}}\\ &=-1+\int_{0}^{1}\mathrm{d}x\,\frac{2}{\left(1+x^{2}\right)\sqrt{1-x^{4}}}\\ &=-1+\int_{0}^{1}\mathrm{d}x\,\frac{1-2x^{2}-x^{4}}{\left(1+x^{2}\right)\sqrt{1-x^{4}}}+\int_{0}^{1}\mathrm{d}x\,\frac{1+x^{2}}{\sqrt{1-x^{4}}}\\ &=-1+\int_{0}^{1}\mathrm{d}x\,\frac{d}{dx}\left[x\sqrt{\frac{1-x^{2}}{1+x^{2}}}\right]+\int_{0}^{1}\mathrm{d}x\,\frac{1+x^{2}}{\sqrt{1-x^{4}}}\\ &=-1+\int_{0}^{1}\mathrm{d}x\,\frac{1+x^{2}}{\sqrt{1-x^{4}}}\\ &=-1+\frac14\left[\operatorname{B}{\left(\frac14,\frac12\right)}+\operatorname{B}{\left(\frac34,\frac12\right)}\right].\\ \end{align}$$

Hence,

$$\mathcal{I}=\pi-\frac{\pi}{8}\left[\operatorname{B}{\left(\frac14,\frac12\right)}+\operatorname{B}{\left(\frac34,\frac12\right)}\right].\blacksquare$$

---

Answer 2

Here's a solution with a bit contour integration, which is, if you ask me, elemantery.

Firstly, substitute $x=\dfrac\pi2-2\arctan t$ to get $$ \int_{0}^{\pi/2}\frac{\sin{x}\log{(\tan{(x/2))}+x}}{\sqrt{\sin{x}}(\sin{x}+1)}dx\\ =\underbrace{2\int_0^1 \left(\sqrt{\frac{1+t^2}{1-t^2}} \cot ^{-1}(t)-\sqrt{\frac{1-t^2}{1+t^2}} \tanh ^{-1}(t)\right) dt}_{J}-\frac\pi2\underbrace{\int_0^1 \sqrt{\frac{1+t^2}{1-t^2}}dt}_{K} $$ $K$ can be simply done via Beta function $$ K=\int_0^1 (1-t^4)^{-1/2}dt+\int_0^1 t^2(1-t^4)^{-1/2}dt=\frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{2 \pi }}+\frac{\pi ^{3/2} \sqrt{2}}{\Gamma \left(\frac{1}{4}\right)^2} $$ One should notice that this contributes to the last two terms.

To show that $J=\pi$ would be the tricky part. Consider the function $$ f(z)=\frac12\log \left(\frac{z+1}{z-1}\right) \exp \left(\frac{1}{2} \log \left(\frac{z^2-1}{z^2+1}\right)\right) $$ and integrate on the contour illustrated below

The branch cuts are marked red, and $f(z)$ is chosen to be real when $z>1$ is real.

$f(z)$ has only four poles, namely $i^k\quad k=0,1,2,3$. Verify that $$ \lim_{z\to i^k}f(z)=0 $$ Thus, small perturbations can be made to bypass them without changing the integral and only the eight segments matter. Arguments should be treated carefully since multiple poles are involved. All eight integrals are long and similar, so only two is shown here. $$ \int_1 f(z)dz=\int_1^0\frac12\left(\log \left(\frac{1+t}{1-t}\right)-\pi i\right) \exp \left(\frac{1}{2} \log \left(\frac{1-t^2}{1+t^2}\right)+\frac{\pi i}2\right)dt\\ =-i\int_0^1 \sqrt{\frac{1-t^2}{1+t^2}} \tanh ^{-1}(t) dt-\frac\pi2\int_0^1 \sqrt{\frac{1-t^2}{1+t^2}}dt $$

$$ \int_2 f(z)dz=\int_0^1\frac12\left(2i\tan^{-1}(t)-\pi i\right) \exp \left(\frac{1}{2} \log \left(\frac{1+t^2}{1-t^2}\right)-\frac{\pi i}2\right)idt\\ =i\int_0^1 \sqrt{\frac{1+t^2}{1-t^2}} \cot ^{-1}(t) dt $$

... (rest six)

After some simplification one arrives that $$ \oint f(z)dz=4i\int_0^1 \left(\sqrt{\frac{1+t^2}{1-t^2}} \cot ^{-1}(t)-\sqrt{\frac{1-t^2}{1+t^2}} \tanh ^{-1}(t)\right) dt=2iJ $$ Note that $z=\infty$ is a simple pole of $f(z)$, so we can apply the residue at infinity $$ \underset{z=\infty}{\text{Res }}f(z)=-\underset{w=0}{\text{Res }}\frac{f(1/w)}{w^2}=-1 $$ Hence, $$ \oint f(z)=-2\pi i \underset{z=\infty}{\text{ Res }}f(z)=2\pi i\\ J=\pi $$ and we are done.