how to find closed form for $\int_0^{\frac{\pi}{2}} \ln(1+\sqrt{\sin x}) dx$?

I tried to solve the integral $$I=\int_0^{\frac{\pi}{2}} \ln(1+\sqrt{\sin x}) dx$$ and by series and beta function I got that $$I=2\sqrt{\pi}\sum_{k=0}^\infty \frac{(-1)^k}{k+1} \frac{\Gamma\left(\frac{k+3}{4}\right)}{\Gamma\left(\frac{k+5}{4}\right)}$$ then by using the rule $$\sum_{k=0}^\infty f(k)=\sum_{k=0}^\infty \left(f(4k)+f(4k+1)+f(4k+2)+f(4k+3)\right)$$ I got $$I=\frac{8\pi\sqrt{2\pi}}{\Gamma^2\left(\frac{1}{4} \right)} {}_3F_2 \left(\frac{1}{4},\frac{3}{4},1;\frac{5}{4},\frac{5}{4};1 \right)+\frac{\sqrt{2}\Gamma^2\left(\frac{1}{4} \right)}{9\sqrt{\pi}} {}_3F_2 \left(\frac{3}{4},\frac{5}{4},1;\frac{7}{4},\frac{7}{4};1 \right)-4C-\pi ln2$$ now since this two hypergeometric function is in the form $${}_3F_2(a,b,c;d,e;1)$$ I think we can get its closed form . I searched here and didn't find any formula help me to find this closed form

So is there a closed form for $$I$$ ?

• Maybe it helps that another form is $\int_0^1\frac{\cos^{-1}(x^2)}{x+1}dx$ Commented Jul 29 at 21:14
• I think that the first summation is exactly four times too large. But this does not change the problem. Using series, I had $$I=2 \sqrt \pi \sum_{n=1}^\infty (-1)^{n+1}\,\frac{\Gamma \left(\frac{n+2}{4}\right)}{n^2 \, \Gamma \left(\frac{n}{4}\right)}$$ Commented Jul 30 at 6:21
• I wonder if this form helps: $\frac{2}{\pi}\int_{0}^{\pi/2}(\sin x)^{2n}\,dx=\frac{1}{4^n}\binom{2n}{n}$ Commented Jul 30 at 7:04
• @whatamidoing. The exponent is $\frac n 2$ and not $2n$ Commented Jul 30 at 7:18
• Starting with a tangent half-angle sub leads to$$I=\frac{3\pi}2\log2+2G-4\int_0^1\frac{\arctan x}{x\sqrt{1-x^4}}\,dx$$Replacing $\arctan$ with a definite integral and changing the order over variables, we can rewrite the remaining term as an integral of an elliptic integral: $$\int_0^1\frac{\arctan x}{x\sqrt{1-x^4}}\,dx=\int_{x=0}^1\int_{y=0}^1\frac{dy\,dx}{\left(1+x^2y^2\right)\sqrt{1-x^4}}\\=\int_0^1\Pi\left(-y^2,i\right)\,dy$$ dlmf.nist.gov/19.13 lists several references that may have more to say about this Commented Jul 31 at 19:06

Let $$\mathcal{I}$$ denote the value of the following trigonometric integral:

$$\mathcal{I}:=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\ln{\left(1+\sqrt{\sin{\left(\theta\right)}}\right)}\approx0.874888.$$

Begin by rewriting the integral and splitting it up as

\begin{align} \mathcal{I} &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\ln{\left(1+\sqrt{\sin{\left(\theta\right)}}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+\sqrt{x}\right)}}{\sqrt{1-x^{2}}};~~~\small{\left[\theta=\arcsin{\left(x\right)}\right]}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{2y\ln{\left(1+y\right)}}{\sqrt{1-y^{4}}};~~~\small{\left[\sqrt{x}=y\right]}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{2y}{\sqrt{1-y^{4}}}\int_{0}^{y}\mathrm{d}x\,\frac{1}{1+x}\\ &=\int_{0}^{1}\mathrm{d}y\int_{0}^{y}\mathrm{d}x\,\frac{1}{1+x}\cdot\frac{2y}{\sqrt{1-y^{4}}}\\ &=\int_{0}^{1}\mathrm{d}x\int_{x}^{1}\mathrm{d}y\,\frac{1}{1+x}\cdot\frac{2y}{\sqrt{1-y^{4}}}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{1}{1+x}\int_{x}^{1}\mathrm{d}y\,\frac{2y}{\sqrt{1-y^{4}}}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\arcsin{\left(1\right)}-\arcsin{\left(x^{2}\right)}}{1+x}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{(1-x)\left[\arcsin{\left(1\right)}-\arcsin{\left(x^{2}\right)}\right]}{1-x^{2}}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\arcsin{\left(1\right)}-\arcsin{\left(x^{2}\right)}}{1-x^{2}}-\int_{0}^{1}\mathrm{d}x\,\frac{x\left[\arcsin{\left(1\right)}-\arcsin{\left(x^{2}\right)}\right]}{1-x^{2}}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\arcsin{\left(1\right)}-\arcsin{\left(x^{2}\right)}}{1-x^{2}}-\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\arcsin{\left(1\right)}-\arcsin{\left(y\right)}}{1-y};~~~\small{\left[x^{2}=y\right]}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\arccos{\left(x^{2}\right)}}{1-x^{2}}-\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\arccos{\left(y\right)}}{1-y}.\\ \end{align}

Integrating by parts, the first remaining integral for $$\mathcal{I}$$ can be rewritten as

\begin{align} \int_{0}^{1}\mathrm{d}x\,\frac{\arccos{\left(x^{2}\right)}}{1-x^{2}} &=\int_{0}^{1}\mathrm{d}x\,\arccos{\left(x^{2}\right)}\frac{d}{dx}\operatorname{artanh}{\left(x\right)}\\ &=\operatorname{artanh}{\left(x\right)}\arccos{\left(x^{2}\right)}\bigg{|}_{0}^{1}-\int_{0}^{1}\mathrm{d}x\,\operatorname{artanh}{\left(x\right)}\frac{d}{dx}\arccos{\left(x^{2}\right)}\\ &=\lim_{x\to1}\operatorname{artanh}{\left(x\right)}\arccos{\left(x^{2}\right)}-0-\int_{0}^{1}\mathrm{d}x\,\operatorname{artanh}{\left(x\right)}\left[-\frac{2x}{\sqrt{1-x^{4}}}\right]\\ &=\left[\lim_{x\to1}\sqrt{1-x^{2}}\operatorname{artanh}{\left(x\right)}\right]\left[\lim_{x\to1}\frac{\arccos{\left(x^{2}\right)}}{\sqrt{1-x^{2}}}\right]+\int_{0}^{1}\mathrm{d}x\,\frac{2x\operatorname{artanh}{\left(x\right)}}{\sqrt{1-x^{4}}}\\ &=0\cdot\sqrt{2}+\int_{0}^{1}\mathrm{d}x\,\frac{2x\operatorname{artanh}{\left(x\right)}}{\sqrt{1-x^{4}}}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2x\operatorname{artanh}{\left(x\right)}}{\sqrt{1-x^{4}}}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\operatorname{artanh}{\left(\sqrt{t}\right)}}{\sqrt{1-t^{2}}};~~~\small{\left[x=\sqrt{t}\right]}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\sqrt{t}}{\sqrt{1-t^{2}}}\cdot\frac{\operatorname{artanh}{\left(\sqrt{t}\right)}}{\sqrt{t}}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\sqrt{t}}{\sqrt{1-t^{2}}}\,{_2F_1}{\left(\frac12,1;\frac32;t\right)}.\\ \end{align}

This integral can be evaluated using integration formula 7.512(9) from Gradshteyn & Ryzhik, which states

\begin{align} \int_{0}^{1}\mathrm{d}t\,\frac{t^{\gamma-1}\left(1-t\right)^{\rho-1}}{\left(1-zt\right)^{\sigma}}\,{_2F_1}{\left(\alpha,\beta;\gamma;t\right)} &=\frac{\Gamma{\left(\gamma\right)}\,\Gamma{\left(\rho\right)}\,\Gamma{\left(\gamma+\rho-\alpha-\beta\right)}}{\left(1-z\right)^{\sigma}\,\Gamma{\left(\gamma+\rho-\alpha\right)}\,\Gamma{\left(\gamma+\rho-\beta\right)}}\\ &~~~~~\times{_3F_2}{\left(\rho,\sigma,\gamma+\rho-\alpha-\beta;\gamma+\rho-\alpha,\gamma+\rho-\beta;\frac{z}{z-1}\right)},\\ \end{align}

where $$\alpha>0\land\gamma>\beta>0\land\rho>0\land\gamma+\rho-\alpha-\beta>0\land z<1$$.

With the appropriate choice of parameters, we then have

$$\int_{0}^{1}\mathrm{d}t\,\frac{\sqrt{t}}{\sqrt{1-t^{2}}}\,{_2F_1}{\left(\frac12,1;\frac32;t\right)}=\frac{\pi}{\sqrt{2}}\,{_3F_2}{\left(\frac12,\frac12,\frac12;\frac32,1;\frac12\right)}.$$

Recalling the inverse trigonometric identity

$$\arccos{\left(1-2x^{2}\right)}=2\arcsin{\left(x\right)};~~~\small{0\le x\le1},$$

the other remaining integral needed for $$\mathcal{I}$$ can be rewritten as the arcsine integral

\begin{align} \frac12\int_{0}^{1}\mathrm{d}y\,\frac{\arccos{\left(y\right)}}{1-y} &=\frac12\int_{0}^{\frac12}\mathrm{d}u\,\frac{\arccos{\left(1-2u\right)}}{u};~~~\small{\left[y=1-2u\right]}\\ &=\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\arccos{\left(1-2x^{2}\right)}}{x};~~~\small{\left[u=x^{2}\right]}\\ &=2\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\arcsin{\left(x\right)}}{x}.\\ \end{align}

In general, the arcsine integral can be expressed (see appendix) in terms of the Clausen function $$\operatorname{Cl}_{2}$$ as

$$\int_{0}^{a}\mathrm{d}x\,\frac{\arcsin{\left(x\right)}}{x}=\ln{\left(2|a|\right)}\arcsin{\left(a\right)}+\frac12\operatorname{Cl}_{2}{\left(2\arcsin{\left(a\right)}\right)};~~~\small{a\in[-1,1]},$$

where

$$\operatorname{Cl}_{2}{\left(\theta\right)}:=-\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|};~~~\small{\theta\in\mathbb{R}}.$$

We then have

\begin{align} 2\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\,\frac{\arcsin{\left(x\right)}}{x} &=\ln{\left(2\right)}\arcsin{\left(\frac{1}{\sqrt{2}}\right)}+\operatorname{Cl}_{2}{\left(2\arcsin{\left(\frac{1}{\sqrt{2}}\right)}\right)}\\ &=\frac{\pi}{4}\ln{\left(2\right)}+\operatorname{Cl}_{2}{\left(\frac{\pi}{2}\right)}\\ &=\frac{\pi}{4}\ln{\left(2\right)}+C,\\ \end{align}

where here $$C$$ denotes the Catalan constant.

Hence,

$$\mathcal{I}=\frac{\pi}{\sqrt{2}}\,{_3F_2}{\left(\frac12,\frac12,\frac12;\frac32,1;\frac12\right)}-\frac{\pi}{4}\ln{\left(2\right)}-C.\blacksquare$$

Appendix.

Given $$a\in(0,1]$$, and setting $$\alpha:=\arcsin{\left(a\right)}\in\left(0,\frac{\pi}{2}\right]$$, we can show that

\begin{align} \int_{0}^{a}\mathrm{d}x\,\frac{\arcsin{\left(x\right)}}{x} &=\int_{0}^{\arcsin{\left(a\right)}}\mathrm{d}\omega\,\frac{\omega\cos{\left(\omega\right)}}{\sin{\left(\omega\right)}};~~~\small{\left[\arcsin{\left(x\right)}=\omega\right]}\\ &=\int_{0}^{\alpha}\mathrm{d}\omega\,\omega\cot{\left(\omega\right)}\\ &=\int_{0}^{\alpha}\mathrm{d}\omega\int_{0}^{\omega}\mathrm{d}\tau\,\cot{\left(\omega\right)}\\ &=\int_{0}^{\alpha}\mathrm{d}\tau\int_{\tau}^{\alpha}\mathrm{d}\omega\,\cot{\left(\omega\right)}\\ &=\int_{0}^{\alpha}\mathrm{d}\tau\int_{\tau}^{\alpha}\mathrm{d}\omega\,\frac{d}{d\omega}\ln{\left(\sin{\left(\omega\right)}\right)}\\ &=\int_{0}^{\alpha}\mathrm{d}\tau\,\left[\ln{\left(\sin{\left(\alpha\right)}\right)}-\ln{\left(\sin{\left(\tau\right)}\right)}\right]\\ &=\int_{0}^{\alpha}\mathrm{d}\tau\,\left[\ln{\left(2\sin{\left(\alpha\right)}\right)}-\ln{\left(2\sin{\left(\tau\right)}\right)}\right]\\ &=\int_{0}^{\alpha}\mathrm{d}\tau\,\ln{\left(2\sin{\left(\alpha\right)}\right)}-\int_{0}^{\alpha}\mathrm{d}\tau\,\ln{\left(2\sin{\left(\tau\right)}\right)}\\ &=\alpha\ln{\left(2\sin{\left(\alpha\right)}\right)}-\int_{0}^{\alpha}\mathrm{d}\tau\,\ln{\left(2\sin{\left(\tau\right)}\right)}\\ &=\alpha\ln{\left(2\sin{\left(\alpha\right)}\right)}-\frac12\int_{0}^{2\alpha}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\varphi}{2}\right)}\right)};~~~\small{\left[\tau=\frac{\varphi}{2}\right]}\\ &=\alpha\ln{\left(2\sin{\left(\alpha\right)}\right)}+\frac12\operatorname{Cl}_{2}{\left(2\alpha\right)}\\ &=\ln{\left(2a\right)}\arcsin{\left(a\right)}+\frac12\operatorname{Cl}_{2}{\left(2\arcsin{\left(a\right)}\right)}.\\ \end{align}

• good solution but isn't $3F2(1)$ more simplify from $3F2(1/2)$ ? Commented Aug 1 at 12:04
• Very nice solution Commented Aug 1 at 12:05
• @Faoler The parameters tend to be more important than the argument in determining how complicated the hypergeometric function is. What’s simpler: $\arctan{(\frac12)}$ or the elliptic integral $E{(1)}$? Commented Aug 1 at 12:39
• of course $E(1)$ is more complicate Commented Aug 1 at 20:13

This is not an answer since it leads to the same problem.

Using what @Тyma Gaidash suggested in comments $$I=\int_0^{\frac{\pi}{2}} \log\bigg(1+\sqrt{\sin (x)}\bigg)\, dx=\int_0^1\frac{\cos^{-1}(x^2)}{x+1}\,dx$$ $$\frac{\cos^{-1}(x^2)}{x+1}=\frac \pi{2(x+1)}-\sum_{n=0}^\infty \frac{ (2 n)!}{4^n \,(2 n+1)\, (n!)^2}\,\frac{x^{4 n+2}}{x+1}$$ Using $$\int_0^1\frac{x^{4 n+2}}{x+1}\,dx=\frac{1}{2} \left(H_{2 n+1}-H_{2 n+\frac{1}{2}}\right)$$ Mathematicaleads to $$I=-\left(C+\frac{1}{4} \pi \log(2)\right)+\frac{25 \pi ^{3/2}}{64 \sqrt{2} \Gamma \left(\frac{9}{4}\right)^2}\, _3F_2\left(\frac{1}{4},\frac{3}{4},1 ;\frac{5}{4},\frac{5}{4};1\right)+$$ $$\frac{\sqrt{\pi } \Gamma\left(\frac{5}{4}\right)}{6 \Gamma\left(\frac{7}{4}\right)}\,_3F_2\left(\frac{3}{4},1,\frac{5}{4};\frac{7}{4},\frac{7}{4};1\right)$$ which could be further simplified but we stay with the same problem (what are the last two terms ?)