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I need help finding an analytical expression for the integral $$I(x)=\int_{0}^{\infty}\frac{dy}{1+y^{2}}\log\left(\sqrt{1+y^{2}}+\sqrt{x+y^{2}}\right), $$ where $0<x<1$. The expression can be written using polylog, elliptic function or any other known integral functions...

Erez Urbach
  • 11

2 Answers2

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OK, I'm not super happy with what I have, but I'm posting in case it helps others find a better formula.

$$\boxed{I(x)=\pi\log 2 -\frac 1 2 \int_x^1 \frac{K(\sqrt{1-t})-\frac \pi 2}{1-t}dt=\pi\log 2-\frac {\pi} 4 \sum_{n\geq 1}\left( \frac{(2n)!}{2^{2n}(n!)^2}\right)^2\frac{(1-x)^{n+1}}{n+1}}$$ where we identify the complete elliptic integral of the first kind $$K(k)=\int_0^{\frac \pi 2} \frac{d\theta}{\sqrt{1 -k^2\sin^2\theta}}$$ Thre might be a way to get another form using hypergeometric functions, but I couldn't. Anyway, here's what I did:

$$\begin{split} I^\prime(x)&= \int_0^{+\infty} \frac{dy}{2(y^2+1)\sqrt{x+y^2}\left(\sqrt{x+y^2}+\sqrt{y^2+1}\right)}\\ &= \int_0^{+\infty} \frac{\sqrt{x+y^2}-\sqrt{y^2+1}}{2(y^2+1)\sqrt{x+y^2}(x-1)}dy\\ &=\frac 1 {2(x-1)}\left(\int_0^{+\infty} \frac{1}{y^2+1}dy - \int_0^{+\infty} \frac{dy}{\sqrt{(y^2+1)(x+y^2)}} \right)\\ &= \frac 1 {2(x-1)}\left(\frac \pi 2 - \int_0^{\frac \pi 2} \frac{d\theta}{\sqrt{\cos^2\theta + x\sin^2\theta}} \right) \text { with }\theta=\arctan \frac y {\sqrt x}\\ &= \frac 1 {2(x-1)}\left(\frac \pi 2 - \int_0^{\frac \pi 2} \frac{d\theta}{\sqrt{1 - (1-x)\sin^2\theta}} \right)\\ &= \frac 1 {2(x-1)}\left(\frac \pi 2 - K(\sqrt{1-x})\right) \end{split}$$ Using the power series for $K$, $$K(\sqrt{1-x})=\frac \pi 2 \sum_{n\geq 0}\left( \frac{(2n)!}{2^{2n}(n!)^2}\right)^2(1-x)^n$$ we obtain $$\begin{split} I(x)-I(1)&=\frac {\pi} 4 \sum_{n\geq 1}\left( \frac{(2n)!}{2^{2n}(n!)^2}\right)^2\int_1^x (1-t)^ndt\\ &=-\frac {\pi} 4 \sum_{n\geq 1}\left( \frac{(2n)!}{2^{2n}(n!)^2}\right)^2\frac{(1-x)^{n+1}}{n+1} \end{split}$$ Finally $$\begin{split} I(1)&=\int_0^{+\infty}\frac{\log(2\sqrt{1+y^2})}{1+y^2}dy\\ &=\frac \pi 2 \log 2+\int_0^{+\infty}\frac{\log(\sqrt{1+y^2})}{1+y^2}dy\\ &=\frac \pi 2 \log 2-\int_0^{\frac \pi 2}\log(\cos \theta)d\theta \\ &= \pi \log 2 \end{split}$$ where we have used this result.

Stefan Lafon
  • 13,728
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Define the function $\mathcal{I}:\mathbb{R}_{\ge0}\rightarrow\mathbb{R}$ via the improper integral

$$\mathcal{I}{\left(p\right)}:=\int_{0}^{\infty}\mathrm{d}t\,\frac{\ln{\left(\sqrt{1+t^{2}}+\sqrt{p+t^{2}}\right)}}{1+t^{2}}.$$

We seek a closed-form expression for $\mathcal{I}{\left(p\right)}$ on the interval $p\in(0,1)$.

Suppose $0<p<1$. We begin by splitting up the integral as

$$\begin{align} \mathcal{I}{\left(p\right)} &=\int_{0}^{\infty}\mathrm{d}t\,\frac{\ln{\left(\sqrt{1+t^{2}}+\sqrt{p+t^{2}}\right)}}{1+t^{2}}\\ &=\int_{0}^{\infty}\mathrm{d}t\,\frac{\ln{\left(\sqrt{1+t^{2}}\right)}+\ln{\left(1+\sqrt{\frac{p+t^{2}}{1+t^{2}}}\right)}}{1+t^{2}}\\ &=\int_{0}^{\infty}\mathrm{d}t\,\frac{\ln{\left(\sqrt{1+t^{2}}\right)}}{1+t^{2}}+\int_{0}^{\infty}\mathrm{d}t\,\frac{\ln{\left(1+\sqrt{\frac{p+t^{2}}{1+t^{2}}}\right)}}{1+t^{2}}.\\ \end{align}$$

The first integral in the last line above can be evaluated using the Clausen function:

$$\begin{align} \int_{0}^{\infty}\mathrm{d}t\,\frac{\ln{\left(\sqrt{1+t^{2}}\right)}}{1+t^{2}} &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\tau\,\ln{\left(\sec{\left(\tau\right)}\right)};~~~\small{\left[t=\tan{\left(\tau\right)}\right]}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\tau\,\ln{\left(\sec{\left(\frac{\pi}{2}-\tau\right)}\right)};~~~\small{\left[\tau\mapsto\frac{\pi}{2}-\tau\right]}\\ &=-\int_{0}^{\frac{\pi}{2}}\mathrm{d}\tau\,\ln{\left(\sin{\left(\tau\right)}\right)}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\tau\,\left[\ln{\left(2\right)}-\ln{\left(2\sin{\left(\tau\right)}\right)}\right]\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\tau\,\ln{\left(2\right)}-\int_{0}^{\frac{\pi}{2}}\mathrm{d}\tau\,\ln{\left(2\sin{\left(\tau\right)}\right)}\\ &=\frac{\pi}{2}\ln{\left(2\right)}-\frac12\int_{0}^{\pi}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\varphi}{2}\right)}\right)};~~~\small{\left[\tau=\frac{\varphi}{2}\right]}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\frac12\operatorname{Cl}_{2}{\left(\pi\right)}\\ &=\frac{\pi}{2}\ln{\left(2\right)}.\\ \end{align}$$

Then,

$$\begin{align} \mathcal{I}{\left(p\right)} &=\int_{0}^{\infty}\mathrm{d}t\,\frac{\ln{\left(\sqrt{1+t^{2}}\right)}}{1+t^{2}}+\int_{0}^{\infty}\mathrm{d}t\,\frac{\ln{\left(1+\sqrt{\frac{p+t^{2}}{1+t^{2}}}\right)}}{1+t^{2}}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\int_{0}^{\infty}\mathrm{d}t\,\frac{\ln{\left(1+\sqrt{\frac{p+t^{2}}{1+t^{2}}}\right)}}{1+t^{2}}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\int_{0}^{\infty}\mathrm{d}u\,\frac{\ln{\left(1+\sqrt{\frac{p+u}{1+u}}\right)}}{2\left(1+u\right)\sqrt{u}};~~~\small{\left[t=\sqrt{u}\right]}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\int_{p}^{1}\mathrm{d}v\,\frac{\left(1-p\right)}{\left(1-v\right)^{2}}\cdot\frac{\ln{\left(1+\sqrt{v}\right)}}{2\left(\frac{1-p}{1-v}\right)\sqrt{\frac{v-p}{1-v}}};~~~\small{\left[\frac{p+u}{1+u}=v\right]}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\int_{p}^{1}\mathrm{d}v\,\frac{\ln{\left(1+\sqrt{v}\right)}}{2\sqrt{\left(1-v\right)\left(v-p\right)}}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\frac{\pi}{2}\ln{\left(2\right)}-\int_{p}^{1}\mathrm{d}v\,\frac{\arcsin{\left(\sqrt{\frac{v-p}{1-p}}\right)}}{2\left(1+\sqrt{v}\right)\sqrt{v}};~~~\small{I.B.P.}\\ &=\pi\ln{\left(2\right)}-\int_{p}^{1}\mathrm{d}v\,\frac{\arcsin{\left(\sqrt{\frac{v-p}{1-p}}\right)}}{2\left(1+\sqrt{v}\right)\sqrt{v}}\\ &=\pi\ln{\left(2\right)}-\int_{0}^{1-p}\mathrm{d}w\,\frac{\arcsin{\left(\sqrt{\frac{w}{1-p}}\right)}}{2\left(1+\sqrt{p+w}\right)\sqrt{p+w}};~~~\small{\left[v-p=w\right]}\\ &=\pi\ln{\left(2\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{(1-p)\arcsin{\left(\sqrt{x}\right)}}{2\left[1+\sqrt{p+(1-p)x}\right]\sqrt{p+(1-p)x}};~~~\small{\left[\frac{w}{1-p}=x\right]}\\ &=\pi\ln{\left(2\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{\left[1-\sqrt{p+\left(1-p\right)x}\right]\arcsin{\left(\sqrt{x}\right)}}{2\left(1-x\right)\sqrt{p+\left(1-p\right)x}},\\ \end{align}$$

and then,

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

where in the last line above we've used the following result:

$$\begin{align} \mathcal{J} &=-\frac12\int_{0}^{1}\mathrm{d}y\,\frac{2y}{1-y^{2}}\left[\arcsin{\left(1\right)}-\arcsin{\left(y\right)}\right]\\ &=-\frac12\int_{0}^{1}\mathrm{d}y\,\frac{2y}{1-y^{2}}\int_{y}^{1}\mathrm{d}x\,\frac{1}{\sqrt{1-x^{2}}}\\ &=-\frac12\int_{0}^{1}\mathrm{d}y\int_{y}^{1}\mathrm{d}x\,\frac{2y}{\left(1-y^{2}\right)\sqrt{1-x^{2}}}\\ &=-\frac12\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\frac{2y}{\left(1-y^{2}\right)\sqrt{1-x^{2}}}\\ &=-\frac12\int_{0}^{1}\mathrm{d}x\,\frac{1}{\sqrt{1-x^{2}}}\int_{0}^{x}\mathrm{d}y\,\frac{2y}{1-y^{2}}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x^{2}\right)}}{2\sqrt{1-x^{2}}}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t\right)}}{4\sqrt{t}\sqrt{1-t}};~~~\small{\left[x=\sqrt{t}\right]}\\ &=\frac14\int_{0}^{1}\mathrm{d}u\,\frac{\ln{\left(u\right)}}{\sqrt{u}\sqrt{1-u}};~~~\small{\left[t=1-u\right]}\\ &=-\frac{\pi}{2}\ln{\left(2\right)}.\\ \end{align}$$

To tackle the remaining integral, we'll need to make use of hypergeometric functions.

Given $z<1$, we can derive an elementary expression for the following integral:

$$\begin{align} \int_{0}^{1}\mathrm{d}t\,\frac{1}{(1-zt)\sqrt{t(1-t)}} &=\int_{0}^{1}\mathrm{d}t\,\frac{1}{(1-zt)(1-t)\sqrt{\frac{t}{1-t}}}\\ &=\int_{0}^{\infty}\mathrm{d}u\,\frac{1}{(1+u)\left(1-\frac{zu}{1+u}\right)\sqrt{u}};~~~\small{\left[\frac{t}{1-t}=u\right]}\\ &=\int_{0}^{\infty}\mathrm{d}u\,\frac{1}{\left(1+u-zu\right)\sqrt{u}}\\ &=\int_{0}^{\infty}\mathrm{d}u\,\frac{1}{\left[1+(1-z)u\right]\sqrt{u}}\\ &=\frac{1}{\sqrt{1-z}}\int_{0}^{\infty}\mathrm{d}v\,\frac{1}{(1+v)\sqrt{v}};~~~\small{\left[(1-z)u=v\right]}\\ &=\frac{2}{\sqrt{1-z}}\int_{0}^{\infty}\mathrm{d}w\,\frac{1}{1+w^{2}};~~~\small{\left[\sqrt{v}=w\right]}\\ &=\frac{\pi}{\sqrt{1-z}}.\\ \end{align}$$

For $-1<z<1$, we then have

$$\begin{align} \frac{1}{\sqrt{1-z}} &=\frac{1}{\pi}\int_{0}^{1}\mathrm{d}t\,\frac{1}{(1-zt)\sqrt{t(1-t)}}\\ &=\frac{1}{\pi}\operatorname{B}{\left(\frac12,\frac12\right)}\,{_2F_1}{\left(1,\frac12;1;z\right)}\\ &={_2F_1}{\left(1,\frac12;1;z\right)}\\ &=\sum_{n=0}^{\infty}\frac{\left(\frac12\right)_{n}\,z^{n}}{n!}.\\ \end{align}$$

Defining the function

$$\mathcal{K}{\left(z\right)}:=\int_{0}^{1}\mathrm{d}t\,\frac{\arcsin{\left(\sqrt{t}\right)}}{2t\sqrt{1-zt}};~~~\small{z\le1},$$

we can then show that for $|z|\le1$,

$$\begin{align} \mathcal{K}{\left(z\right)} &=\int_{0}^{1}\mathrm{d}t\,\frac{\arcsin{\left(\sqrt{t}\right)}}{2t\sqrt{1-zt}}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{\arcsin{\left(u\right)}}{u\sqrt{1-zu^{2}}};~~~\small{\left[\sqrt{t}=u\right]}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{\arcsin{\left(u\right)}}{u}\sum_{n=0}^{\infty}\frac{\left(\frac12\right)_{n}\,(zu^{2})^{n}}{n!}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{\arcsin{\left(u\right)}}{u}\left[1+\sum_{n=1}^{\infty}\frac{\left(\frac12\right)_{n}\,z^{n}u^{2n}}{n!}\right]\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{\arcsin{\left(u\right)}}{u}+\int_{0}^{1}\mathrm{d}u\,\frac{\arcsin{\left(u\right)}}{u}\sum_{n=1}^{\infty}\frac{\left(\frac12\right)_{n}\,z^{n}u^{2n}}{n!}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\sum_{n=1}^{\infty}\int_{0}^{1}\mathrm{d}u\,\frac{\left(\frac12\right)_{n}\,z^{n}u^{2n-1}}{n!}\arcsin{\left(u\right)}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\sum_{n=1}^{\infty}\frac{\left(\frac12\right)_{n}\,z^{n}}{n!\,(2n)}\int_{0}^{1}\mathrm{d}u\,(2n)u^{2n-1}\arcsin{\left(u\right)}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\sum_{n=1}^{\infty}\frac{\left(\frac12\right)_{n}\,z^{n}}{n!\,(2n)}\left[\frac{\pi}{2}-\int_{0}^{1}\mathrm{d}u\,\frac{u^{2n}}{\sqrt{1-u^{2}}}\right];~~~\small{I.B.P.}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\sum_{n=1}^{\infty}\frac{\left(\frac12\right)_{n}\,z^{n}}{n!\,(2n)}\left[\frac{\pi}{2}-\int_{0}^{1}\mathrm{d}v\,\frac{v^{n}}{2\sqrt{v}\sqrt{1-v}}\right];~~~\small{\left[u=\sqrt{v}\right]}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\frac14\sum_{n=1}^{\infty}\frac{\left(\frac12\right)_{n}\,z^{n}}{n!\,n}\left[\pi-\int_{0}^{1}\mathrm{d}v\,\frac{v^{n-1/2}}{\sqrt{1-v}}\right]\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\frac14\sum_{n=1}^{\infty}\frac{\left(\frac12\right)_{n}\,z^{n}}{n!\,n}\left[\pi-\operatorname{B}{\left(n+\frac12,\frac12\right)}\right]\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\frac{\pi}{4}\sum_{n=1}^{\infty}\frac{\left(\frac12\right)_{n}\,z^{n}}{n!\,n}-\frac14\sum_{n=1}^{\infty}\frac{\left(\frac12\right)_{n}\,z^{n}}{n!\,n}\operatorname{B}{\left(n+\frac12,\frac12\right)}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\frac{\pi}{4}\sum_{n=1}^{\infty}\frac{\left(\frac12\right)_{n}}{n!}\int_{0}^{z}\mathrm{d}y\,y^{n-1}\\ &~~~~~-\frac14\sum_{n=1}^{\infty}\frac{\left(\frac12\right)_{n}\,z^{n}}{n!\,n}\cdot\frac{\Gamma{\left(n+\frac12\right)}\,\Gamma{\left(\frac12\right)}}{\Gamma{\left(n+1\right)}}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\frac{\pi}{4}\int_{0}^{z}\mathrm{d}y\,\sum_{n=1}^{\infty}\frac{\left(\frac12\right)_{n}\,y^{n-1}}{n!}-\frac{\pi}{4}\sum_{n=1}^{\infty}\frac{\left(\frac12\right)_{n}\,\left(\frac12\right)_{n}\,z^{n}}{\left(1\right)_{n}\,\left(1\right)_{n}\,n}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\frac{\pi}{4}\int_{0}^{z}\mathrm{d}y\,\frac{1}{y}\left[-1+\sum_{n=0}^{\infty}\frac{\left(\frac12\right)_{n}\,y^{n}}{n!}\right]\\ &~~~~~-\frac{\pi}{4}\sum_{n=0}^{\infty}\frac{\left(\frac12\right)_{n+1}\,\left(\frac12\right)_{n+1}\,z^{n+1}}{\left(1\right)_{n+1}\,\left(1\right)_{n+1}\,\left(n+1\right)}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\frac{\pi}{4}\int_{0}^{z}\mathrm{d}y\,\frac{1}{y}\left[-1+\frac{1}{\sqrt{1-y}}\right]\\ &~~~~~-\frac{\pi}{16}z\sum_{n=0}^{\infty}\frac{\left(1\right)_{n}\,\left(1\right)_{n}\,\left(\frac32\right)_{n}\,\left(\frac32\right)_{n}\,z^{n}}{\left(2\right)_{n}\,\left(2\right)_{n}\,\left(2\right)_{n}\,n!}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\frac{\pi}{4}\int_{0}^{z}\mathrm{d}y\,\frac{1-\sqrt{1-y}}{y\sqrt{1-y}}-\frac{\pi}{16}z\,{_4F_3}{\left(1,1,\frac32,\frac32;2,2,2;z\right)}\\ &=\frac{\pi}{2}\ln{\left(2\right)}+\frac{\pi}{2}\left[\ln{\left(2\right)}-\ln{\left(1+\sqrt{1-z}\right)}\right]-\frac{\pi}{16}z\,{_4F_3}{\left(1,1,\frac32,\frac32;2,2,2;z\right)}\\ &=\frac{\pi}{2}\ln{\left(\frac{4}{1+\sqrt{1-z}}\right)}-\frac{\pi}{16}z\,{_4F_3}{\left(1,1,\frac32,\frac32;2,2,2;z\right)}.\\ \end{align}$$

Hence, for $p\in(0,1)$ we finally obtain

$$\begin{align} \mathcal{I}{\left(p\right)} &=\frac{\pi}{2}\ln{\left(1+\sqrt{p}\right)}+\int_{0}^{1}\mathrm{d}y\,\frac{\arcsin{\left(\sqrt{y}\right)}}{2y\sqrt{1-\left(1-p\right)y}}\\ &=\frac{\pi}{2}\ln{\left(1+\sqrt{p}\right)}+\mathcal{K}{\left(1-p\right)}\\ &=\frac{\pi}{2}\ln{\left(1+\sqrt{p}\right)}+\frac{\pi}{2}\ln{\left(\frac{4}{1+\sqrt{p}}\right)}-\frac{\pi}{16}\left(1-p\right)\,{_4F_3}{\left(1,1,\frac32,\frac32;2,2,2;1-p\right)}\\ &=\pi\ln{\left(2\right)}-\frac{\pi}{16}\left(1-p\right)\,{_4F_3}{\left(1,1,\frac32,\frac32;2,2,2;1-p\right)}.\blacksquare\\ \end{align}$$

David H
  • 32,536