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For $a,b\in\mathbb{R}\land0<a\le1\land0\le b$, define $\mathcal{I}{\left(a,b\right)}$ by the integral $$\mathcal{I}{\left(a,b\right)}:=\int_{0}^{a}\frac{\arcsin{\left(2x-1\right)}\,\mathrm{d}x}{\sqrt{\left(a-x\right)\left(b+x\right)}}.\tag{1}$$

The integral $(1)$ above has closed forms in the following special cases:

$$\mathcal{I}{\left(a,0\right)}=4\,\chi_{2}{\left(\sqrt{a}\right)}-\frac{\pi^2}{2};~~~\small{0\le a\le1},\tag{2}$$

$$\mathcal{I}{\left(1,b\right)}=4\,\chi_{2}{\left(\frac{1}{\sqrt{1+b}}\right)}-\pi\operatorname{arccot}{\left(\sqrt{b}\right)};~~~\small{0<b},\tag{3}$$

where $\chi_{2}{\left(z\right)}$ is the Legendre chi function of order 2, which may be defined as

$$\chi_{2}{\left(z\right)}:=\int_{0}^{z}\frac{\operatorname{arctanh}{\left(t\right)}}{t}\,\mathrm{d}t=\frac{\operatorname{Li}_{2}{\left(z\right)}-\operatorname{Li}_{2}{\left(-z\right)}}{2};~~~\small{\left[\left|z\right|\le1\right]}.$$

I was also able to find a closed form for the special case where $b=a$ involving ${_4F_3}$ generalized hypergeometric functions. This leads me to believe that if the integral $\mathcal{I}{(a,b)}$ possesses a closed form at all, it will likely be in terms of hypergeometrics instead of simpler functions like the standard polylogarithms and elliptic integrals.

Question: Can $(1)$ be evaluated in terms of familiar special functions in the general case where $0<a<1\land0<b$? If not, can we at least find a nice hypergeometric function representation?


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  • $\begingroup$ I think that your methodology is a little above the regulars who generally answer these sorts of questions. There's really only a handful of people on this site that could hope to answer this question. For instance the square root in the denominator means that complex analysis would be superficial at best. If you haven't tried series expansion, I'd go for it at this point. At worst it'll illuminate some of the difficulty inherent to this problem. Best of luck! :) $\endgroup$ – Zach466920 Jun 11 '15 at 23:31
  • $\begingroup$ Just out of interest, have you solved the special case $a=b$ already? It looks not too bad after all. $\endgroup$ – tired Jun 15 '15 at 14:29
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For $0<a<1\land0<b$, $$\small{\mathcal{I}{\left(a,b\right)}=-\pi\arcsin{\left(\sqrt{\frac{a}{a+b}}\right)}+\frac{\pi a}{\sqrt{a+b}}F_{3}{\left(\frac12,\frac12;\frac12,\frac12;2;a,\frac{a}{a+b}\right)}.}$$


Proof:

The inverse sine function may be defined on the complex unit circle via the integral representation

$$\arcsin{\left(z\right)}:=\int_{0}^{z}\frac{\mathrm{d}t}{\sqrt{1-t^{2}}};~~~\small{z\in\mathbb{C}\land\left|z\right|\le1}.$$

Using the integral representation of $\arcsin{\left(z\right)}$ above, we may express $\mathcal{I}{\left(a,b\right)}$ equivalently as a double integral with an algebraic integrand. It has come to my attention that the resulting double integral has a natural expression an Appell hypergeometric function of the third kind:

$$F_{3}{\left(\alpha,\alpha^{\prime};\beta,\beta^{\prime};\gamma;z,w\right)}=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\alpha^{\prime}\right)_{k}\left(\beta\right)_{n}\left(\beta^{\prime}\right)_{k}}{\left(\gamma\right)_{n+k}n!\,k!}z^{n}w^{k},$$

where $\left|z\right|<1\land\left|w\right|<1$.

An integral representation for $F_{3}$ is derived in the appendix below.

Now, assuming $0<a<1\land0<b$, we find

$$\begin{align} \mathcal{I}{\left(a,b\right)} &=\int_{0}^{a}\frac{\arcsin{\left(2t-1\right)}}{\sqrt{\left(a-t\right)\left(b+t\right)}}\,\mathrm{d}t\\ &=\int_{0}^{a}\mathrm{d}t\,\frac{1}{\sqrt{\left(a-t\right)\left(b+t\right)}}\int_{\frac12}^{t}\mathrm{d}u\,\frac{1}{\sqrt{u\left(1-u\right)}}\\ &=-\int_{0}^{a}\mathrm{d}t\,\frac{1}{\sqrt{\left(a-t\right)\left(b+t\right)}}\int_{0}^{\frac12}\mathrm{d}u\,\frac{1}{\sqrt{u\left(1-u\right)}}\\ &~~~~~+\int_{0}^{a}\mathrm{d}t\,\frac{1}{\sqrt{\left(a-t\right)\left(b+t\right)}}\int_{0}^{t}\mathrm{d}u\,\frac{1}{\sqrt{u\left(1-u\right)}}\\ &=-\frac{\pi}{2}\int_{0}^{a}\frac{\mathrm{d}t}{\sqrt{\left(a-t\right)\left(b+t\right)}}\\ &~~~~~+\int_{0}^{a}\mathrm{d}t\,\frac{1}{\sqrt{\left(a-t\right)\left(b+t\right)}}\int_{0}^{t}\mathrm{d}u\,\frac{1}{\sqrt{u\left(1-u\right)}}\\ &=-\frac{\pi}{2}\int_{0}^{a}\frac{\mathrm{d}u}{\sqrt{u\left(a+b-u\right)}};~~~\small{\left[a-t=u\right]}\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{a}{\sqrt{ax\left(a+b-ax\right)}}\int_{0}^{a-ax}\mathrm{d}u\,\frac{1}{\sqrt{u\left(1-u\right)}};~~~\small{\left[\frac{a-t}{a}=x\right]}\\ &=-\frac{\pi}{2}\int_{0}^{\frac{a}{a+b}}\frac{\mathrm{d}v}{\sqrt{v\left(1-v\right)}};~~~\small{\left[u=\left(a+b\right)v\right]}\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{a}{\sqrt{ax\left(a+b-ax\right)}}\int_{0}^{1-x}\mathrm{d}y\,\frac{a}{\sqrt{ay\left(1-ay\right)}};~~~\small{\left[u=ay\right]}\\ &=-\pi\int_{0}^{\sqrt{\frac{a}{a+b}}}\frac{\mathrm{d}w}{\sqrt{1-w^2}};~~~\small{\left[\sqrt{v}=w\right]}\\ &~~~~~+\frac{a}{\sqrt{a+b}}\int_{0}^{1}\mathrm{d}x\int_{0}^{1-x}\mathrm{d}y\,\frac{1}{\sqrt{yx\left(1-ay\right)\left(1-\frac{a}{a+b}x\right)}}\\ &=-\pi\arcsin{\left(\sqrt{\frac{a}{a+b}}\right)}+\frac{\pi a}{\sqrt{a+b}}F_{3}{\left(\frac12,\frac12;\frac12,\frac12;2;a,\frac{a}{a+b}\right)}.\\ \end{align}$$


Appendix:

Given the parameter assumptions

$$\small{z,w\in\mathbb{D}\land\alpha,\alpha^{\prime},\beta,\beta^{\prime},\gamma\in\mathbb{C}\setminus\mathbb{Z}^{\le0}\land0<\Re{\left(\beta\right)}\land0<\Re{\left(\beta^{\prime}\right)}\land0<\Re{\left(\gamma-\beta-\beta^{\prime}\right)}}$$

an integral representation for the Appell hypergeometric function of the third kind may be derived as follows:

$$\begin{align} F_{3} &=F_{3}{\left(\alpha,\alpha^{\prime};\beta,\beta^{\prime};\gamma;z,w\right)}\\ &=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\alpha^{\prime}\right)_{k}\left(\beta\right)_{n}\left(\beta^{\prime}\right)_{k}}{\left(\gamma\right)_{n+k}n!\,k!}z^{n}w^{k}\\ &=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\alpha^{\prime}\right)_{k}\left(\beta\right)_{n}\left(\beta^{\prime}\right)_{k}}{\left(\gamma\right)_{n}\left(\gamma+n\right)_{k}n!\,k!}z^{n}w^{k}\\ &=\sum_{n=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\beta\right)_{n}z^{n}}{\left(\gamma\right)_{n}n!}\sum_{k=0}^{\infty}\frac{\left(\alpha^{\prime}\right)_{k}\left(\beta^{\prime}\right)_{k}w^{k}}{\left(\gamma+n\right)_{k}\,k!}\\ &=\sum_{n=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\beta\right)_{n}z^{n}}{\left(\gamma\right)_{n}n!}\,{_2F_1}{\left(\alpha^{\prime},\beta^{\prime};\gamma+n;w\right)}\\ &=\small{\sum_{n=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\beta\right)_{n}z^{n}}{\left(\gamma\right)_{n}n!}\cdot\frac{1}{\operatorname{B}{\left(\beta^{\prime},\gamma+n-\beta^{\prime}\right)}}\int_{0}^{1}\frac{t^{\beta^{\prime}-1}\left(1-t\right)^{\gamma+n-\beta^{\prime}-1}}{\left(1-wt\right)^{\alpha^{\prime}}}\,\mathrm{d}t}\\ &=\small{\sum_{n=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\beta\right)_{n}z^{n}}{\left(\gamma\right)_{n}n!}\cdot\frac{\Gamma{\left(\gamma+n\right)}}{\Gamma{\left(\beta^{\prime}\right)}\,\Gamma{\left(\gamma+n-\beta^{\prime}\right)}}\int_{0}^{1}\frac{t^{\beta^{\prime}-1}\left(1-t\right)^{\gamma+n-\beta^{\prime}-1}}{\left(1-wt\right)^{\alpha^{\prime}}}\,\mathrm{d}t}\\ &=\small{\frac{\Gamma{\left(\gamma\right)}}{\Gamma{\left(\beta^{\prime}\right)}\,\Gamma{\left(\gamma-\beta^{\prime}\right)}}\sum_{n=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\beta\right)_{n}z^{n}}{\left(\gamma-\beta^{\prime}\right)_{n}n!}\int_{0}^{1}\frac{t^{\beta^{\prime}-1}\left(1-t\right)^{\gamma+n-\beta^{\prime}-1}}{\left(1-wt\right)^{\alpha^{\prime}}}\,\mathrm{d}t}\\ &=\small{\frac{1}{\operatorname{B}{\left(\beta^{\prime},\gamma-\beta^{\prime}\right)}}\int_{0}^{1}\mathrm{d}t\,\frac{t^{\beta^{\prime}-1}\left(1-t\right)^{\gamma-\beta^{\prime}-1}}{\left(1-wt\right)^{\alpha^{\prime}}}\sum_{n=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\beta\right)_{n}z^{n}\left(1-t\right)^{n}}{\left(\gamma-\beta^{\prime}\right)_{n}n!}}\\ &=\small{\frac{1}{\operatorname{B}{\left(\beta^{\prime},\gamma-\beta^{\prime}\right)}}\int_{0}^{1}\mathrm{d}t\,\frac{t^{\beta^{\prime}-1}\left(1-t\right)^{\gamma-\beta^{\prime}-1}}{\left(1-wt\right)^{\alpha^{\prime}}}{_2F_1}{\left(\alpha,\beta;\gamma-\beta^{\prime};z\left(1-t\right)\right)}}\\ &=\frac{1}{\operatorname{B}{\left(\beta^{\prime},\gamma-\beta^{\prime}\right)}\cdot\operatorname{B}{\left(\beta,\gamma-\beta-\beta^{\prime}\right)}}\times\\ &~~~~~\int_{0}^{1}\mathrm{d}t\,\frac{t^{\beta^{\prime}-1}\left(1-t\right)^{\gamma-\beta^{\prime}-1}}{\left(1-wt\right)^{\alpha^{\prime}}}\int_{0}^{1}\mathrm{d}u\,\frac{u^{\beta-1}\left(1-u\right)^{\gamma-\beta-\beta^{\prime}-1}}{\left(1-z\left(1-t\right)u\right)^{\alpha}}\\ &=\frac{\Gamma{\left(\gamma\right)}}{\Gamma{\left(\beta\right)}\,\Gamma{\left(\beta^{\prime}\right)}\,\Gamma{\left(\gamma-\beta-\beta^{\prime}\right)}}\times\\ &~~~~~\small{\int_{0}^{1}\mathrm{d}t\,\frac{t^{\beta^{\prime}-1}}{\left(1-wt\right)^{\alpha^{\prime}}}\int_{0}^{1-t}\mathrm{d}y\,\frac{y^{\beta-1}\left(1-t-y\right)^{\gamma-\beta-\beta^{\prime}-1}}{\left(1-zy\right)^{\alpha}}};~~~\small{\left[\left(1-t\right)u=y\right]}\\ &=\frac{\Gamma{\left(\gamma\right)}}{\Gamma{\left(\beta\right)}\,\Gamma{\left(\beta^{\prime}\right)}\,\Gamma{\left(\gamma-\beta-\beta^{\prime}\right)}}\times\\ &~~~~~\int_{0}^{1}\mathrm{d}x\int_{0}^{1-x}\mathrm{d}y\,\frac{x^{\beta^{\prime}-1}y^{\beta-1}\left(1-x-y\right)^{\gamma-\beta-\beta^{\prime}-1}}{\left(1-wx\right)^{\alpha^{\prime}}\left(1-zy\right)^{\alpha}}.\blacksquare\\ \end{align}$$


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Again, this is not going to be a full answer but since this problem is very similar to A tough series related with a hypergeometric function with quarter integer parameters I would like to sketch my approach. As usual in these kind of questions the tactic is to squeeze out certain parts that are given through elementary functions of Euler sums and then classify the remaining part and leave it un-evaluated for the time being. Denote $b_0=-b$. Then we have: \begin{eqnarray} &&-\imath {\mathcal I}(a,b_0)= \int\limits_0^a \frac{\arcsin(2x-1)}{\sqrt{(a-x)(b-x)}} dx=\\ &&\mbox{arcsin}(1-2 a) \log (b-a)-\pi \log \left(\sqrt{a}+\sqrt{b}\right)+2 \int\limits_0^a \frac{\log\left[\sqrt{a-x}+\sqrt{b-x}\right]}{\sqrt{(1-x)x}}dx=\\ &&\mbox{arcsin}(1-2 a) \log (b-a)-\pi \log \left(\sqrt{a}+\sqrt{b}\right)+2 \int\limits_0^a \frac{1/2\log\left[a-x\right]+\log\left[1+\sqrt{\frac{b-x}{a-x}}\right]}{\sqrt{(1-x)x}}dx=\\ &&\mbox{arcsin}(1-2 a) \log (b-a)-\pi \log \left(\sqrt{a}+\sqrt{b}\right)+2 \log(a) \mbox{arcsin}(\sqrt{a})+2 \sqrt{a} {\mathcal J}(a)+\\ &&4(a-b) \int\limits_0^{\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}} \frac{(1-u)u}{1-2 u}\cdot \frac{\log(u)}{\sqrt{\left(a (u-1)^2-b u^2\right) \left(u^2 (b-a)+(2 a-2) u-a+1\right)}}du \end{eqnarray} In the top line we took out the minus out of the square root in the denominator. In the second line we integrated by parts and in the third line we factored the expression in the log appropriately. In the fourth line we split the remaining integral into two parts and we expressed the first part via the following function: \begin{eqnarray} {\mathcal J}(a)&:=& -\sum\limits_{l=0}^\infty a^l \binom{l-1/2}{l} \frac{H_{1/2+l}}{2l+1} \\ &=& \int\limits_0^1 \frac{\log[1-x^2]}{\sqrt{1-a x^2}} dx\\ &=&-\frac{1}{\sqrt{a}} \sum\limits_{\xi=\pm} \int\limits_{\pi/2}^{\arccos(\sqrt{a})} \log(1+\frac{\xi}{\sqrt{a}} \cos(\theta)) d\theta\\ &=&\frac{\imath}{6 \sqrt{a}} \left[ \pi^2+3 \imath \pi \log(4 a) - 6 \arccos(\sqrt{a})\left( \pi-\arccos(\sqrt{a})+\imath \log(4 a)\right)+ 6 \sum\limits_{\xi_1=\pm,\xi_2=\pm} Li_2(\xi_1 \sqrt{1-a}+\imath \xi_1 \sqrt{a}) -6 \sum\limits_{\xi_1=\pm} Li_2(\xi_1(1-2 a-2 \imath \sqrt{a(1-a)})) \right] \end{eqnarray} and in the second part we substituted for $u:= (1+\sqrt{(b-x)/(a-x)})^{-1}$. We found the closed form for the function ${\mathcal J}(a)$ above using Anti-derivative of a function containing a log and a sine. .

Now the last integral is clearly very similar to the one in my answer to A tough series related with a hypergeometric function with quarter integer parameters. The integrans is a product of a rational function , a log and an inverse square root of a fourth order polynomial. Now, if there was no logarithm in the integrand it could have been always reduced to elliptic functions by following the recipe from here http://mathworld.wolfram.com/EllipticIntegral.html . However the logarithm makes thinks harder. Yet still maybe the methods for the Wolfram's site above make give an incling how to tackle such integrals.

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  • $\begingroup$ Thank you for taking an interest in my question. I've come to have tremendous respect for your abilities and follow your posts with great interest! $\endgroup$ – David H Oct 1 '17 at 7:14
  • $\begingroup$ @DavidH : Thank you for the nice words. I just like mathematics I practice a lot. As for integration I have the impression that a large class of integrals you can come up with reduces to very few cases . The case you quote is non-trivial because the remaining integral certainly cannot be expressed by elementary functions only and on the face of it it seems to be going beyond the realm of elliptic functions(??). Again, I would love to master calculating such integrals because from my experiences that stuff pops up everywhere in analysis. $\endgroup$ – Przemo Oct 4 '17 at 12:44

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