# Is there a closed form for $\int_a^b\frac{{\rm arccosh}x}{\sqrt{(x-a)(b-x)}}$?

I have solved $$\int^3_1\dfrac{{\rm arccosh}x}{\sqrt{(x-1)(3-x)}}{\rm d}x=4G$$where $$G$$ is Catalan constant, by rewriting it into double integral $$\iint_{[0,\pi]\times[0,\pi]}\ln(2-\cos x-\cos y){\rm d}x{\rm d}y$$ and computing it using Leibniz integral rule.

Now my question here is:

Is there a closed form for integral $$\int_a^b\dfrac{{\rm arccosh}x}{\sqrt{(x-a)(b-x)}}{\rm d}x$$ as $$b>a>0$$? Or it is just a coincidence?

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• Your question is interesting, I would like to work on it, but I don't have time. I tried partial integration and substitution, it looks like it could work with lots of transformations. – stocha Aug 14 at 8:02
• Can you show me how did you write to arrive at this double integral: $\iint_{[0,\pi]\times[0,\pi]}\ln(2-\cos x-\cos y){\rm d}x{\rm d}y$? Btw, we have: $$\int_a^b\dfrac{{\rm arccosh}x}{\sqrt{(x-a)(b-x)}}{\rm d}x=2\int_0^1 \frac{\operatorname{arccosh}(b-(b-a)x^2)}{\sqrt{1-x^2}}dx$$ – Zacky Aug 14 at 8:25
• substitute $x=2-\cos t$ and use $\int_0^{\pi}\ln(a-\cos y)dy=\pi {\rm arccosh}a-\pi\ln 2$ – Alice Aug 14 at 9:26
• Note that if ${\rm arccosh}\left(x\right)$ is supposed to be a real function, then its domain needs to be restricted to $x\in\mathbb{R}: x\ge1$. For your integral, this would mean changing the conditions on $a$ and $b$ to $b>a\ge1$. – David H Aug 16 at 0:41

In order to give some insight I will show a derivation for $$a=1$$, where a closed form in terms of the Inverse Tangent Integral is: $$\bbox[10pt, border:2px, lightblue]{\int_1^s \frac{\operatorname{arccosh}x}{\sqrt{(x-1)(s-x)}}dx=4\operatorname{Ti}_2\left(\sqrt{\frac{s-1}{2}}\right),\ s>1}$$ We'll start off with the substitution $$\frac{s-x}{s-1}=t$$ to get: $$\int_1^s \frac{\operatorname{arccosh}x}{\sqrt{(x-1)(s-x)}}dx=\int_0^1 \frac{\operatorname{arccosh}(s-(s-1)t)}{\sqrt{t}\sqrt{1-t}}dt\overset{t=x^2}=2\int_0^1\frac{\operatorname{arccosh}(s-(s-1)x^2)}{\sqrt{1-x^2}}dx$$ $$\overset{IBP}=4\int_0^1 \frac{x\arcsin x}{\sqrt{(1-x^2)\left(\frac{s+1}{s-1}-x^2\right)}}dx\overset{x=\sin t}=4\int_0^\frac{\pi}{2}\frac{t\sin t }{\sqrt{\frac{2}{s-1}+\cos^2 t}}dt\overset{\cos t=x}=4\int_0^1\frac{\arccos x}{\sqrt{\frac{2}{s-1}+x^2}}dx$$ $$\overset{IBP}=4\int_0^1 \frac{\operatorname{arcsinh} \left(x\sqrt{\frac{s-1}{2}}\right)}{\sqrt{1-x^2}}dx\overset{x=\sin t}=4\int_0^\frac{\pi}{2}\operatorname{arcsinh} \left(\sqrt{\frac{s-1}{2}}\sin t\right)dt$$ $$=4\sum_{n=0}^\infty \frac{(-1)^n\binom{2n}{n}}{(2n+1)4^n}\left(\sqrt{\frac{s-1}{2}}\right)^{2n+1}\int_0^\frac{\pi}{2}\sin^{2n+1}t dt=4\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}\left(\sqrt{\frac{s-1}{2}}\right)^{2n+1}$$ Above was used the Maclaurin series for $$\operatorname{arcsinh} z$$ and the Wallis integral. Finally, using the definition of the inverse tangent integral we can rewrite the result as announced.
Some steps may be simplified in the above, but I don't think it really matters. For a general case it doesn't seem promising because, by the same approach we get: $$\int_a^b \frac{\operatorname{arccosh} x}{\sqrt{(x-a)(b-x)}}dx=4\int_0^1 \frac{x\arcsin x}{\sqrt{\left(\frac{b-1}{b-a}-x^2\right)\left(\frac{b+1}{b-a}-x^2\right)}}dx$$ And we can't simplify that thing inside the square root via $$x=\sin t$$.