evaluation of a definite integral involving inverse hyperbolic functions

Question

The following integral appears during the analysis of statistical mechanical models.

$$\int\limits_0^{\pi /2}{\operatorname{arctanh} \left[ {\sqrt {1 - \sin (x)\sin (x){a^2}} } \right]dx} $$

Is there a closed form solution for the integral?

Answer 1

Consider $$I(b)=\int_0^{\frac \pi 2} \tanh ^{-1}\left(\sqrt{1-b \sin ^2(x)}\right)\,dx$$ $$I'(b)=-\frac 1{2b}\int_0^{\frac \pi 2} \frac{dx}{\sqrt{1-b \sin ^2(x)}}=-\frac{K\left(\frac{b}{b-1}\right)}{2 b\sqrt{1-b}}$$ and $I(1)=2 C$. So $$I(a^2)=2C+\frac 12 \int_{a^2}^1\frac{1}{ b\sqrt{1-b}}K\left(\frac{b}{b-1}\right)\,db$$

You could use the series $$I'(b)=\frac{\pi }{4 b}+ \pi \sum_{n=0}^\infty \frac {\alpha_n}{2^{\beta_n}} b^n $$

The $\alpha_n$ correspond to sequence $A038534$ in $OEIS$ and the $\beta_n$ form the sequence $$\{4,8,10,16,18,22,24,32,34,38,40,46,48,52,54,64,66,70,72,78,80,84,86 ,94,96,100,\cdots\}$$

Integrating termwise leads to more than decent approximations.

$$J=\int_0^{\frac \pi 2} \tanh ^{-1}\left(\sqrt{1-a^2 \sin ^2(x)}\right)\,dx=2C-\frac{1}{2} \pi \log (a)+\pi \sum_{n=0}^\infty \frac {\alpha_n}{2^{\beta_n}\,(n+1)}(1-a^{2(n+1)})$$

Edit

Reworking $$J=\frac 12 \int_{a^2}^1\frac{1}{ b\sqrt{1-b}}K\left(\frac{b}{b-1}\right)\,db$$

let $b=\sin^2(t)$ to make $$J=\int_{\sin ^{-1}(a)}^{\frac \pi 2} \csc (t) \,K\left(-\tan ^2(t)\right)\,dt$$

Using $$K\left(-z^2\right)=\pi \sum_{n=0}^\infty (-1)^n\, \frac {\alpha_n}{2^{\gamma_n}}\, z^{2n} $$ the $\gamma_n$ forming the sequence $$\{1,3,7,9,15,17,21,23,31,33,37,39,45,47,51,53,63,65,69,71,77,79,8 3,85,93,\cdots\}$$ $$J=-\frac{1}{2} \pi \log \left(\tan \left(\frac{\sin ^{-1}(a)}{2} \right)\right)+$$ $$\pi \sum_{n=0}^\infty (-1)^n\, \frac {\alpha_n}{2^{\gamma_n}}\,2^{2 n-1} \left(B_1(n,1-2 n)-B_{\tan ^2\left(\frac{\sin ^{-1}(a)}{2} \right)}(n,1-2 n)\right)$$

Answer 2

Given: $$I(a)=\int\limits_0^{\pi /2}{\operatorname{arctanh} ( {\sqrt {1 - a^2\sin^2 x}}) \;dx}$$

The inverse hyperbolic tangent is given by

$$\operatorname{arctanh}z=\frac{1}{2}\ln\frac{1+z}{1-z}$$

So we can after some transformations present the integrand in the form

$$\ln(1+\sqrt {1 - a^2\sin^2 x})-\ln a -\ln \sin x $$

Using well known result

$$\int\limits_0^{\pi /2}\ln \sin x\;dx=-\frac{\pi}{2}\ln 2$$

we get after integrating:

$$I(a)=\frac{\pi}{2}\ln\frac{2}{a}+\int\limits_0^{\pi /2}\ln(1+\sqrt {1 - a^2\sin^2 x})\;dx$$

We can already get useful information from this result.

Asymptotic behavior of $I(a)$ as $a$ tends to zero:

$$I(a)\to \frac{\pi}{2}\ln\frac{2}{a}$$

At $a=1$ we get:

$$I(1)=\frac{\pi}{2}\ln 2+\int\limits_0^{\pi /2}\ln(1+\cos x)\;dx=2G$$

where $G$ is Catalan's constant.

Well known result

$$\int\limits_0^{\pi /2}\ln(1+\cos x)\;dx=2G-\frac{\pi}{2}\ln 2$$

is used.

For values around $a=\frac{1}{2}$ simply expand $I(a)$ in Taylor's series about $a$ as given by Mariusz Iwaniuk.

Some of the first terms are already giving an excellent accuracy.

Answer 3

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Define the function $\mathcal{I}:\left(0,1\right]\rightarrow\mathbb{R}$ via the definite integral

$$\mathcal{I}{\left(b\right)}:=\int_{0}^{\pi/2}\mathrm{d}\theta\,\operatorname{artanh}{\left(\sqrt{1-b\sin^{2}{\left(\theta\right)}}\right)},$$

where here the real inverse hyperbolic tangent is defined by

$$\operatorname{artanh}{\left(x\right)}:=\int_{0}^{x}\mathrm{d}t\,\frac{1}{1-t^{2}}=\frac12\ln{\left(\frac{1+x}{1-x}\right)};~~~\small{x\in\left(-1,1\right)}.$$

Consider the following specific value for the generalized hypergeometric function $_3F_2$:

$${_3F_2}{\left(1,1,\frac32;2,2;z\right)}=-\frac{4}{z}\ln{\left(\frac{1+\sqrt{1-z}}{2}\right)};~~~\small{z\in(-\infty,1]}.$$

We can use this fact to rewrite the most complicated part of the integrand of $\mathcal{I}$ as a hypergeometric function.

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For any $b\in\left(0,1\right]$,

$$\begin{align} \mathcal{I}{\left(b\right)} &=\int_{0}^{\pi/2}\mathrm{d}\theta\,\operatorname{artanh}{\left(\sqrt{1-b\sin^{2}{\left(\theta\right)}}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{artanh}{\left(\sqrt{1-bx^{2}}\right)}}{\sqrt{1-x^{2}}};~~~\small{\left[\theta=\arcsin{\left(x\right)}\right]}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\operatorname{artanh}{\left(\sqrt{1-bt}\right)}}{2\sqrt{t}\sqrt{1-t}};~~~\small{\left[x=\sqrt{t}\right]}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(\frac{1+\sqrt{1-bt}}{1-\sqrt{1-bt}}\right)}}{4\sqrt{t}\sqrt{1-t}}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(\frac{(1+\sqrt{1-bt})^{2}}{bt}\right)}}{4\sqrt{t}\sqrt{1-t}}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(1+\sqrt{1-bt}\right)}-\ln{\left(bt\right)}}{4\sqrt{t}\sqrt{1-t}}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(1+\sqrt{1-bt}\right)}-2\ln{\left(2\right)}+2\ln{\left(2\right)}-\ln{\left(b\right)}-\ln{\left(t\right)}}{4\sqrt{t}\sqrt{1-t}}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(\frac{1+\sqrt{1-bt}}{2}\right)}-\ln{\left(\frac{b}{4}\right)}-\ln{\left(t\right)}}{4\sqrt{t}\sqrt{1-t}}\\ &=-\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(\frac{b}{4}\right)}}{4\sqrt{t}\sqrt{1-t}}-\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}}{4\sqrt{t}\sqrt{1-t}}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(\frac{1+\sqrt{1-bt}}{2}\right)}}{2\sqrt{t}\sqrt{1-t}}\\ &=\pi\ln{\left(2\right)}-\frac{\pi}{4}\ln{\left(b\right)}+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(\frac{1+\sqrt{1-bt}}{2}\right)}}{\sqrt{t}\sqrt{1-t}}\\ &=\pi\ln{\left(2\right)}-\frac{\pi}{4}\ln{\left(b\right)}-\frac18\int_{0}^{1}\mathrm{d}t\,\frac{bt}{\sqrt{t}\sqrt{1-t}}\left[-\frac{4}{bt}\ln{\left(\frac{1+\sqrt{1-bt}}{2}\right)}\right]\\ &=\pi\ln{\left(2\right)}-\frac{\pi}{4}\ln{\left(b\right)}-\frac{b}{8}\int_{0}^{1}\mathrm{d}t\,t^{1/2}\left(1-t\right)^{-1/2}\,{_3F_2}{\left(1,1,\frac32;2,2;bt\right)}\\ &=\pi\ln{\left(2\right)}-\frac{\pi}{4}\ln{\left(b\right)}-\frac{b}{8}\operatorname{B}{\left(\frac32,\frac12\right)}\,{_4F_3}{\left(1,1,\frac32,\frac32;2,2,2;b\right)}\\ &=\pi\ln{\left(2\right)}-\frac{\pi}{4}\ln{\left(b\right)}-\frac{\pi}{16}b\,{_4F_3}{\left(1,1,\frac32,\frac32;2,2,2;b\right)},\\ \end{align}$$

where the final integration comes from Euler's integral formula for higher-order hypergeometric functions:

$$\int_{0}^{1}\mathrm{d}t\,t^{d-1}\left(1-t\right)^{r-d-1}\,{_3F_2}{\left(a,b,c;p,q;zt\right)}=\operatorname{B}{\left(d,r-d\right)}\,{_4F_3}{\left(a,b,c,d;p,q,r;z\right)};~~~\small{0<d<r}.$$

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