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When I came across the integral $$J=\int_0^{\infty} \frac{\operatorname{artanh}\left(\frac{1}{\sqrt{1+x^2}}\right)}{\sqrt{1+x^2}} d x=\frac{\pi^2}{4} $$ whose answer is surprisingly decent, I, as usual, put $x=\tan \theta$ and transform the integral into

$$ \begin{aligned} J & =\int_0^{\frac{\pi}{2}} \frac{\operatorname{artanh}\left(\frac{1}{\sec \theta}\right)}{\sec \theta} \sec ^2 \theta d \theta \\ & =\int_0^{\frac{\pi}{2}} \sec \theta \operatorname{artanh}\left(\frac{1}{\sec \theta}\right) d \theta \end{aligned} $$

Feynman’s trick reminds me to deal with its parametrized integral

$$ J(a)= \int_0^{\infty} \frac{\operatorname{artanh}\left(\frac{a}{\sqrt{1+x^2}}\right)}{\sqrt{1+x^2}} d x =\int_0^{\frac{\pi}{2}} \sec \theta \operatorname{artanh} \left(\frac{a}{\sec \theta}\right) d \theta $$ with $|a|<1$ and $J(0)=0$.

Consequently, differentiation under integral make our life easier as $$ \begin{aligned} J^{\prime}(a)&=\int_0^{\frac{\pi}{2}} \frac{1}{1-\frac{a^2}{\sec ^2 \theta}} d \theta \\& =\int_0^{\frac{\pi}{2}} \frac{\sec ^2 \theta}{\sec ^2 \theta-a^2} d \theta \\ & =\int_0^{\frac{\pi}{2}} \frac{d(\tan \theta)}{\left(1-a^2\right)+\tan ^2 \theta} \\ & =\frac{1}{\sqrt{1-a^2}}\left[\operatorname{artan}\left(\frac{\tan \theta}{\sqrt{1-a^2}}\right)\right]_0^{\frac{\pi}{2}} \\ & =\frac{\pi}{2 \sqrt{1-a^2}} \\ & \end{aligned} $$

Integrating back with $J(0)=0$ yields

$$ \boxed{J(a)= \int_0^{\infty} \frac{\operatorname{artanh}\left(\frac{a}{\sqrt{1+x^2}}\right)}{\sqrt{1+x^2}} d x=\frac{\pi}{2} \int \frac{1}{\sqrt{1-a^2}} d a=\frac{\pi}{2} \operatorname{arcsin}a \,} \tag*{(1)} $$

In particular, let $a$ approach to $1$, we get

$$ J= \lim _{a \rightarrow 1} J(a)= \lim _{a \rightarrow 1}\frac{\pi}{2} \operatorname{arcsin} 1=\frac{\pi^2}{4} $$


Inspired by $J(a)$, I believe that whenever $\left|\frac ab \right| <1$, we can further evaluate

$$ \int_0^{\infty} \frac{\operatorname{artanh}\left(\frac{a}{\sqrt{b^2+x^2}}\right)}{\sqrt{b^2+x^2}} d x $$ by simply letting $x\mapsto bx$ which transforms

$$ \boxed{\int_0^{\infty} \frac{\operatorname{artanh}\left(\frac{a}{\sqrt{b^2+x^2}}\right)}{\sqrt{b^2+x^2}} d x=\int_0^{\infty} \frac{\operatorname{artanh}\left(\frac{\frac{a}{b}}{\sqrt{1+x^2}}\right)}{\sqrt{1+x^2}} d x=J\left(\frac{a}{b}\right )= \frac{\pi}{2} \operatorname{arcsin}\left(\frac{a}{b}\right) \,} $$



Latest Edit :After submitting the question, I found that there is a simpler version with Feynman’s trick to share with you.

Considering $b$ is a constant and let $$\displaystyle J(a)=\int_0^{\infty} \frac{\operatorname{artanh}\left(\frac{a}{\sqrt{b^2+x^2}}\right)}{\sqrt{b^2+x^2}} d x\tag*{} $$ where $J(0)=0$. Differentiating $J(a)$ w.r.t. $a$ yields $$\displaystyle \begin{aligned}J^{\prime}(a) & =\int_0^{\infty} \frac{1}{\left(1-\frac{a^2}{b^2+x^2}\right)\left(b^2+x^2\right)} d x \\& =\int_0^{\infty} \frac{1}{b^2-a^2+x^2} d x \\& =\frac{1}{\sqrt{b^2-a^2}}\left[\tan ^{-1}\left(\frac{x}{\sqrt{b^2-a^2}}\right)\right]_0^{\infty} \\& =\frac{\pi}{2 \sqrt{b^2-a^2}}\end{aligned}\tag*{} $$ Integrating back w.r.t. $a$ with $J(0)=0$ yields $$\displaystyle \boxed{ \begin{aligned}J(a) & =\frac{\pi}{2} \int \frac{d a}{\sqrt{b^2-a^2}} =\frac{\pi}{2} \arcsin \left(\frac{a}{b}\right)\end{aligned}}\tag*{} $$



Can it be done without Feynman’s trick?

Your comments and alternative methods are highly appreciated.

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  • $\begingroup$ Thoughts on using the residue theorem: substituting $\tanh y=\frac{a}{\sqrt{a^2+x^2}}$, the integral becomes a function of $a/b$. The case $a=b$ is easily handled with a semicircular contour in the upper half-plane, using heat kernel regularization. I suspect the general case is harder even with a dogbone contour, because of a surd in the integrand. $\endgroup$
    – J.G.
    Commented Apr 21, 2023 at 10:26
  • $\begingroup$ Thank you for your suggestion which is too hard to me. $\endgroup$
    – Lai
    Commented Apr 22, 2023 at 2:50
  • $\begingroup$ We also have the rather neat equivalent form$$J=\int_0^1K'(k)\,dk$$where $K(k)$ is the complete elliptic integral of the first kind and $K'(k)=K(\sqrt{1-k^2})$. (Easily shown by converting $\tanh^{-1}$ to $\sinh^{-1}$, then $\sinh^{-1}$ to a definite integral.) I'm not familiar enough with elliptic integrals to know where to go from here, though, at least not without retreading steps from the posted answers. $\endgroup$
    – user170231
    Commented Apr 24, 2023 at 6:55
  • $\begingroup$ when you introduced a parameter $a$ you wrote $|a| < 1$ but to get $J(1)=\frac{\pi^2}{4}$ you put $a=1$ how?? $\endgroup$ Commented Sep 26, 2023 at 4:48
  • $\begingroup$ You are right. We can consider the limit instead. $\endgroup$
    – Lai
    Commented Sep 26, 2023 at 7:54

7 Answers 7

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Substituting $x = \operatorname{csch} t$ and noting that $ \frac{1}{\sqrt{x^2+1}} = \tanh t ,$ the integral reduces to

\begin{align*} J &= \int_{0}^{\infty} \frac{t}{\sinh t} \, \mathrm{d} t \\ &= 2 \sum_{n=0}^{\infty} \int_{0}^{\infty} t e^{-(2n+1)t} \, \mathrm{d}t \\ &= 2 \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} \\ &= \frac{\pi^2}{4}. \end{align*}


Also, for $|\alpha| < 1$, OP's substitution shows that

\begin{align*} J(\alpha) &= \int_{0}^{\frac{\pi}{2}} \frac{\operatorname{artanh}(\alpha \sin\theta)}{\sin \theta} \, \mathrm{d} \theta \\ &= \sum_{n=0}^{\infty} \frac{\alpha^{2n+1}}{2n+1} \int_{0}^{\frac{\pi}{2}} \sin^{2n}\theta \, \mathrm{d} \theta \\ &= \sum_{n=0}^{\infty} \frac{\alpha^{2n+1}}{2n+1} \cdot (-1)^n \frac{\pi}{2} \binom{-1/2}{n} \\ &= \frac{\pi}{2} \int_{0}^{\alpha} \frac{\mathrm{d}t}{\sqrt{1 - t^2}} \\ &= \frac{\pi}{2} \arcsin \alpha. \end{align*}

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  • $\begingroup$ Wonderful use of power series. Thank you. $\endgroup$
    – Lai
    Commented Apr 21, 2023 at 12:11
  • $\begingroup$ @Lai, Thank you :) By the way, we have $$\operatorname{artanh}(x)=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}, \quad|x|<1$$ and $$\int_{0}^{\frac{\pi}{2}} \sin^{2n}\theta \, \mathrm{d} \theta = \frac{\pi}{2}\cdot\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)} = (-1)^n \frac{\pi}{2} \binom{-1/2}{n}.$$ So the term $(-1)^n$ should only appear in the third line of the last equation. $\endgroup$ Commented Apr 21, 2023 at 12:28
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    $\begingroup$ Yes, I am so careless. Thank you for your creative method! $\endgroup$
    – Lai
    Commented Apr 21, 2023 at 12:35
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    $\begingroup$ Elegant and skillful solution to my question. $\endgroup$
    – Lai
    Commented Apr 22, 2023 at 1:41
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Substitute $x= \frac{2y}{y^2-1}$ to simplify the integral \begin{align} \int_0^{\infty} \frac{{\tanh^{-1}}\frac{1}{\sqrt{1+x^2}}}{\sqrt{1+x^2}} d x =& \int_1^\infty \frac{2\ln y}{y^2-1}dy =\int_0^\infty \frac{\ln y}{y^2-1}dy\\ =&\int_0^\infty \int_0^1 \frac{z}{1+(y^2-1)z^2}dz\>dy\\ =&\ \frac\pi2 \int_0^1 \frac1{\sqrt{1-z^2}}dz=\frac{\pi^2}4 \end{align}

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  • $\begingroup$ Simple and elegant as usual! Can you help find the general case $\int_0^{\infty} \frac{{\tanh^{-1}}\frac{a}{\sqrt{1+x^2}}}{\sqrt{1+x^2}} d x?$ $\endgroup$
    – Lai
    Commented Apr 22, 2023 at 2:29
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This nice integral can be evaluated using some power series. Start with a substitution $u = \frac{1}{\sqrt{x^2+1}}$, so ${\rm d}x = -\frac{1}{u^2\sqrt{1-u^2}}{\rm d}u$ and $$ J = \int_0^{\infty} \frac{1}{\sqrt{1+x^2}}\operatorname{arctanh}\left(\frac{1}{\sqrt{1+x^2}}\right){\rm d} x = \int_0^1 \frac{\operatorname{arctanh}(u)}{u\sqrt{1-u^2}}{\rm d}u. $$ As $\left(\operatorname{arctanh}(u)\right)' = \frac{1}{1-u^2}$, one can obtain a series expansion $$ \operatorname{arctanh}(u) = \sum_{k=0}^{\infty}\frac{u^{2k+1}}{2k+1}, \quad |u|<1, $$ so (take a pause here to think why integration and summation can be swapped): $$ \int_0^1 \frac{\operatorname{arctanh}(u)}{u\sqrt{1-u^2}}{\rm d}u = \sum_{k=0}^{\infty}\frac{1}{2k+1} \int_0^1 \frac{u^{2k+1}}{u\sqrt{1-u^2}}{\rm d}u = \sum_{k=0}^{\infty}\frac{1}{2k+1} \int_0^1 \frac{u^{2k}}{\sqrt{1-u^2}}{\rm d}u. $$ Using $s = u^2$ for the integral, $$ \int_0^1 \frac{u^{2k}}{\sqrt{1-u^2}}{\rm d}u = \frac{1}{2}\int_0^1 s^{k-\frac{1}{2}} (1-s)^{-\frac{1}{2}} {\rm d} s = \frac{1}{2} \cdot B\left(k+\frac{1}{2}, \frac{1}{2}\right) = \\ = \frac{1}{2} \frac{\Gamma\left(k+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(k+1)} = \frac{1}{2}\cdot\frac{1}{k!}\cdot\frac{(2k-1)!!}{2^k} \sqrt{\pi} \cdot \sqrt{\pi} = \frac{\pi}{2}\frac{(2k-1)!!}{2^k k!}, $$ so $$ J = \frac{\pi}{2} \sum_{k=0}^{\infty}\frac{(2k-1)!!}{(2k+1)2^k k!}. $$ Integrating the series expansion for $\left(\arcsin (x)\right)' = \frac{1}{\sqrt{1-x^2}} = \sum\limits_{k=0}^\infty\frac{(2k-1)!!}{2^k k!}x^{2k}$, one can get $$ \arcsin (x) =\sum_{k=0}^{\infty}\frac{(2k-1)!!}{(2k+1)2^k k!} x^{2k+1}. $$ It means that $J = \frac{\pi}{2} \cdot \arcsin(1) = \left(\frac{\pi}{2}\right)^2$.

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By IBP, \begin{eqnarray} I &=& \int_0^{\infty} \text{arctanh}\left(\frac{1}{\sqrt{1+x^2}}\right) \frac{1}{\sqrt{1+x^2}}\,dx \\ &=& \int_0^{\infty} \text{arctanh}\left(\frac{1}{\sqrt{1+x^2}}\right)\mathrm{d}\;\text{arcsinh}(x)\\ &=& \int_0^{\infty} \text{arcsinh}(x)\frac{1}{x\sqrt{1+x^2}}\mathrm d\,x\\ &\overset{x=\sinh(u)}=& \int_0^{\infty} \frac{u}{\sinh(u)}\mathrm d\,u=2\int_0^{\infty}\frac{ue^{-u}}{1-e^{-2u}}\mathrm d\,u\\ &=&2\int_0^{\infty}\sum_{n=0}^\infty ue^{-(2n+1)u}\mathrm d\,u\\ &=&2\sum_{n=0}^\infty \frac1{(2n+1)^2}=\frac{\pi^2}{4}. \end{eqnarray}

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  • $\begingroup$ Nice to see IBP works. Thank you. $\endgroup$
    – Lai
    Commented Apr 21, 2023 at 23:43
  • $\begingroup$ Should the third step be: $\int_0^{\infty} \text{arcsinh}(x)\frac{1}{x\sqrt{1+x^2}}\mathrm d\,x?$ $\endgroup$
    – Lai
    Commented Apr 22, 2023 at 1:23
  • $\begingroup$ You're right. See the update. Thanks. $\endgroup$
    – xpaul
    Commented Apr 22, 2023 at 15:33
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$$\frac 1{\sqrt{1+x^2}}=t \quad \implies J=\int_0^1 \frac{\tanh ^{-1}(t)}{t \,\sqrt{1-t^2}}\,dt$$ One integration by parts $$\int\frac{\tanh ^{-1}(t)}{t \,\sqrt{1-t^2}}\,dt=\text{Li}_2\left(-e^{-\tanh ^{-1}(t)}\right)-\text{Li}_2\left(e^{-\tanh ^{-1}(t)}\right)+$$ $$\tanh ^{-1}(t) \left(\log \left(1-e^{-\tanh ^{-1}(t)}\right)-\log \left(1+e^{-\tanh ^{-1}(t)}\right)\right)$$ Using the limits, the result.

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  • $\begingroup$ Interesting solution! $\endgroup$
    – Lai
    Commented May 7, 2023 at 0:20
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So let me try, I used the identity $\text{arctanh}(x) = \frac{1}{2}\ \ln\left(\frac{1+x}{1-x}\right)$.

Further, $$\text{arctanh}\left(\frac{1}{\sqrt{1+x^2}}\right) = \frac{1}{2}\ \ln\left(\frac{1+\sqrt{1+x^2}}{\sqrt{1+x^2}-1}\right)$$

Let $$I = \int_0^{\infty} \text{arctanh}\left(\frac{1}{\sqrt{1+x^2}}\right) \frac{1}{\sqrt{1+x^2}}\,dx $$

$$= \frac{1}{2}\int_0^{\infty} \ln\left(\frac{1+\sqrt{1+x^2}}{\sqrt{1+x^2}-1}\right) \frac{1}{\sqrt{1+x^2}}\,dx$$.

Then, on evaluating $I$ (without bounds), we get

\begin{align*} I &= \mathrm{Li}_2(-e^{-\text{arcsinh(x)}}) - \mathrm{Li}_2(e^{-\text{arcsinh(x)}}) + \frac{1}{2}\text{arcsinh(x)}\left(\ln\left(\frac{1+\sqrt{1+x^2}}{\sqrt{1+x^2}-1}\right)\right) \\ &+\ln\left(1-e^{-\text{arcsinh(x)}}\right)-\ln\left(1+e^{-\text{arcsinh(x)}}\right) \end{align*}

Further it simplies approximately to $$\frac{ π^2}{4} = 2.4674....$$

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    $\begingroup$ Constructive idea! $\endgroup$
    – Lai
    Commented Apr 22, 2023 at 2:33
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    $\begingroup$ You said On evaluting $I$ without bounds so did you mean $$ J = \frac{1}{2} \int \ln \left( \frac{1+\sqrt{1+x^2}}{\sqrt{1+x^2}-1} \right) \frac{1}{\sqrt{1+x^2}} dx $$ And $$ I =J\bigg{|}_0^\infty \, \, ? $$ $\endgroup$ Commented Sep 26, 2023 at 4:39
  • $\begingroup$ yes, that's right. $\endgroup$ Commented Sep 26, 2023 at 12:46
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Using the integral $$ \int \frac{1}{x^2-k^2} d x=-\frac{\tanh ^{-1}\left(\frac{x}{k}\right)}{k}+\text { constant } $$ to transform the itegral into a double integral $$ \int_0^{\infty} \frac{\operatorname{artanh}\left(\frac{a}{\sqrt{1+x^2}}\right)}{\sqrt{b^2+x^2}}dx=-\int_0^{\infty} \int_0^a \frac{1}{y^2-\left(b^2+x^2\right)} d y d x $$ Interchange the variables yields $$ \begin{aligned} \int_0^{\infty} \frac{\operatorname{artanh}\left(\frac{a}{\sqrt{1+x^2}}\right)}{\sqrt{b^2+x^2}}dx & =\int_0^a \int_0^{\infty} \frac{1}{x^2+\left(b^2-y^2\right)} d x d y \\ & =\int_0^a\left[\frac{1}{\sqrt{b^2-y^2}} \tan ^{-1} \frac{x}{\sqrt{b^2-y^2}}\right]_0^{\infty} d y \\ & =\frac{\pi}{2} \int_0^a \frac{1}{\sqrt{b^2-y^2}} d y \\ & =\frac{\pi}{2} \arcsin \left(\frac{a}{b}\right) \end{aligned} $$

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