# Can the integral be found without Feynman’s trick?

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?

• 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.
– J.G.
Commented Apr 21, 2023 at 10:26
• Thank you for your suggestion which is too hard to me.
– Lai
Commented Apr 22, 2023 at 2:50
• 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. Commented Apr 24, 2023 at 6:55
• when you introduced a parameter $a$ you wrote $|a| < 1$ but to get $J(1)=\frac{\pi^2}{4}$ you put $a=1$ how?? Commented Sep 26, 2023 at 4:48
• You are right. We can consider the limit instead.
– Lai
Commented Sep 26, 2023 at 7:54

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*}

• Wonderful use of power series. Thank you.
– Lai
Commented Apr 21, 2023 at 12:11
• @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. Commented Apr 21, 2023 at 12:28
• Yes, I am so careless. Thank you for your creative method!
– Lai
Commented Apr 21, 2023 at 12:35
• Elegant and skillful solution to my question.
– Lai
Commented Apr 22, 2023 at 1:41

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}

• Simple and elegant as usual! Can you help find the general case $\int_0^{\infty} \frac{{\tanh^{-1}}\frac{a}{\sqrt{1+x^2}}}{\sqrt{1+x^2}} d x?$
– Lai
Commented Apr 22, 2023 at 2:29

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 $$J = \frac{\pi}{2} \cdot \arcsin(1) = \left(\frac{\pi}{2}\right)^2$$.

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

• Nice to see IBP works. Thank you.
– Lai
Commented Apr 21, 2023 at 23:43
• Should the third step be: $\int_0^{\infty} \text{arcsinh}(x)\frac{1}{x\sqrt{1+x^2}}\mathrm d\,x?$
– Lai
Commented Apr 22, 2023 at 1:23
• You're right. See the update. Thanks. Commented Apr 22, 2023 at 15:33

$$\frac 1{\sqrt{1+x^2}}=t \quad \implies J=\int_0^1 \frac{\tanh ^{-1}(t)}{t \,\sqrt{1-t^2}}\,dt$$ 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.

• Interesting solution!
– Lai
Commented May 7, 2023 at 0:20

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 $$\frac{ π^2}{4} = 2.4674....$$

• Constructive idea!
– Lai
Commented Apr 22, 2023 at 2:33
• 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 $$I =J\bigg{|}_0^\infty \, \, ?$$ Commented Sep 26, 2023 at 4:39
• yes, that's right. Commented Sep 26, 2023 at 12:46

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}