Feynman´s trick to solve $\int_0^\infty \frac{\arctan(x)}{\sqrt{x}(1+x^2)}\,dx$

Question

I wanted to evaluate the integral

\begin{align*} \int_0^\infty \frac{\arctan(x)}{\sqrt{x}(1+x^2)}\,dx=\frac{\pi^2}{4\sqrt{2}}-\frac{\pi \ln(2)}{2\sqrt{2}} \tag{1} \end{align*}

I thought of using Feynman´s trick by considering the integral

$$ \begin{align*} I(a)&=\int_0^\infty \frac{\arctan(ax)}{\sqrt{x}(1+x^2)}\,dx \tag{2} \end{align*} $$

Differentiating $(2)$ w.r. to $a$ we obtain:

\begin{align*} I^\prime(a)&=\int_0^\infty \frac{x}{\sqrt{x}(1+x^2)(1+a^2x^2)}\,dx\\ &=\int_0^\infty \frac{x^{1/2}}{(1+x^2)(1+a^2x^2)}\,dx\\ &=\frac{1}{a^2-1}\left(a^2 \int_0^\infty \frac{x^{1/2}}{1+a^2x^2}\,dx-\int_0^\infty \frac{x^{1/2}}{1+x^2}\,dx\right)\\ &=\frac{1}{a^2-1}\left(\sqrt{a} \int_0^\infty \frac{x^{1/2}}{1+x^2}\,dx-\int_0^\infty \frac{x^{1/2}}{1+x^2}\,dx\right)\\ &=\frac{1}{a^2-1}\left(\frac{\sqrt{a}}{2} \int_0^\infty \frac{x^{-1/4}}{1+x}\,dx-\frac12\int_0^\infty \frac{x^{-1/4}}{1+x}\,dx\right)\\ &=\frac{\pi}{\sqrt{2}}\left( \frac{\sqrt{a}}{a^2-1}-\frac{1}{a^2-1}\right)\\ \end{align*}

Integrating back

\begin{align*} I(a)&=\frac{\pi}{\sqrt{2}}\left( \int\frac{\sqrt{a}}{a^2-1}\,da-\int\frac{1}{a^2-1}\,da\right)\\ &=\frac{\pi}{\sqrt{2}}\left( \int\frac{1}{(1-a)(1+a)}\,da-\int\frac{\sqrt{a}}{(1-a)(1+a)}\,da\right)\\ &=\frac{\pi}{\sqrt{2}}\left(\frac12 \int\frac{da}{1-a}+\frac12 \int\frac{da}{1+a}-\frac12 \int\frac{\sqrt{a}}{1-a}\,da-\frac12 \int\frac{\sqrt{a}}{1+a}\,da\right)\\ &=\frac{\pi}{\sqrt{2}}\left(\frac12\ln\left(\frac{1+a}{1-a} \right)- \int\frac{u^2}{1-u^2}\,du- \int\frac{u^2}{1+u^2}\,du\right)\\ &=\frac{\pi}{\sqrt{2}}\left(\frac12\ln\left(\frac{1+a}{1-a} \right)+ \int\frac{u^2-1+1}{u^2-1}\,du- \int\frac{u^2+1-1}{u^2+1}\,du\right)\\ &=\frac{\pi}{\sqrt{2}}\left(\frac12\ln\left(\frac{1+a}{1-a} \right)- \int\frac{du}{1-u^2}+ \int\frac{du}{u^2+1}\right)\\ &=\frac{\pi}{\sqrt{2}}\left(\frac12\ln\left(\frac{1+a}{1-a} \right)- \frac12\ln\left(\frac{1+u}{1-u} \right)+ \arctan(u)+C\right)\\ &=\frac{\pi}{\sqrt{2}}\left(\frac12\ln\left(\frac{1+a}{1-a} \right)- \frac12\ln\left(\frac{1+\sqrt{a}}{1-\sqrt{a}} \right)+ \arctan\left(\sqrt{a}\right)+C\right)\\ &=\frac{\pi}{\sqrt{2}}\left(\operatorname{arctanh}(a)- \operatorname{arctanh}\left(\sqrt{a} \right)+ \arctan\left(\sqrt{a}\right)+C\right)\\ \end{align*}

Now, if we set $a=0$ in $(2)$ we find that $C=0$. Therefore

\begin{align*} \int_0^\infty \frac{\arctan(ax)}{\sqrt{x}(1+x^2)}\,dx=\frac{\pi}{\sqrt{2}}\left(\operatorname{arctanh}(a)- \operatorname{arctanh}\left(\sqrt{a} \right)+ \arctan\left(\sqrt{a}\right)\right) \tag{3} \end{align*}

Supposedly, we should now let $a \to 1$ in $(3)$ to find $(1)$, but for $\lim_{a \to 1} \operatorname{arctanh}(a) \to \infty$. Is there a way to fix this problem so Feynman´s trick is still applicable?

To add a backgroud to the question. I was originally trying to solve the integral $\int_0^{\pi/2}\frac{x}{\sqrt{\tan(x)}}\,dx$. By a obvious change of variable we find that $\int_0^{\pi/2}\frac{x}{\sqrt{\tan(x)}}\,dx=\int_0^\infty \frac{\arctan(x)}{\sqrt{x}(1+x^2)}\,dx$. Now observe that

\begin{align*} \int_0^{\pi/2}\frac{x}{\sqrt{\tan(x)}}\,dx&=\frac{\pi}{2}\int_0^{\pi/2}\sqrt{\tan(x)}\,dx-\int_0^{\pi/2}x\sqrt{\tan(x)}\,dx & \left(x \to \frac{\pi}{2}-x \right)\\ &=\frac{\pi}{2}\frac{\pi}{\sqrt{2}}-\int_0^{\pi/2}x\sqrt{\tan(x)}\,dx & ( \text{by beta function})\\ &=\frac{\pi^2}{2\sqrt{2}}-\int_0^\infty \frac{\sqrt{x}\arctan(x)}{1+x^2}\,dx\\ &=\frac{\pi^2}{2\sqrt{2}}-\left(\frac{\pi}{2\sqrt{2}}\ln(2)+\frac{{\pi}^2}{4\sqrt{2}}\right) \end{align*}

last line result follows from this post

Answer 1

Note that, as $a\to 1$, $$\begin{align} \operatorname{arctanh}(a)- \operatorname{arctanh}\left(\sqrt{a} \right)&= \frac12\ln\left(\frac{1+a}{1-a} \right)- \frac12\ln\left(\frac{1+\sqrt{a}}{1-\sqrt{a}}\right)\\ &=\frac12\ln\left(\frac{1+a}{1-a}\cdot\frac{1-\sqrt{a}}{1+\sqrt{a}} \right)\\&=\frac12\ln\left(\frac{1+a}{(1+\sqrt{a})^2} \right)\to-\frac{\ln\left(2 \right)}{2}. \end{align}$$ Therefore, from your work, we find $$ \int_0^\infty \frac{\arctan(ax)}{\sqrt{x}(1+x^2)}\,dx=\frac{\pi}{\sqrt{2}}\left(\operatorname{arctanh}(a)- \operatorname{arctanh}\left(\sqrt{a} \right)+ \arctan\left(\sqrt{a}\right)\right) \to -\frac{\pi \ln(2)}{2\sqrt{2}}+\frac{\pi^2}{4\sqrt{2}}.$$

Answer 2

A more direct way to do it is to note that your formula for $I^\prime(a)$ has a removable singularity at $a=1,$ and simplifies to

$\frac{\pi}{\sqrt{2}} \frac{1}{(\sqrt{a}+1)(a+1)}.$

Which would also indicate that you might have wanted to use $a^2$ instead of $a$ to begin with. By the way, that trick predates Feinman by about two hundred years.

Answer 3

Using what you wrote

\begin{align*} \int_0^\infty \frac{\arctan(ax)}{\sqrt{x}(1+x^2)}\,dx=\frac{\pi}{\sqrt{2}}\left(\operatorname{arctanh}(a)- \operatorname{arctanh}\left(\sqrt{a} \right)+ \arctan\left(\sqrt{a}\right)\right) \tag{3} \end{align*}

let $a=b^2$ and expand as a series around $b=1$ $$\operatorname{arctanh}(b^2)- \operatorname{arctanh}\left(b \right)+ \arctan\left(b\right)=\frac{1}{4} (\pi -2 \log (2))+\frac{1}{2}(b-1)+O\left((b-1)^2\right)$$

It could look paradoxal but using \begin{align*} I(b)&=\int_0^\infty \frac{\arctan(b^2x)}{\sqrt{x}(1+x^2)}\,dx\implies I'(b)=\int_0^\infty \frac{2 b \sqrt{x}}{\left(1+x^2\right) \left(1+b^4 x^2\right)} \,dx \end{align*} does not change anything except that $$I(b)=\pi \sqrt 2 \int_0^1\frac{b}{b^3+b^2+b+1}\,db=\pi \sqrt 2 \int_0^1\frac{b}{(b+1) \left(b^2+1\right)}\,db$$ $$I(b)=\frac\pi {\sqrt 2} \int_0^1 \Bigg[\frac{b+1}{b^2+1}-\frac{1}{b+1} \Bigg]\,db=$$ is simpler