Evaluate $\int_0^{\infty } \frac{\arctan \left(\sqrt{a^2+x^2}\right)}{\left(x^2+1\right)\sqrt{a^2+x^2}} \, dx$

Question

J. Borwein's review on Experimental Mathematics gives the following \begin{align} & \int_0^{\infty } \frac{\arctan \left(\sqrt{a^{2} + x^{2}}\right)}{\left(x^{2} + 1\right)\sqrt{a^{2} + x^{2}}} \, dx \\[2mm] = & \ \frac{\pi\left[2\arctan\left(\sqrt{a^{2} - 1}\right) - \arctan\left(\sqrt{a^{4} - 1}\right)\right]}{2\sqrt{a^{2} - 1}}, \quad a > 1 \end{align} How can we establish it ?. Any help will be appreciated.

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Update: The original problem has a typo and the current one is easy.

Answer 1

As suggested by Ali Shather in the comments, we can write the inverse tangent as an integral to obtain \begin{align} \int \limits_0^\infty \frac{\arctan\left(\sqrt{a^2+x^2}\right)}{(1+x^2)\sqrt{a^2+x^2}} \, \mathrm{d} x &= \int \limits_0^\infty \int \limits_0^1 \frac{\mathrm{d} y}{1+y^2 (a^2+x^2)} \,\frac{\mathrm{d} x}{1+x^2} \\ &= \int \limits_0^1 \int \limits_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+y^2(a^2+x^2))} \, \mathrm{d} y \\ &= \int \limits_0^1 \frac{1}{1 + (a^2-1)y^2} \int \limits_0^\infty \left(\frac{1}{1+x^2} - \frac{1}{a^2+\frac{1}{y^2} + x^2}\right) \, \mathrm{d} x \, \mathrm{d} y \\ &= \frac{\pi}{2} \int \limits_0^1 \frac{\mathrm{d} y}{\sqrt{1+a^2 y^2}(y + \sqrt{1+a^2 y^2})} \\&\!\!\!\!\!\!\stackrel{y = \frac{2t}{a(1-t^2)}}{=} \pi \int \limits_0^{\frac{a}{1+\sqrt{1+a^2}}}\frac{\mathrm{d} t}{a(1+t^2)+2t} \\ &= \frac{\pi a}{a^2 - 1} \int \limits_0^{\frac{a}{1+\sqrt{1+a^2}}} \frac{\mathrm{d}t}{1 + \left(\frac{1+at}{\sqrt{a^2 - 1}}\right)^2} \\ &\!\!\!\!\!\!\!\stackrel{\frac{1+at}{\sqrt{a^2 - 1}} = \frac{1}{u}}{=} \frac{\pi}{\sqrt{a^2-1}} \int \limits_{\sqrt{\frac{a^2-1}{a^2+1}}}^{\sqrt{a^2-1}} \frac{\mathrm{d} u}{1+u^2} \\ &= \frac{\pi}{\sqrt{a^2-1}}\left[\arctan\left(\sqrt{a^2-1}\right) - \arctan\left(\sqrt{\frac{a^2-1}{a^2+1}}\right)\right] \\ &= \frac{\pi}{\sqrt{a^2-1}}\left[\arctan\left(\sqrt{a^2-1}\right) - \frac{1}{2}\arctan\left(\sqrt{a^4-1}\right)\right] \end{align} for $a > 1$. The last step follows from $\arctan(x) = \frac{1}{2} \arctan\left(\frac{2 x}{1-x^2}\right)$ for $x^2 < 1$. Using the more general argtangent addition formula this result can also be written as $$ \int \limits_0^\infty \frac{\arctan\left(\sqrt{a^2+x^2}\right)}{(1+x^2)\sqrt{a^2+x^2}} \, \mathrm{d} x = \frac{\pi}{\sqrt{a^2 - 1}} \arctan \left(\frac{\sqrt{a^2-1}}{2 + \sqrt{a^2+1}}\right) \, , \, a > 1 \, .$$

Answer 2

Substitute $x=a\tan t$ to remove the square-roots\begin{align} &\int_0^\infty \frac{\tan^{-1}\sqrt{a^2+x^2}}{(1+x^2)\sqrt{a^2+x^2}}dx =\int_0^{\pi/2}\frac{\tan^{-1}(a\sec t) \sec t}{1+a^2\tan^2t}dt\\ =&\int_0^{\pi/2}\int_0^{\tan^{-1}a} \frac{\sec^2t\sec^2s}{(1+a^2\tan^2t)(1+ \tan^2s\sec^2t)}ds\ dt\\ = &\ \int_0^{\tan^{-1}a}\frac {\frac\pi2}{a+\sin s}ds =\frac\pi{\sqrt{a^2-1}}\left(\sec^{-1}a -\frac12\sec^{-1}a^2\right) \end{align}

Answer 3

\begin{align} J(a)&=\int_0^\infty \frac{\arctan\left(x^2+a^2\right)}{(1+x^2)\sqrt{x^2+a^2}}dx,a>1\\ F(x,y,a)&=\frac{1}{(1+x^2)(x^2+y^2+a^2)}\\ \int_0^1 \left(\int_0^\infty F(x,y,a)dx\right)dy&=\int_0^1\left[\frac{\arctan x}{y^2+a^2-1}-\frac{\arctan\left(\frac{x}{\sqrt{y^2+a^2}}\right)}{(y^2+a^2-1)\sqrt{y^2+a^2}}\right]_0^\infty dy\\ &=\frac{\pi}{2}\int_0^1 \left(\frac1{y^2+a^2-1}-\frac1{(y^2+a^2-1)\sqrt{y^2+a^2}}\right)dy\\ &=\frac{\pi}{2}\left[\frac{\arctan\left(\frac{y}{\sqrt{a^2-1}}\right)}{\sqrt{a^2-1}}-\frac{\arctan\left(\frac{y}{\sqrt{a^2-1}\sqrt{y^2+a^2}}\right)}{\sqrt{a^2-1}}\right]_0^1\\ &=\frac{\pi}{2\sqrt{a^2-1}}\left(\arctan\left(\frac{1}{\sqrt{a^2-1}}\right)-\arctan\left(\frac{1}{\sqrt{a^4-1}}\right)\right)\\ &=\boxed{\frac{\pi}{2\sqrt{a^2-1}}\left(\arctan\left(\sqrt{a^4-1}\right)-\arctan\left(\sqrt{a^2-1}\right)\right)}\\ \int_0^\infty \left(\int_0^1 F(x,y,a)dy\right)dx&=\int_0^\infty \left[\frac{\arctan\left(\frac{y}{\sqrt{x^2+a^2}}\right)}{(1+x^2)\sqrt{x^2+a^2}}\right]_0^1dx=\int_0^\infty \frac{\arctan\left(\frac{1}{\sqrt{x^2+a^2}}\right)}{(1+x^2)\sqrt{x^2+a^2}}dx\\ &=\frac{\pi}{2}\int_0^\infty \frac{1}{(1+x^2)\sqrt{x^2+a^2}}dx-J(a)\\ &=\frac{\pi}{2}\left[\frac{\arctan\left(\frac{x\sqrt{a^2-1}}{\sqrt{x^2+a^2}}\right)}{\sqrt{a^2-1}}\right]_0^\infty-J(a)\\ &=\boxed{\frac{\pi\arctan\left(\sqrt{a^2-1}\right)}{2\sqrt{a^2-1}}-J(a)} \end{align} Therefore, \begin{align}\boxed{J(a)=\frac{\pi}{2\sqrt{a^2-1}}\left(2\arctan\left(\sqrt{a^2-1}\right)-\arctan\left(\sqrt{a^4-1}\right)\right)}\end{align}