Other integral related to Ahmed's integral

I have a doubt regarding the evaluation of the following integral :

$$ \int_0^\frac{1}{\sqrt{5}} \frac{\tan^{-1}\left({\sqrt{(1 + x^2)/2}}\right)} {(1 + 3x^2)\sqrt{1 + x^2}}\,du = \frac{\pi^2\sqrt{2}}{60}. $$

Could anybody please help by offering useful hints or solutions? I think very difficult to prove.

Solution 1:

Following the notation of this paper, we define the generalized Ahmed integral as $$A\left(p,q,r\right)=\int_{0}^{1}\frac{\tan^{-1}\left(\sqrt{q\left(px^{2}+1\right)}\right)}{\sqrt{q\left(px^{2}+1\right)}\left(\left(r+1\right)px^{2}+1\right)}dx$$ so if we put $$x=\frac{u}{\sqrt{5}}$$ in your integral we obtain $$I_{1}=\int_{0}^{1/\sqrt{5}}\frac{\tan^{-1}\left(\sqrt{\frac{1}{2}\left(x^{2}+1\right)}\right)}{\sqrt{x^{2}+1}\left(3x^{2}+1\right)}dx=\frac{1}{\sqrt{2}}\int_{0}^{1/\sqrt{5}}\frac{\tan^{-1}\left(\sqrt{\frac{1}{2}\left(x^{2}+1\right)}\right)}{\sqrt{\frac{1}{2}\left(x^{2}+1\right)}\left(3x^{2}+1\right)}dx=\frac{1}{\sqrt{10}}\int_{0}^{1}\frac{\tan^{-1}\left(\sqrt{\frac{1}{2}\left(\frac{u^{2}}{5}+1\right)}\right)}{\sqrt{\frac{1}{2}\left(\frac{u^{2}}{5}+1\right)}\left(\frac{3u^{2}}{5}+1\right)}du=\frac{1}{\sqrt{10}}A\left(\frac{1}{5},\frac{1}{2},2\right).$$ Now following the same paper, we note that $$A\left(\frac{1}{5},\frac{1}{2},2\right)=\frac{\sqrt{5}}{4}\int_{0}^{\pi/3}\cos^{-1}\left(\frac{\cos\left(x\right)}{1+2\cos\left(x\right)}\right)dx=\frac{\sqrt{5}}{2}\int_{0}^{\pi/3}\tan^{-1}\left(\sqrt{\frac{\cos\left(x\right)+1}{1+3\cos\left(x\right)}}\right)dx$$ by half-angle formula. So$$I_{1}=\frac{1}{2\sqrt{2}}\int_{0}^{\pi/3}\tan^{-1}\left(\sqrt{\frac{\cos\left(x\right)+1}{1+3\cos\left(x\right)}}\right)dx=\frac{1}{2\sqrt{2}}I_{2}.$$ Put $x=2u.$ We have $$I_{2}=2\int_{0}^{\pi/6}\tan^{-1}\left(\frac{\cos\left(u\right)}{\sqrt{2-3\sin^{2}\left(u\right)}}\right)du$$ and using $$\frac{1}{a}\tan^{-1}\left(\frac{1}{a}\right)=\int_{0}^{1}\frac{1}{x^{2}+a^{2}}dx,\, a\neq0$$ we have $$I_{2}=2\int_{0}^{\pi/6}\frac{\sqrt{2-3\sin^{2}\left(u\right)}}{\cos\left(u\right)}\int_{0}^{1}\frac{\cos^{2}\left(u\right)}{x^{2}\cos^{2}\left(u\right)+2-3\sin^{2}\left(u\right)}dxdu=2\int_{0}^{\pi/6}\int_{0}^{1}\frac{\cos\left(u\right)\sqrt{2-3\sin^{2}\left(u\right)}}{x^{2}\cos^{2}\left(u\right)+2-3\sin^{2}\left(u\right)}dxdu=2\int_{0}^{\pi/6}\int_{0}^{1}\frac{\cos\left(u\right)\sqrt{2-3\sin^{2}\left(u\right)}}{x^{2}+2-(3+x^{2})\sin^{2}\left(u\right)}dxdu.$$ Put $u=\sin^{-1}\left(\sqrt{2/3}\sin\left(w\right)\right)$ , $du=\frac{\sqrt{2}\cos\left(w\right)}{\sqrt{2-3\cos^{2}\left(w\right)}}dw$ ,$$I_{2}=4\sqrt{3}\int_{0}^{\sin^{-1}\left(\sqrt{3}/2\sqrt{2}\right)}\int_{0}^{1}\frac{\cos^{2}\left(w\right)}{3x^{2}+6-(3+x^{2})2\sin^{2}\left(w\right)}dxdw$$ now put $w=\tan^{-1}\left(z\right),$ $dw=\frac{1}{1+z^{2}}dz$ $$I_{2}=4\sqrt{3}\int_{0}^{\sqrt{15}/5}\int_{0}^{1}\frac{1}{\left(z^{2}+1\right)\left(x^{2}z^{2}+3x^{2}+6\right)}dxdz.$$ Now integrate respecto to $z$ $$2\sqrt{3}\tan^{-1}\left(\frac{\sqrt{15}}{5}\right)\int_{0}^{1}\frac{1}{x^{2}+3}dx-2\int_{0}^{1}\frac{x\tan^{-1}\left(\frac{x}{\sqrt{5x^{2}+10}}\right)}{\sqrt{x^{2}+2}\left(x^{2}+3\right)}dx$$ hence $$\frac{\pi}{3}\tan^{-1}\left(\frac{\sqrt{15}}{5}\right)-2\int_{0}^{1}\frac{x\tan^{-1}\left(\frac{x}{\sqrt{5x^{2}+10}}\right)}{\sqrt{x^{2}+2}\left(x^{2}+3\right)}dx.$$ Now if we integrate by parts we have $$\frac{\pi}{3}\tan^{-1}\left(\frac{\sqrt{15}}{5}\right)-\frac{2\pi}{3}\tan^{-1}\left(\frac{1}{\sqrt{15}}\right)+2\sqrt{5}\int_{0}^{1}\frac{\tan^{-1}\left(\sqrt{t^{2}+2}\right)}{\sqrt{t^{2}+2}\left(3x^{2}+5\right)}dt$$ hence $$I_{1}=\frac{1}{2\sqrt{2}}\left(\frac{\pi}{3}\tan^{-1}\left(\frac{\sqrt{15}}{5}\right)-\frac{2\pi}{3}\tan^{-1}\left(\frac{1}{\sqrt{15}}\right)+2\sqrt{5}\int_{0}^{1}\frac{\tan^{-1}\left(\sqrt{t^{2}+2}\right)}{\sqrt{t^{2}+2}\left(3x^{2}+5\right)}dt\right)=\frac{1}{2\sqrt{2}}\left(\frac{\pi}{3}\tan^{-1}\left(\frac{\sqrt{15}}{5}\right)-\frac{2\pi}{3}\tan^{-1}\left(\frac{1}{\sqrt{15}}\right)+2\sqrt{5}J_{1}\right).$$ It remain to studies $J_{1}.$ Let $$I=I\left(a\right)=\int_{0}^{1}\frac{\tan^{-1}\left(a\sqrt{t^{2}+2}\right)}{\sqrt{t^{2}+2}\left(3x^{2}+5\right)}dt$$ and differentate respect to $a$ $$J_{1}=\int_{0}^{1}\frac{1}{\left(a^{2}\left(t^{2}+2\right)+1\right)\left(3t^{2}+5\right)}dt=\frac{\sqrt{15}\tan^{-1}\left(\frac{\sqrt{15}}{5}\right)}{5}\int_{0}^{1}\frac{1}{a^{2}+3}da-\int_{0}^{1}\frac{a\tan^{-1}\left(\frac{a}{\sqrt{2a^{2}+1}}\right)}{\sqrt{2a^{2}+1}\left(a^{2}+3\right)}da.$$ Using the identity $\tan^{-1}\left(x\right)+\tan^{-1}\left(1/x\right)=\pi/2$ we have $$J_{1}=\frac{\sqrt{5}\pi\tan^{-1}\left(\frac{\sqrt{15}}{5}\right)}{30}-\frac{\pi}{2}\int_{0}^{1}\frac{a}{\sqrt{2a^{2}+1}\left(a^{2}+3\right)}da+\int_{0}^{1}\frac{a\tan^{-1}\left(\frac{\sqrt{2a^{2}+1}}{a}\right)}{\sqrt{2a^{2}+1}\left(a^{2}+3\right)}da=\frac{\sqrt{5}\pi\tan^{-1}\left(\frac{\sqrt{15}}{5}\right)}{30}-\frac{\pi\sqrt{5}}{10}\left(\tan^{-1}\left(\frac{\sqrt{15}}{5}\right)-\tan^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)+\int_{0}^{1}\frac{a\tan^{-1}\left(\frac{\sqrt{2a^{2}+1}}{a}\right)}{\sqrt{2a^{2}+1}\left(a^{2}+3\right)}da.$$ Let $a=1/y$ . We have $$J_{1}=\frac{\sqrt{5}\pi\tan^{-1}\left(\frac{\sqrt{15}}{5}\right)}{30}-\frac{\pi\sqrt{5}}{10}\left(\tan^{-1}\left(\frac{\sqrt{15}}{5}\right)-\tan^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)+\int_{1}^{\infty}\frac{\tan^{-1}\left(\sqrt{y^{2}+2}\right)}{\sqrt{y^{2}+2}\left(3y^{2}+1\right)}dy.$$ Let $$J_{2}=\int_{1}^{\infty}\frac{\tan^{-1}\left(\sqrt{y^{2}+2}\right)}{\sqrt{y^{2}+2}\left(3y^{2}+1\right)}dy$$ so we have $$J_{1}-J_{2}=\frac{-\sqrt{5}\pi}{30}\left(2\tan^{-1}\left(\frac{\sqrt{15}}{5}\right)-3\tan^{-1}\left(\frac{1}{\sqrt{15}}\right)\right).\,\,\,\,\,\,(1)$$ Now we observe that, diff respect to $a$ again, $$\int_{0}^{\infty}\frac{\tan^{-1}\left(\sqrt{y^{2}+2}\right)}{\sqrt{y^{2}+2}\left(3y^{2}+1\right)}dy=\int_{0}^{\infty}\int_{0}^{1}\left(\frac{\tan^{-1}\left(a\sqrt{y^{2}+2}\right)}{\sqrt{y^{2}+2}\left(3y^{2}+1\right)}\right)^{'}dadx=\frac{\sqrt{5}\pi}{30}\left(-3\tan^{-1}\left(\frac{\sqrt{15}}{5}\right)+3\tan^{-1}\left(\frac{1}{\sqrt{5}}\right)+\frac{\pi}{2}\right)$$ then $$J_{1}+J_{2}=-\frac{\sqrt{5}\pi}{30}\left(4\tan^{-1}\left(\frac{\sqrt{15}}{5}\right)-3\tan^{-1}\left(\frac{1}{\sqrt{5}}\right)+\frac{2\pi}{5}\right)\,\,\,\,\,\,(2)$$ so if we sum $(1)$ and $(2)$, $$J_{1}=-\frac{\sqrt{5}\pi}{30}\left(\tan^{-1}\left(\frac{\sqrt{15}}{5}\right)-2\tan^{-1}\left(\frac{1}{\sqrt{5}}\right)+\frac{\pi}{5}\right).$$ Finally we have $$I_{1}=\frac{1}{2\sqrt{2}}\left(\frac{\pi}{3}\tan^{-1}\left(\frac{\sqrt{15}}{5}\right)-\frac{2\pi}{3}\tan^{-1}\left(\frac{1}{\sqrt{15}}\right)+2\sqrt{5}J_{1}\right)=\frac{\sqrt{2}\pi^{2}}{60}.$$