Does anyone have any idea on how to evaluate the following generalized Ahmed integral?
$$I(t):=\int_{0}^{1}\frac{\arctan(t\sqrt{2+x^2})}{\sqrt{2+x^2}(1+x^2)}\,dx$$
Here is my attempt:
$$I’(t)=\int_{0}^{1}\frac{1}{1+x^2}\frac{1}{1+2t^2+t^2x^2}\,dx$$
$$=\frac{1}{1+t^2}[\int_{0}^{1}\frac{1}{1+x^2}\,dx-t^2\int_{0}^{1}\frac{1}{1+2t^2+t^2x^2}\,dx]$$
$$=\frac{1}{1+t^2}[\arctan(x)-\frac{t}{\sqrt{1+2t^2}}\arctan(\frac{tx}{\sqrt{1+2t^2}})]_{0}^{1}$$
$$I’(t)=\frac{1}{1+t^2}[\frac{\pi}{4}-\frac{t\arctan(\frac{t}{\sqrt{1+2t^2}})}{\sqrt{1+2t^2}}]$$
$$I(t)=I(t)-I(t\rightarrow{\infty})+\frac{\pi^2}{12}=\frac{\pi^2}{12}-\int_{t}^{\infty}I’(x)\,dx$$
And so
$$I(t)=\frac{\pi^2}{12}-\int_{t}^{\infty}\frac{1}{1+x^2}[\frac{\pi}{4}-\frac{x}{\sqrt{1+2x^2}}\arctan(\frac{x}{\sqrt{1+2x^2}})]\,dx$$
$$=\frac{\pi^2}{12}-\frac{\pi}{4}\arctan(\frac{1}{t})+\int_{t}^{\infty}\frac{x}{(1+x^2)\sqrt{1+2x^2}}\arctan(\frac{x}{\sqrt{1+2x^2}})\,dx$$
Let $$x\longrightarrow{\frac{1}{x}}$$
$$=\frac{\pi^2}{12}-\frac{\pi}{4}\arctan(\frac{1}{t})+\int_{0}^{\frac{1}{t}}\frac{\frac{\pi}{2}-\arctan(\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}}\,dx$$
I’m too lazy to go further because it doesn’t look like it’ll do any good unless $t=1$ which of course is Ahmed’s integral. It seems I have hit a dead end. Does anyone know what to do?
