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Does anyone have any idea on how to evaluate the following generalized Ahmed integral?

$$I(t):=\int_{0}^{1}\frac{\arctan(t\sqrt{2+x^2})}{\sqrt{2+x^2}(1+x^2)}\,dx$$

My Attempt:

\begin{aligned} I’(t) &=\int_{0}^{1}\frac{1}{1+x^2}\frac{1}{1+2t^2+t^2x^2}\,dx \\ &=\frac{1}{1+t^2}\bigg[\int_{0}^{1}\frac{1}{1+x^2}\,dx-t^2\int_{0}^{1}\frac{1}{1+2t^2+t^2x^2}\,dx\bigg] \\ &=\frac{1}{1+t^2}\bigg[\arctan(x)-\frac{t}{\sqrt{1+2t^2}}\arctan\bigg(\frac{tx}{\sqrt{1+2t^2}}\bigg)\bigg]_{0}^{1} \\ &=\frac{1}{1+t^2}\bigg[\frac{\pi}{4}-\frac{t\arctan(\frac{t}{\sqrt{1+2t^2}})}{\sqrt{1+2t^2}}\bigg] \end{aligned}

And this implies

$$I(t)=I(t)-I(t\rightarrow{\infty})+\frac{\pi^2}{12}=\frac{\pi^2}{12}-\int_{t}^{\infty}I’(x)\,dx$$

And so

\begin{aligned} I(t) &=\frac{\pi^2}{12}-\int_{t}^{\infty}\frac{1}{1+x^2}\bigg[\frac{\pi}{4}-\frac{x}{\sqrt{1+2x^2}}\arctan\bigg(\frac{x}{\sqrt{1+2x^2}}\bigg)\bigg]\,dx \\ &=\frac{\pi^2}{12}-\frac{\pi}{4}\arctan\bigg(\frac{1}{t}\bigg)+\int_{t}^{\infty}\frac{x}{(1+x^2)\sqrt{1+2x^2}}\arctan\bigg(\frac{x}{\sqrt{1+2x^2}}\bigg)\,dx \end{aligned}

Then let $x\longrightarrow{\frac{1}{x}}$, and we have

$$I(t)=\frac{\pi^2}{12}-\frac{\pi}{4}\arctan\bigg(\frac{1}{t}\bigg)+\int_{0}^{\frac{1}{t}}\frac{\frac{\pi}{2}-\arctan(\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}}\,dx$$

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?

Yunxuan Zhang
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1 Answers1

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\begin{align} I(a)=&\int_{0}^{1}\frac{\arctan\left(a\sqrt{2+x^2}\right)}{(1+x^2) \sqrt{2+x^2}}\,dx\\ =& \int_0^1 \int_0^x \frac{ax}{(1+x^2)[x^2+a^2(x^2+2)y^2]}dy\ dx\\ =& \int_0^1 \int_y^1 \frac{ax}{(1+x^2)[x^2+a^2(x^2+2)y^2]}dx\ dy\\ =& \ \frac12\int_0^1 \frac 1{\frac1a-a y^2}\ln \frac{(1+y^2)(1+3a^2y^2)}{2y^2(1+2a^2+a^2y^2)}dy \end{align} Utilize \begin{align} &H(p,q)= \int_0^1 \frac{\ln\frac{p^2+y^2}{p^2+1}}{q-\frac1q y^2}dy \\&=\Re \left[\text{Li}_2 \left( \frac{q-1}{q+i p}\right)-\text{Li}_2 \left( \frac{q+1}{q-i p} \right)\right] -\frac14\ln\frac{1+p^2}{q^2+p^2}\ln\bigg(\frac{q-1}{q+1} \bigg)^2 \end{align} to obtain \begin{align} I(a)=&\ \frac12\bigg[H\bigg(1,\frac1a\bigg)-H\bigg(0,\frac1a\bigg)+ H\bigg(\frac1{\sqrt3a},\frac1a\bigg)-H\bigg(\frac{\sqrt{1+2{a^2}}}a,\frac1a\bigg)\bigg]\\ =& \ \frac12\Re \bigg[\text{Li}_2 \left( \frac{1-a}{1+i a}\right)-\text{Li}_2 \left( \frac{1+a}{1-i a} \right)\bigg]\\ &\> -\frac12\Re \bigg[\text{Li}_2 \left( {1-a}\right)-\text{Li}_2 \left( {1+a}\right)\bigg]\\ &\> +\frac12\Re \bigg[\text{Li}_2 \left( \frac{1-a}{1+i /\sqrt3}\right)-\text{Li}_2 \left( \frac{1+a}{1-i /\sqrt3} \right)\bigg]\\ &\> -\frac12\Re \bigg[\text{Li}_2 \bigg( \frac{1-a}{1+i \sqrt{1+2a^2}}\bigg)-\text{Li}_2 \bigg( \frac{1+a}{1-i \sqrt{1+2a^2}} \bigg)\bigg]\\ \end{align} of which the Ahmed integral is a special case, i.e. \begin{align} &I(1)=\int_{0}^{1}\frac{\arctan\sqrt{2+x^2}}{(1+x^2) \sqrt{2+x^2}}\,dx\\ =& \ \frac12 \Re\bigg[-\text{Li}_2 \left( \frac{2}{1-i} \right)+ \text{Li}_2 \left(2\right) -\text{Li}_2 \bigg( \frac{2}{1-\frac i{\sqrt3}}\bigg)+ \text{Li}_2 \bigg( \frac{2}{1-i\sqrt3} \bigg)\bigg]\\ =&\ \frac12\bigg( -\frac{\pi^2}{16}+ \frac{\pi^2}{4}-\frac{\pi^2}{9} + \frac{\pi^2}{36}\bigg)=\frac{5\pi^2}{96} \end{align}

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