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Ahmed’s Integral

Prove that

    \[\int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{\mathrm{d}x}{x^{2}+1} = \frac{5\pi^{2}}{96}\]

Solution

Consider the function \displaystyle{f(t) = \int_{0}^{1} \frac{\arctan \left(t\sqrt{2+x^2} \right)}{(1+x^2)\sqrt{2+x^2}} \, \mathrm{d}x}. Differentiating with respect to t we have that:

    \begin{align*} f'(t) &= \frac{\mathrm{d} }{\mathrm{d} t} \int_{0}^{1} \frac{\arctan \left(t\sqrt{2+x^2} \right)}{(1+x^2)\sqrt{2+x^2}} \, \mathrm{d}x \\ &= \int_{0}^{1} \frac{\partial }{\partial t} \frac{\arctan \left(t\sqrt{2+x^2} \right)}{(1+x^2)\sqrt{2+x^2}} \, \mathrm{d}x \\ &= \int_{0}^{1} \frac{\mathrm{d}x}{\left ( 1+x^2 \right ) \left ( 1+2t^2 + t^2 x^2 \right )}\\ &= \int_{0}^{1} \frac{\mathrm{d}x}{\left ( 1+t^2 \right )\left ( 1+x^2 \right )} - \int_{0}^{1} \frac{t^2}{\left ( 1+t^2 \right ) \left ( 1+2t^2+t^2 x^2 \right )} \, \mathrm{d}x \\ &= \frac{\pi}{4} \cdot \frac{1}{1+t^2} - \frac{t}{\left ( 1+t^2 \right ) \sqrt{1+2t^2}} \arctan \frac{t}{\sqrt{1+2t^2}} \end{align*}

We integrate the last equation from 1 to \infty. Thus,

    \begin{align*} \int_{1}^{\infty} f'(t) \, \mathrm{d}t &= \int_1^\infty \left (\frac{\pi}{4} \cdot \frac{1}{1+t^2} - \frac{t}{\left ( 1+t^2 \right ) \sqrt{1+2t^2}} \arctan \frac{t}{\sqrt{1+2t^2}} \right ) \, \mathrm{d}t \\ &=\frac{\pi}{4} \left ( \frac{\pi}{2} - \frac{\pi}{4} \right ) - \int_1^\infty \frac{t}{\left ( 1+t^2 \right ) \sqrt{1+2t^2}} \arctan \frac{t}{\sqrt{1+2t^2}} \, \mathrm{d}t \\ &= \frac{\pi^2}{16} - \int_1^\infty \frac{t}{\left ( 1+t^2 \right ) \sqrt{1+2t^2}} \arctan \frac{t}{\sqrt{1+2t^2}} \, \mathrm{d}t \end{align*}

However,

    \begin{align*} \lim_{t \rightarrow +\infty} f(t) &= \lim_{t \rightarrow +\infty} \int_{0}^{1} \frac{\arctan \left(t\sqrt{2+x^2} \right)}{(1+x^2)\sqrt{2+x^2}} \, \mathrm{d}x \\ &= \int_{0}^{1} \lim_{t \rightarrow +\infty} \frac{\arctan \left(t\sqrt{2+x^2} \right)}{(1+x^2)\sqrt{2+x^2}} \, \mathrm{d}x \\ &= \frac{\pi}{2} \int_{0}^{1} \frac{\mathrm{d}x}{\left ( 1+x^2 \right )\sqrt{2+x^2}}\\ &\!\!\!\!\!\!\!\overset{\text{Lemma 2}}{=\! =\! =\! =\! =\! =\! } \frac{\pi}{2} \cdot \frac{\pi}{6} \\ &= \frac{\pi^2}{12} \end{align*}

Hence the last equation gives

(1)   \begin{equation*} \frac{\pi^2}{12} - f(1) = \frac{\pi^2}{16} - \int_1^\infty \frac{t}{\left ( 1+t^2 \right ) \sqrt{1+2t^2}} \arctan \frac{t}{\sqrt{1+2t^2}} \, \mathrm{d}t \end{equation*}

Suffice to calculate the integral. Applying the change of variables t \mapsto \frac{1}{t} we have:

    \begin{align*} \mathcal{J} &= \int_1^\infty \frac{t}{\left ( 1+t^2 \right ) \sqrt{1+2t^2}} \arctan \frac{t}{\sqrt{1+2t^2}} \, \mathrm{d}t \\ &= \int_{1}^{\infty} \frac{1}{1+\left ( \frac{1}{t} \right )^2 \sqrt{2+\frac{1}{t^2}}} \arctan \frac{1}{\sqrt{2 + \frac{1}{t^2}}} \frac{\mathrm{d}t}{t^2} \\ &\!\!\!\!\overset{t \mapsto 1/t}{=\! =\! =\! =\!} \int_{0}^{1} \frac{1}{\left (1+t^2 \right )\sqrt{2+t^2}} \arctan \frac{1}{\sqrt{2+t^2}} \, \mathrm{d}t\\ &= \int_{0}^{1}\frac{1}{\left (1+t^2 \right )\sqrt{2+t^2}} \left ( \frac{\pi}{2} - \arctan \sqrt{2+t^2}\right ) \, \mathrm{d}t \\ &=\frac{\pi}{2} \int_{0}^{1} \frac{\mathrm{d}t}{\left (1+t^2 \right )\sqrt{2+t^2}} - f(1) \\ &= \frac{\pi}{2} \cdot \frac{\pi}{6} - f(1) \\ &= \frac{\pi^2}{12} - f(1) \end{align*}

Going back at (1) we have that:

    \begin{align*} \frac{\pi^2}{12} - f(1) = \frac{\pi^2}{16} - \left ( \frac{\pi^2}{12} - f(1) \right ) &\Leftrightarrow \frac{\pi^2}{12} - f(1) = \frac{\pi^2}{16} - \frac{\pi^2}{12} + f(1) \\ &\Leftrightarrow 2f(1) = \frac{5\pi^2}{48} \\ &\Leftrightarrow \boxed{f(1) = \frac{5 \pi^2}{96}} \end{align*}

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