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Another integral

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#1 h Mar 9, 2025, 10:16 AM

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$I = \int_{0}^{\frac{1}{\sqrt{3}}} \frac{u \arctan(u)}{(1 - u^2) \sqrt{1 - 2 u^2}} \, du$

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#2 h Tuesday at 8:51 AM

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#3 h Tuesday at 12:43 PM • 2 Y

Y by Mathzeus1024, Alphaamss

Martin.s wrote:

$I = \int_{0}^{\frac{1}{\sqrt{3}}} \frac{u \arctan(u)}{(1 - u^2) \sqrt{1 - 2 u^2}} \, du$

Let $I_t= \int_0^{\frac 1{\sqrt 3}} \frac{x \arctan tx}{(1 - x^2) \sqrt{1 - 2x^2}}\mathrm dx$ , with $I_0=0, I_1=I.$
$\begin{align*} &\ I_t'=\int_0^{\frac 1{\sqrt 3}} \frac x{(1 - x^2)(1+t^2x^2)\sqrt{1 - 2x^2}}\mathrm dx\\ \xRightarrow{\hspace{25pt}}&\ I_t'= \frac{\arctan\frac x{\sqrt{1 - 2x^2}}}{t^2+ 1} - \frac{\arctan\frac{x\sqrt{t^2 + 2}}{\sqrt{1 - 2x^2}}}{(t^2 + 1) \sqrt{t^2 + 2}}\Bigg|_0^{\frac 1{\sqrt3}}=\frac{\pi}{4t^2+4}-\frac{\arctan\sqrt{t^2 + 2}}{(t^2+1)\sqrt{t^2 + 2}}.\\ \xRightarrow{\hspace{25pt}}&\ I_1=\int_0^1 \frac{\pi}{4t^2+4}-\underbrace{\frac{\arctan\sqrt{t^2 + 2}}{(t^2+1)\sqrt{t^2 + 2}}}_{J} \mathrm dt\\ \color{blue}\xRightarrow{\hspace{25pt}}&\ \color{blue}I=\frac{\pi^2}{16}-\frac{5\pi^2}{96}=\frac{\pi^2}{96}. \end{align*}$ Note: $J$ is the well-known Ahmed's Integral.

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