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I did a further investigating into this look-alike's Ahmed's integral on my previous post

I found the following results with no proof.

$$\int_{0}^{1}{\arctan^2(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{dx\over x^2+1}={\pi^3\over162}\tag1$$

$$\int_{0}^{1}{\arctan^3(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{dx\over x^2+1}={\pi^{4}\over 648}\tag2$$

$$\int_{0}^{1}{\arctan^4(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{dx\over x^2+1}={\pi^5\over 2430}\tag3$$ I was able to find the simple closed form of the variation of Ahmed's integral but not a general type.

The general form in term of $k$ $$\int_{0}^{1}{\arctan^k(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{2\over x^2+1}dx=\left({\pi\over 3}\right)^{k+1}\cdot{1\over k+1}\tag4$$

Where $k\ge0$(valid for any real number)

I am reluctant to say, but I am not able to prove $(4)$

On my previous post I made an attempt substitution $u=\arctan(x\sqrt{x^2+2})$ but I got it wrong somewhere during my calculation. So there is no point of me trying another attempt at this point. So can any users help to give a hand of proving $(4)$?

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  • $\begingroup$ try induction to prove this $\endgroup$
    – Dr. Sonnhard Graubner
    Jan 8, 2017 at 16:14
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    $\begingroup$ By substituting $u=\arctan\left(x\sqrt{x^2+2}\right)$ you just get $$\int_{0}^{\arctan(\sqrt{3})}{u^k}\cdot{du}=\left[\frac{u^k}{k+1}\right]^{\frac{\pi}{3}}_0=\left({\pi\over 3}\right)^{k+1}\cdot{1\over k+1}.$$ $\endgroup$
    – Olivier Oloa
    Jan 8, 2017 at 16:19
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    $\begingroup$ Generalisation should be straight forward from ur previous post. $\endgroup$
    – Zaid Alyafeai
    Jan 8, 2017 at 16:20

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