An $\arctan$ integral
- Tolaso J Kos
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An $\arctan$ integral
Let $a \in \mathbb{R} \setminus \{0\}$. Prove that:
$$\int_0^\infty\frac{x^{a-1}\arctan x}{(x^a+1)^2}\, {\rm d}x=\frac{\pi}{4|a|}$$
$$\int_0^\infty\frac{x^{a-1}\arctan x}{(x^a+1)^2}\, {\rm d}x=\frac{\pi}{4|a|}$$
Imagination is much more important than knowledge.
Re: An $\arctan$ integral
If we use parts on this critter by letting
$u=\tan^{-1}(x), \;\ dv=\frac{x^{a-1}}{(1+x^{a})^{2}}dx$, then the resulting integral is the famous:
$$\frac{1}{|a|}\int_{0}^{\infty}\frac{1}{(x^{2}+1)(x^{a}+1)}dx$$
This integral resolves to $\frac{\pi}{4}$ regardless of the value of 'a'.
It is related to the other famous:
$$\int_{0}^{\frac{\pi}{2}}\frac{1}{\tan^{a}(x)+1}dx=\frac{\pi}{4}$$
$u=\tan^{-1}(x), \;\ dv=\frac{x^{a-1}}{(1+x^{a})^{2}}dx$, then the resulting integral is the famous:
$$\frac{1}{|a|}\int_{0}^{\infty}\frac{1}{(x^{2}+1)(x^{a}+1)}dx$$
This integral resolves to $\frac{\pi}{4}$ regardless of the value of 'a'.
It is related to the other famous:
$$\int_{0}^{\frac{\pi}{2}}\frac{1}{\tan^{a}(x)+1}dx=\frac{\pi}{4}$$
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