prove-0-arctan-x-2-2-x-2-1-x-2-2-dx-pi-2-12-

Question Number 210820 by Ghisom last updated on 19/Aug/24

prove

\underset{0}{\int^{\infty}} \frac{\arctan \sqrt{x^{2} + 2}}{(x^{2} + 1) \sqrt{x^{2} + 2}} d x = \frac{π^{2}}{12}

Answered by BHOOPENDRA last updated on 20/Aug/24

I = \int_{0}^{\infty} \frac{a r c t a n \sqrt{x^{2} + 2}}{(x^{2} + 1) (\sqrt{x^{2} + 2)}} d x

U s i n g F e y m a n i n t e g r a t i o n

I (α) = \int_{0}^{\infty} \frac{a r c t a n α (\sqrt{x^{2} + 2)}}{(x^{2} + 1) (\sqrt{x^{2} + 2}} d x

I^{'} (α) = \int_{0}^{\infty} \frac{1}{(x^{2} + 1) (\sqrt{x^{2} + 2})} \partial_{α} a r c t a n (α \sqrt{x^{2} + 2})

= \int_{0}^{\infty} \frac{1}{(x^{2} + 1) (α^{2} x^{2} + 2 α^{2} + 1)} d x

= \int_{0}^{1} [\int_{0}^{\infty} \frac{\frac{1}{1 + α^{2}}}{(1 + x^{2})} d x - \int_{0}^{\infty} \frac{α^{2} / (1 + α^{2})}{(α^{2} x^{2} + x^{2} + (2 α^{2} + 1)} d x] d α

= \int_{0}^{1} \frac{1}{(1 + α^{2})} [\int_{0}^{\infty} \frac{1}{1 + x^{2}} d x - \frac{α^{2}}{α^{2}} \int_{0}^{\infty} \frac{d x}{x^{2} + {(\sqrt{\frac{2 α^{2} + 1}{α^{2}}})}^{2}}] d α

= \int_{0}^{1} \frac{1}{1 + α^{2}} [\frac{π}{2} - (\frac{π}{2} \times \frac{α}{\sqrt{2 α^{2} + 1}})]

= \frac{π}{2} [\int_{0}^{1} \frac{1}{1 + α^{2}} d α - \int_{0}^{1} \frac{α}{(1 + α^{2}) (\sqrt{2 α^{2} + 1})} d α]

= \frac{π}{2} [\frac{π}{4} - \frac{π}{12}]

= \frac{π \times 2 π}{2 \times 12}

= \frac{π^{2}}{12}

Commented by Ghisom last updated on 20/Aug/24

thank you

Answered by BHOOPENDRA last updated on 20/Aug/24

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