solution-with-residu-theorem-0-x-2-x-4-2x-2-2-dx-

Question Number 163854 by amin96 last updated on 11/Jan/22

solution with residu theorem

\int_{0}^{\infty} \frac{x^{2}}{x^{4} + 2 x^{2} + 2} dx = ?

Answered by Ar Brandon last updated on 11/Jan/22

I = \int_{0}^{\infty} \frac{x^{2}}{x^{4} + 2 x^{2} + 2} d x = \frac{1}{2} \int_{0}^{\infty} \frac{(x^{2} + \sqrt{2}) + (x^{2} - \sqrt{2})}{x^{4} + 2 x^{2} + 2} d x

= \frac{1}{2} \int_{0}^{\infty} \frac{x^{2} + \sqrt{2}}{x^{4} + 2 x^{2} + 2} d x + \frac{1}{2} \int_{0}^{\infty} \frac{x^{2} - \sqrt{2}}{x^{4} + 2 x^{2} + 2} d x

= \frac{1}{2} \int_{0}^{\infty} \frac{1 + \frac{\sqrt{2}}{x^{2}}}{x^{2} + 2 + \frac{2}{x^{2}}} d x + \frac{1}{2} \int_{0}^{\infty} \frac{1 - \frac{\sqrt{2}}{x^{2}}}{x^{2} + 2 + \frac{2}{x^{2}}} d x

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

= \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{d u}{u^{2} + (2 + 2 \sqrt{2})} + \frac{1}{2} \int_{+ \infty}^{+ \infty} \frac{d v}{v^{2} + 2 - 2 \sqrt{2}}

= \frac{1}{2} \cdot \frac{1}{\sqrt{2 + 2 \sqrt{2}}} {[\arctan (\frac{u}{\sqrt{2 + 2 \sqrt{2}}})]}_{- \infty}^{+ \infty} = \frac{π}{2 \sqrt{2 + 2 \sqrt{2}}}

Commented by peter frank last updated on 11/Jan/22

great

Commented by amin96 last updated on 11/Jan/22

g r e a t s i r . c o r r e c t a n s w e r

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