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Improper Integrals

∫ 0 ∞ x 1+x 1...

Question

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

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Solution

$We have, I = \int_{0}^{\infty} \frac{x}{(1 + x) (1 + x^{2})} d x Putting x = \tan θ \Rightarrow d x = \sec^{2} θ d θ When x \to 0; θ \to 0 and x \to \infty; θ \to \frac{π}{2} Now, integral becomes$

$I = \int_{0}^{\frac{π}{2}} \frac{\tan θ}{(1 + \tan θ) \sec^{2} θ} \sec^{2} θ d θ = \int_{0}^{\frac{π}{2}} \frac{\tan θ}{1 + \tan θ} d θ = \int_{0}^{\frac{π}{2}} \frac{\frac{\sin θ}{co s θ}}{1 + \frac{\sin θ}{\cos θ}} d θ \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sin θ}{\sin θ + \cos θ} d θ . . . . . (1) \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sin (\frac{π}{2} - θ)}{\sin (\frac{π}{2} - θ) + \cos (\frac{π}{2} - θ)} d θ [∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\cos θ}{\cos θ + \sin θ} d θ \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\cos θ}{\sin θ + \cos θ} d θ . . . . . (2)$

$Adding (1) and (2), we get 2 I = \int_{0}^{\frac{π}{2}} \frac{\sin θ + \cos θ}{\sin θ + \cos θ} d θ \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} d θ \Rightarrow 2 I = \frac{π}{2} \Rightarrow I = \frac{π}{4} ∴ \int_{0}^{\infty} \frac{x}{(1 + x) (1 + x^{2})} d x = \frac{π}{4}$

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