I'm not able to find the value of:$ \int_a^\infty \frac{1}{x^2+1}dx, a>0 $ What I can do?
Convergence of an improper integral - II
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integration
convergence-divergence
improper-integrals
4 Answers
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$I = \int_a^\infty \dfrac{dx}{x^2+1}$ Set $x = \tan(t)$. This gives us that $dx = \sec^2(t) \, dt$. Also, recall that $\tan^2(t) + 1 = \sec^2(t)$. Hence, $I = \int_{\arctan(a)}^{\pi/2} \, dt = \dfrac{\pi}2 - \arctan(a) = \text{arccot}(a)$
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The answer is $ \int_{0}^{\infty}\frac{1}{1+x^2}dx=\frac{\pi}{2}-\tan^{-1}a. $
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$ \int_a^\infty \frac{1}{x^2+1}dx=\tan^{-1}x\mid_{a}^\infty =\frac{\pi}2-\tan^{-1}a=\cot^{-1}a$
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Try a trigonometric substitution $x=\tan\theta$. That should allow you to get an antiderivative. Specifically, $\tan^{-1}x$. Apply Fundamental Theorem of Calculus, and take the limit of $\tan^{-1}b-\tan^{-1}a$ as $b\to\infty$.