# Convergence of an improper integral - II

I'm not able to find the value of:$$\int_a^\infty \frac{1}{x^2+1}dx, a>0$$ What I can do?

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## 4 Answers

$$I = \int_a^\infty \dfrac{dx}{x^2+1}$$ 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 $b\to\infty$.

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