How large should $a$ be so that $\int_a^{\infty} \frac{dx}{1+x^2} < \frac{1}{1000}$

Question

I want to solve this without using calculator.

Answer 1

Hint: The integral is a standard trigonometric integral.

Extra hint: The arctan of infinity is $\pi/2$.

Last hint: at small values of $x$, $\cos(x)$ becomes close to 1, and $\sin(x)$ is close to $x$.

Answer 2

Use that $$ \frac{1}{1+x^2}<\frac{1}{x^2}. $$

Answer 3

$\displaystyle\int_a^{+\infty}\dfrac{1}{1+x^2}=\dfrac{\pi}{2}-\tan^{-1}(a)$, so the inequality becomes: $\dfrac{\pi}{2}-\tan^{-1}(a)\lt 10^{-3}$, so: $a\gt\cot\left(\dfrac{1}{1000}\right)$

Answer 4

Ok... I not sure if this is right, if anyone points out the flaws that would help me as well! Took so long to type up using this new math formatting!

The curve $$ 1/{1+x^2} $$

gets closer to $$ \frac {1}{x^2} $$ as x gets larger, so if you want an approximation this might work $$ \int^{\infty} _a \frac {1}{x^2}dx = \frac{-1}{\infty} - \frac{-1}a $$ $$ \frac{-1}{\infty} - \frac{-1}a < \frac1{1000} $$ $$ \frac1a<\frac1{1000} $$ $$ a>1000 $$