Determine how large the number a has to be?

Question

This is what i've done so far:

1. i converted to limit notation lim as t goes to infinity of integral from a to t of 1/t^2+1 dt

2. lim as to goes to infinity [arctan(t)] from a to t

3. (lim as t goes to infinity of arctan(t))-arctan(a)

4. tan(pi/2)=infinity arctan(infinity) = pi/2

5. (pi/2-arctan(a))<0.007

6. arctan(a)>pi/2-0.007

7. a>tan(pi/2-0.007)

8. a>143

Is my answer correct or have i done something wrong? I couldnt really figure out the codes to make this a little organized, sorry about that

Answer 1

You'r answer is correct. If it tells you are wrong, try to add a digit or two. However I want to present two different (perhaps easier?) solutions below.

Solution 1

For simplicity sake let $\varepsilon = 0.007\ll1$. Now we have that $$ \int_a^\infty \frac{\mathrm{d}x}{1+x^2}< \int_a^\infty \frac{\mathrm{d}x}{x^2} = \frac{1}{a}\,. $$ Hence $$ \int_a^\infty \frac{\mathrm{d}x}{1+x^2}< \varepsilon \ \Rightarrow \ \frac{1}{a} < \varepsilon \ \Rightarrow \ a > \frac{1}{\varepsilon }\,. $$

Solution 2

A quick integration using $\int \frac{\mathrm{d}x}{1+x^2}=\arctan x$. Gives $$ \int_a^\infty \frac{\mathrm{d}x}{1+x^2} = \arctan(\infty) - \arctan(a) = \frac{\pi}{2} - \arctan \,.\tag{1} $$ This was the same result you obtained. However we can write it in a slightly easier fashion using the following equation $$ \arctan x + \arctan \frac1x = \frac\pi2\,. \tag{2} $$ Which for all $x \in \mathbb{R}$. This can be proven by noting that the derivative is a constant, a geometric argument, or $-$ as we will see shortly $-$ integration. Using $(2)$ in $(1)$ gives now $$ \int_a^\infty \frac{\mathrm{d}x}{1+x^2} = \arctan \frac1a < \varepsilon\,. $$ In other words $$ a > \frac1{\tan \varepsilon}\,. $$ By using the taylor expansion $\tan x \approx x+\frac13x^3+\frac2{15}x^5+O(x^7)$ we get the same result as above. Eg let $\tan x \sim x$ whenever $x\ll1$.

Addendum

Proof: Here is one way to prve $(2)$ using integration. Let us use the substitution $x \mapsto 1/t$, then $\mathrm{d}x = -\mathrm{d}t/t^2$. When $x \to \infty$ then $t\to 0$ and $x \to a$ then $t \to 1/a$. Hence

$$ \int_a^{\infty} \frac{\mathrm{d}x}{1+x^2} = \int_{1/a}^0 \frac{-\mathrm{d}t/t^2}{1+(1/t)^2} = \int_{0}^{1/a} \frac{\mathrm{d}t}{1+t^2} = \arctan \frac1a \tag{3}\,. $$

Since $\arctan 0 = 0$ because $\tan 0 = 0$. We have already shown that $\int_a^{\infty} \frac{\mathrm{d}x}{1+x^2}=\frac\pi2-\arctan a$, see $(1)$. Comparing equation $(1)$ and $(3)$ completes the proof. $\square$

Note that we can generalize the result as

$$ \int_0^{1/t} \frac{\mathrm{d}x}{1+x^2} = \int_t^{\infty} \frac{\mathrm{d}x}{1+x^2}\,, $$

for every $t$. But here this is more a fun fact than a useful proposition.