# Thread: Improper Integral - # 2

1. ## Improper Integral - # 2

$\displaystyle \int_{0}^{\infty} \dfrac{dx}{x^{2} + 1}$

$\displaystyle x = (1)\tan\theta$

$\displaystyle dx = \sec^{2}\theta d\theta$

$\displaystyle (\tan\theta)^{2} + 1$

$\displaystyle \tan^{2}\theta + 1 = sec^{2}\theta$

$\displaystyle \int_{0}^{\infty} \dfrac{\sec^{2}\theta}{\sec^{2}\theta}$ - ??? Simplify further or take integral? Simplification seems to just make a 1 Maybe trig substitution wasn't the way to go.

2. ## Re: Improper Integral - # 2

Hey Jason76.

If you use a trig substitution, you need to remember to change the limits when you do this. Note that your limits should change to [0,pi/2) if you use these kinds of substitutions.

3. ## Re: Improper Integral - # 2

Surely you can see that \displaystyle \begin{align*} \frac{\sec^2{\left( \theta \right)}}{\sec^2{ \left( \theta \right) }} = 1 \end{align*}...

4. ## Re: Improper Integral - # 2

In this case the answer would be

$\displaystyle x$ evaluated at $\displaystyle 0$ (lower bound) and $\displaystyle 2 \pi$ upper bound

$\displaystyle [[x] - [x]]$

$\displaystyle [2\pi - 0 ]= 2\pi$

However, I think this problem is another $\displaystyle \arctan(x)$ one as the integral of $\displaystyle \dfrac{1}{x^{2} + 1}$ is $\displaystyle \arctan(x)$

If your going the $\displaystyle \arctan(x)$ route

$\displaystyle [\arctan(b)] - [\arctan(0)]$

$\displaystyle [\arctan(b)] - [0]$

$\displaystyle \lim b \rightarrow \infty[\arctan(b)] - 0$ indeterminate

Using L hopitals

$\displaystyle \lim b \rightarrow\infty [\dfrac{1}{(\infty)^{2} + 1}]= 0$

5. ## Re: Improper Integral - # 2

The answer in the back of the book, though, is $\displaystyle \dfrac{\pi}{2}$

They must have went the trig substitution route.

$\displaystyle \int_{0}^{\infty} \dfrac{dx}{x^{2} + 1}$

$\displaystyle \int_{0}^{\pi/2\dfrac{dx}{x^{2} + 1}$

$\displaystyle x = (1)\tan\theta$

$\displaystyle dx = \sec^{2}\theta d\theta$

$\displaystyle (\tan\theta)^{2} + 1$

$\displaystyle \tan^{2}\theta + 1 = sec^{2}\theta$

$\displaystyle \int_{0}^{\infty} \dfrac{\sec^{2}\theta}{\sec^{2}\theta}$

$\displaystyle \int_{0}^{\pi/2} dx$

$\displaystyle = x$ evaluated at $\displaystyle 0$ (lower bound) and $\displaystyle \pi/2$ (upper bound)

$\displaystyle [(\pi/2) - (0)] = \pi/2$ - correct answer on homework.

6. ## Re: Improper Integral - # 2

fyi ... $\displaystyle \frac{d}{dx} \arctan{x} = \frac{1}{1+x^2}$ ... you should know this fact.

so, $\displaystyle \int_0^\infty \frac{dx}{1+x^2} = \lim_{b \to \infty} \left[\arctan{x}\right]_0^b = \lim_{b \to \infty} \left[\arctan(b) - \arctan(0)\right] = \frac{\pi}{2}$

btw, you are expected to be familiar with the graph of the arctangent function and its end behavior.

7. ## Re: Improper Integral - # 2

Originally Posted by skeeter
fyi ... $\displaystyle \frac{d}{dx} \arctan{x} = \frac{1}{1+x^2}$ ... you should know this fact.

so, $\displaystyle \int_0^\infty \frac{dx}{1+x^2} = \lim_{b \to \infty} \left[\arctan{x}\right]_0^b = \lim_{b \to \infty} \left[\arctan(b) - \arctan(0)\right] = \frac{\pi}{2}$

btw, you are expected to be familiar with the graph of the arctangent function and its end behavior.
$\displaystyle \dfrac{\pi}{2}$ would have came as the answer using either method.

The limit of $\displaystyle \arctan(x)$ as it approaches infinity is $\displaystyle \pi/2$