# How do you evaluate the integral int 1/(x^2+16)dx from -oo to oo?

Mar 18, 2017

The answer is $= \frac{\pi}{4}$

#### Explanation:

We need

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

$\left(\tan \theta\right) ' = {\sec}^{2} \theta$

We apply the substitution

$x = 4 \tan \theta$

$\mathrm{dx} = 4 {\sec}^{2} \theta d \theta$

Therefore,

We compute the indefinite integral first

$\int \frac{\mathrm{dx}}{{x}^{2} + 16}$

$= \int \frac{4 {\sec}^{2} \theta d \theta}{16 {\tan}^{2} \theta + 16}$

$= \int \frac{4}{16} d \theta$

$= \frac{1}{4} \theta$

$= \frac{1}{4} \arctan \left(\frac{x}{4}\right) + C$

Now,

we compute the boundaries

${\int}_{- \infty}^{+ \infty} \frac{\mathrm{dx}}{{x}^{2} + 16} = {\left[\frac{1}{4} \arctan \left(\frac{x}{4}\right)\right]}_{-} {\infty}^{+ \infty}$

$= \frac{1}{4} \left({\lim}_{x \to + \infty} \arctan \left(\frac{x}{4}\right) - {\lim}_{x \to - \infty} \arctan \left(\frac{x}{4}\right)\right)$

${\lim}_{x \to + \infty} \left(\frac{1}{4} \arctan \left(\frac{x}{4}\right)\right) = \frac{1}{4} {\lim}_{x \to + \infty} \left(\arctan \left(\frac{x}{4}\right)\right)$

$= \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8}$

${\lim}_{x \to - \infty} \left(\frac{1}{4} \arctan \left(\frac{x}{4}\right)\right) = \frac{1}{4} {\lim}_{x \to - \infty} \left(\arctan \left(\frac{x}{4}\right)\right)$

$= - \frac{1}{4} \cdot \frac{\pi}{2} = - \frac{\pi}{8}$

Therefore,

${\int}_{- \infty}^{+ \infty} \frac{\mathrm{dx}}{{x}^{2} + 16} = \frac{\pi}{8} - \left(- \frac{\pi}{8}\right) = \frac{\pi}{4}$

Mar 18, 2017

$\frac{\pi}{4}$

#### Explanation:

$\frac{1}{{x}^{2} + 16} = \frac{i}{8} \left(\frac{1}{x + i 4} - \frac{1}{x - i 4}\right)$ so

$\int \frac{\mathrm{dx}}{{x}^{2} + 16} = \frac{i}{8} \int \left(\frac{1}{x + i 4} - \frac{1}{x - i 4}\right) \mathrm{dx}$

$= \frac{i}{8} \left(\log \left(4 i + x\right) - \log \left(4 i - x\right)\right) = \log {\left(\frac{4 i + x}{4 i - x}\right)}^{\frac{i}{8}}$

but $4 i \pm x = \sqrt{16 + {x}^{2}} {e}^{\pm i \phi}$ where $\phi = \arctan \left(\frac{4}{x}\right)$ so

$\log {\left(\frac{4 i + x}{4 i - x}\right)}^{\frac{i}{8}} = \log \left({\left({e}^{2 i \phi}\right)}^{\frac{i}{8}}\right) = \log \left({e}^{- \frac{\phi}{4}}\right) = - \frac{\phi}{4}$

now

${\int}_{0}^{\infty} \frac{\mathrm{dx}}{{x}^{2} + 16} = {\lim}_{x \to \infty} \left(- \frac{\phi}{4}\right) - {\lim}_{x \to 0} \left(- \frac{\phi}{4}\right) = 0 - \frac{1}{4} \left(- \frac{\pi}{2}\right) = \frac{\pi}{8}$

so

${\int}_{\infty}^{\infty} \frac{\mathrm{dx}}{{x}^{2} + 16} = \frac{\pi}{4}$