Simple looking integral #2

We shall make a simple substitution here.

Let $x = \tan(\theta)$

Hence:

$dx = \sec^2(\theta) d\theta \Rightarrow \frac{dx}{1+x^2} = d\theta$

The limits correspondingly change to $\frac{-\pi}{2} \text{ \& } \frac{\pi}{2}$

$I = \displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{d\theta}{(\sec^2{\theta})^2} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta \cos^4\theta$

$I = \displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta \cos^4\theta = \dfrac{3\pi}{8}$

Note

If you want to evaluate $I = \displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta \cos^4\theta$ take $\cos^4\theta = ( \dfrac{1 + \cos2\theta}{2} )^2$

Well, I will use Contour integration here.

Let's just assume a contour $C$ in the argand plane as a semicircle with radius $a$ with positive complex axis in the circle, then the following will be our integral

$\displaystyle \int_{-a}^{a}{\frac{dz}{{({z}^{2}+1)}^{3}}}$

By Cauchy's theorem(as in the contour they all will homotopic)-

$\displaystyle \int_{C}{\frac{dz}{{({z}^{2}+1)}^{3}}} = \int_{-a}^{a}{\frac{dz}{{({z}^{2}+1)}^{3}}} + \int_{arc}{\frac{dz}{{({z}^{2}+1)}^{3}}}$

$\displaystyle \int_{-a}^{a}{\frac{dz}{{({z}^{2}+1)}^{3}}} = \int_{C}{\frac{dz}{{({z}^{2}+1)}^{3}}} - \int_{arc}{\frac{dz}{{({z}^{2}+1)}^{3}}}$

First, let's compute our line integral first

By Cauchy's General Formula

$\displaystyle {f}^{(n)}({z}_{0}) = \frac{n!}{2\pi\iota}\int_{C}{\frac{f(z)}{{(z-{z}_{0})}^{n+1}}}$

Now, let's try to make our integral in this form

We can write it as

$\displaystyle \int_{C}{\frac{\frac{1}{{(z+i)}^{3}}}{{(z-i)}^{3}}}$ , Then $f(z) = \frac{1}{{(z+i)}^{3}}, {z}_{0} = i$

Why we write like this? There is one more possibility, why we didn't chose that? Because the function $f(z)$ is not analytic in the contour we have chosen then.

As a result of substituting,

$\displaystyle \int_{C}{\frac{dz}{{({z}^{2}+1)}^{3}}} = \frac{3\pi}{8}$

Just substitute in the Cauchy's Formula, differentiate twice then put $z=i$ , you are done!

Hence,

$\displaystyle \int_{-a}^{a}{\frac{dz}{{({z}^{2}+1)}^{3}}} = \frac{3\pi}{8} - \int_{arc}{\frac{dz}{{({z}^{2}+1)}^{3}}}$

For our 2nd integral, use Estimation Lemma or ML inequality,

$\displaystyle \left|\int_{arc}{\frac{dz}{{({z}^{2}+1)}^{3}}}\right| \le \frac{a\pi}{{({a}^{2}-1)}^{3}}$

As $\displaystyle a \Rightarrow \infty, \frac{a\pi}{{({a}^{2}-1)}^{3}} \Rightarrow 0$

Therefore, $\displaystyle \int_{arc}{\frac{dz}{{({z}^{2}+1)}^{3}}} \Rightarrow 0$

As a result, we are done with our answer,

$\displaystyle \int_{-\infty}^{\infty}{\frac{dx}{{({x}^{2}+1)}^{3}}} =$

$\displaystyle \frac{3\pi}{8}$