Evaluate the integral of 1/(2 + 2x^2)^2 from minus infinity to infinity.

Step-by-step solution

1. 1

   Pull the constant out of the bracket first. Since $2+2x^2=2(1+x^2)$,

   $\left(2+2x^2\right)^{2}=4\left(1+x^2\right)^{2}\ \Longrightarrow\ \int_{-\infty}^{\infty}\frac{dx}{\left(2+2x^2\right)^2}=\frac14\int_{-\infty}^{\infty}\frac{dx}{\left(1+x^2\right)^{2}}$

   Doing this before substituting keeps the standard integral recognisable.

2. 2

   Check that the improper integral converges. For large $|x|$ the integrand behaves like $x^{-4}$, and $\int^{\infty}x^{-4}\,dx$ converges, so both tails are finite and the value is well defined.

3. 3

   Substitute $x=\tan\theta$. Then $dx=\sec^2\theta\,d\theta$ and $1+x^2=\sec^2\theta$, so

   $\int\frac{dx}{(1+x^2)^2}=\int\frac{\sec^2\theta}{\sec^4\theta}\,d\theta=\int\cos^2\theta\,d\theta=\frac{\theta}{2}+\frac{\sin 2\theta}{4}$

4. 4

   Take the limits over the whole line. As $x$ runs from $-\infty$ to $\infty$, $\theta$ runs from $-\tfrac\pi2$ to $\tfrac\pi2$, and $\sin 2\theta$ vanishes at both ends:

   $\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^2}=\left[\frac{\theta}{2}+\frac{\sin 2\theta}{4}\right]_{-\pi/2}^{\pi/2}=\frac{\pi}{2}$

5. 5

   Multiply by the constant.

   $\frac14\cdot\frac{\pi}{2}=\frac{\pi}{8}\approx 0.3926991$

6. 6

   Confirm numerically. A midpoint rule over $[-2000,2000]$ with four million subintervals returns $0.39269908168$, against $\pi/8=0.39269908170$ — agreement to ten decimals, and the truncated tails beyond $\pm 2000$ contribute less than $10^{-9}$.