0

Where is the error in evaluating this integration? $$\int^{\infty}_{-\infty} \frac{(-1)^{2x}}{(1+x^2)^2} dx$$

I know $$(-1)^{2x}=1$$

therfore: $$I=\int^{\infty}_{-\infty} \frac{(-1)^{2x}}{(1+x^2)^2} dx=\int^{\infty}_{-\infty} \frac{1}{(1+x^2)^2} dx=\frac{\pi}{2}$$

But using wolfram alpha $$I = 0.0213... $$

Delta
  • 1

2 Answers2

1

Your error is assuming that $x$ is an integer. If $x = 1.5$, we have $-1^3 = -1$. $$\int^{\infty}_{-\infty} \frac{1}{(1+x^2)^2} dx=\frac{\pi}{2}$$ The integral above which you found is correct though :)

wasu
  • 109
0

The integral in question is $$\int_{-\infty}^{\infty}\frac{\cos(2\pi x)+i\sin(2\pi x)}{(1+x^2)^2}dx.$$

The imaginary part vanishes due to oddity of this integral.

The real part, $$\int_{-\infty}^{\infty}\frac{\cos(2\pi x)}{(1+x^2)^2}dx \approx 0.02,$$

according to WA.

  • 1
    I believe it's $2\pi$ because of $-1 = \cos(\pi i)$ and DeMoivre's formula. Does that make sense? –  Apr 25 '24 at 20:06