I came across the following integral:
$$\int_{-\infty}^\infty \frac{\sin (2x)}{x^3} \mathrm dx$$
In the textbook, this integral is done using residue theorem by replacing $\sin (2x)$ as the imaginary part of $e^{i2x}$. And the result is $-2\pi$ !
I wonder how the result can be interpreted. This is how the graph of the function look like:
Clearly the integral is divergent. Even if it exist, it should be positive, since the major area is above x axis.
Then what does the -ve value of integral mean ? Or what is it's geometrical interpretation?
Best Answer
It's the finite part integral: $$[x^{-2}] \frac {\sin 2 x} {x^3} = 2, \\ \operatorname{FP} \int_{\mathbb R} \frac {\sin 2x} {x^3} dx = \int_{\mathbb R} \left( \frac {\sin 2x} {x^3} - \frac 2 {x^2} \right) dx = -2 \pi.$$