# Change of Integral Limits for Even Functions

For fun I’m enrolled in an online computational finance certificate at UW. In one of my homework problems I wanted to use the following fact about the integral of single variable, even functions: $\int_{-\infty}^{-x} f(s)ds = \int_{x}^{\infty} f(s)ds$

If it’s been a few years since you’ve taken calculus that may not make much sense, but trust me when I tell you that it’s analytically obvious, especially when looking at functions graphically, as this terrible hand drawn image shows: Intuitively, we know the two red areas are the same, so it seems we should be able to interchange the limits as I described above. Indeed, playing around in Mathematica suggests that this is true. However, I could not find a proof or theorem for this online so perhaps it is rarely used. I decided to prove it myself:

Original equation: $\int_{s=-\infty}^{s=-x} f(s)ds$

Use u-substitution with $-u=s$ and $-du=ds$ : $= \int_{u=\infty}^{u=x} -f(-u)du$

Bring minus sign outside integral: $= -\int_{u=\infty}^{u=x} f(-u)du$

Use the fact that $-\int_{b}^{a} f(x)dx = \int_{a}^{b} f(x)dx$ : $= \int_{u=x}^{u=\infty} f(-u)du$

By assumption $f$ is even so $f(-u)=f(u)$ : $= \int_{u=x}^{u=\infty} f(u)du$

Rewrite improper integral: $= \lim_{t \to \infty} \int_{x}^{t} f(u)du$

By Fundamental Theorem of Calculus: $= \lim_{t \to \infty} [F(t)] - F(x)$

By Fundamental Theorem of Calculus: $= \lim_{t \to \infty} \int_{x}^{t} f(s)ds$ $= \int_{x}^{\infty} f(s)ds$
Which is exactly the result we were trying to obtain.