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:

FullSizeRender (1)

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.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s