Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f(x)$ be non-negative and decreasing for $ x > 0$. Suppose that $\int_0^\infty f(x)dx < \infty$. Let $g(x) = \sum_{n=1}^\infty f(2^nx)$. How do I prove that $\int_0^\infty f(x) dx = \int_0^\infty g(x) dx$?

share|cite|improve this question
up vote 2 down vote accepted

The conditions imposed on $f$ mean that we can swap the sum and integral signs to get

$$\int_0^{\infty} g(x)\, dx = \sum_{n=1}^{\infty} \int_0^{\infty} f(2^nx)\, dx$$

Substituting $y=2^nx$ into the integrals on the RHS gives

$$\int_0^{\infty} g(x)\, dx = \sum_{n=1}^{\infty} \int_0^{\infty} 2^{-n}f(y)\, dy$$

We can then use linearity properties of the integral and sum on the RHS to show that this is equal to $\int_0^{\infty} f(x)\, dx$, as required.

[Hint: think geometric series]

share|cite|improve this answer

By Fubini for non-negative functions, we have $$\int_0^{+\infty}g(x)dx=\sum_{n=1}^{+\infty}\int_0^{+\infty}f(2^nx)dx.$$ In each integral, do the substitution $t=2^nx$, $dt=2^ndx$ to get $$\int_0^{+\infty}g(x)dx=\sum_{n=1}^{+\infty}2^{-n}\int_0^{+\infty}f(x)dx,$$ and we are done.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.