Originally Posted by

**redsoxfan325** I don't believe this is necessarily true. There are some weird and very contrived functions that are unbounded but still Riemann integrable on the real line, such as:

$\displaystyle f(x)=\left\{\begin{array}{lr}0:&0\leq x<1\\-4n^7(x-n)(x-(n+n^{-3})):&x\in[n,n+n^{-3}]\\0:& otherwise\end{array}\right\}$ for $\displaystyle n\in\mathbb{N}$.

The maximum value of $\displaystyle f$ in the interval $\displaystyle [n,n+n^{-3}]$ is $\displaystyle f\left(n+\frac{n^{-3}}{2}\right)=n$, so it's unbounded.

$\displaystyle \int_0^{\infty} f(x)\,dx=\frac{2}{3}\sum_{n=1}^{\infty}\frac{1}{n^ 2}=\frac{\pi^2}{9}$